if you're returning, then please press the "ctrl" and "F5" keys together to update your cache.

just click on the headings above to navigate to the several different pages of this website.

...

my efforts here are to use well-formed formulas to demonstrate the solutions to a variety of

(formerly) unsolved math problems; I didn't use a type-setting format, so you'll just have to

put your pencil to work to follow along. I've also interjected with reasons for the solutions.

...

many people have problably wondered why Leonhard Euler failed to find a closed form for

Apery's constant or for any zeta(2n+1). the explanation would have been difficult for anyone

to understand. after seeing the form for zeta(3), I can explain to anyone "why it is impossible

to find a closed form for zeta(3)", and the best part about the answer??--- membership is free!

just go to the bottom of this page to find it; it'd be like taking a refreshing drink of water...

...

someone argued that the following proof was so elementary that D. H. Lehmer would have

presented it himself! I took it as a compliment, since it meant that he/she actually loved it.

...

the third time's a charm (it took me roughly 3 drafts to prove the following conjecture):

D. H. Lehmer's Totient Problem (1932): phi(n) divides (n -1) iff 'n' is prime.

(proof, backward)

if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n) divides (n -1) as desired.

(proof, forward)

let phi(n) divide (n -1) such that k*phi(n)= (n -1). assume that n= b^j*C is 'composite' where

gcd(b^j, C)= 1, 'b' is a 'prime' factor, k <> 1. if C=1 when n=b^j*C, then we'd have 'n' equal

to the power of a prime, and phi(b^j)=b^j -b^(j-1). that's not equal to n−1. so, if C<>1, then

k*phi(b^j*C)= b^j*C -1 implies k*phi(b^j)*phi(C)= b^j*C -1 implies (k*b^j -k*b^(j-1))*phi(C)=

b^j*C -1, phi(C)= [b^j*C -1]/[k*b^j -k*b^(j-1)] implies that k*b^(j-1)= 1, and an integral 'phi'

function assumes that m*[k*b^j -k*b^(j-1)] = [b^j*C -1].

...

see the argument! let W -X = Y -Z where W= m*k*b^j, X= m*k*b^(j-1), Y= b^j*C, and Z= 1.

now, W has to equal either Y or -Z, but not their difference, or -X has to do the same, and

the only possibility

it takes a little bit of the fun out of it... if I have to explain the limiting constraint of Z= 1

along with the fact that each of the terms of the 2 expressions contain 'm's, 'k's, and 'b's.

...

or m*k*b^(j-1) = 1, or /*phi*/ would be undefined. so, phi(C) = (bC -1) / (b -1), b= C, and

phi(C)= C +1. either /*phi*/ is undefined, or 'C' cannot have more co-prime divisors than

it has a total number of possible divisors. thus, 'n' must be prime. enjoy! this idea took me

25 minutes on the back of an old receipt for my car repairs using a black sharpie marker!

...

*QED

Bill Bouris

05/26/2013 (I created some new /*phi*/ formulas. they appear after my No OPNs proof.)

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

...

the solutions to both "Ramsey Theory" and the Erdős–Szekeres conjecture can be viewed

by clicking on the Ramsey Theory link. I'm limited to only 6 pages on FatCow. Bill Bouris.

...

there won't be a proof for the Riemann Hypothesis; please click on the Reiki link above to

review how it can ONLY be found through logical deduction; it CANNOT be solved using

algebra, since any/all analysis of the geometry will be flawed!

...

if you'd like to count the number of normal magic squares, for n= a, then go a little furtherdown this page. I hope you enjoy it!

...

...

it was only after a math coach from IMSA e-mailed me several times using the pseudonym

of Bob Smith from a yahoo e-mail account to tell me that every attempt of mine was NOT

a solution to the problem. we fought for several weeks, back and forth, until I was rude to

him for his using several spoken languages in his salutations/greetings. pisica_t_cat [at]

yahoo.com was responsible for making me look much deeper into how very delicate the

situation actually was... until I noticed the real argument! the integral of a form as alluded

to by Euler with his formula over the boundary (seen by taking the derivative of the para-

metric equivalent) was equal to the integral of the differentiated form over the region). I'm

indebted to an unknown... Bob Smith for pushing over the edge... to the inevitable proof!

12/21/2013**

...

for those of you that've visited my website, I've spent the last 10 years (almost 100 drafts)

searching for a reasonable explanation for why odd-perfect numbers don't exist; here it is...

...

(a perfect number) = (a square) *(1 +r +r^2 + ... +r^n);

...

(an even-perfect number), EP(N) = (2^(n-1))*(2^n -1), but please dismiss the first EP(N) = 2*

(1 +2) = 6 such that EP(N)'s = 2^2*(1 +2 +4)= 28, 2^4*(1 +2 +4 +8 +16)= 496, etc.

...

(an odd-perfect number), (OPN), M = Q^2 *p^q= (4z+1)^2 *(4x+1)^(4y +1) for any/all x, y, z.

giving Q^2 *(1 +r +r^2 + ... +r^n)= Q^2 *(1-r^(n+1))/(1-r), or (4x+1)^(4y+1) = (1-r^(n+1))/(1-r);

...

the first derivative is symmetric around sigma's unity at n = 0: (EP(N) = (2^0)*(2^0 -1) = 1*1,

and for x, y, & z = 0: (OPN), M = (4*0+1)^2*(4*0+1)^(4*0+1) = 1*1 such that Q^2 = (4z+1)^2.

...

(d/dr) [(1-r)* (4x+1)^(4y+1)] = [1-r^(n+1)] (d/dr); (the derivative acts as an operator applied

to a tensor whereby we discover the restriction or boundary of Euler's summation formula!)

...

-(4x+1)^(4y+1) = -(n+1)* r^n; n = 4x; r = 4x+1; -(4x+1)^(4y+1) = -(4x+1)^(4x+1) and x = y;

-4x* (4x+1)^(4x+1) = 1 - (4x+1)^(4x+1); -1 -4x* (4x+1)^(4x+1) = -(4x+1)^(4x+1) iff x = 0.

...

<><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>

mathematically speaking...

the summation of the 'form' over the boundary of a region (== Euler's formula) is equal to the

summation of the derivative of the 'form' over the region (== a perfect square). this becomes

visible only after taking the derivative of the parametric portion of Euler's formula. hence, the

latter summation is a perfect square over the entire region, and according to the generalized

Stokes Theorem, and someone's proof that OPN's can't be perfect squares, there can be "NO"

odd-perfect numbers as noted by Euler's exceptional attempt. no new information is needed.

<><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>

...

thus, odd-perfect numbers don't exist!

...

what's this stuff ??? just cereal,... supposed to be good for you,... you try it,... I'm not gonna try

it,... let's get Mikey,... yeah, he won't eat it,... he hates everything!... ... he likes it... hey, Mikey!

...

(OR) you can expect to comprehend the heuristic and believe that it's unlikely that they exist!

...

obviously, when someone as prolific as Euler stated that this problem was too complicated

to be easily solved, every mathematician was easily persuaded to believe it; and if you (OR)

I decided to retain Euler's already-proven formula for an odd-perfect number, it's surprising

to me that we could have been easily mislead by his prior conclusion! there are further con-

clusions from the fact that odd-perfect numbers don't exist; read on...

...

*QED

Bill Bouris

05/26/2012

...

...

when you first take a look at the "algebra" for Euler's /*phi*/ function, you'll say... "they don't

know much about it", then you'll say, "we don't know much about it", and then you'll realize that

"I don't know much at all about the /*phi*/ function", and that sets the stage for learning some-

thing about Euler's /*phi*/ function: if and only if... phi(n^2 -1) = 2*phi(n +1)*phi(n -1) can be

proved, then because phi(n^k) = n^(k -1)*phi(n), we can surely understand the following two

equations: phi(n+1) = [(n -1)*phi(n^2 -1)] / [2*phi((n -1)^2)] & phi(n -1) = [phi((n-1)^2) / (n -1)].

now, I understand them; we... and everyone can understand them! they must be written in the

subtractive format, and it must be noted that 'n' is prime. Also, I can see how the -1's could be

replaced by -k's such that we'd come to know that phi(n +k) & phi(n -k) can be meaningful, iff

gcd(n, k) =1 is closely monitored, and 'n' is prime.

...

...

C_5 must be prime! let's take a little excursion...

remember... C_5 is a 51,217,599,719,369,681,875,006,054,625,051,616,350-digit number!

...

if p= 4*k +1, and q= 2*p +3 are both prime, then if [(M_r)^p -p] mod q == N, and q mod N ==

+/-1, then (M_r), the base... is prime. also, if (M_r) mod p = 1, then choose a different 'p' or if N

is a square, then (M_r) is prime. (someone would have to prove this conjecture.)

...

let 2^127 -1 = 170141183460469231731687303715884105727

...

C_5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1

...

let p = 5, and q = 13 such that [(C_5)^5 -5] mod 13 = N, but 2^(2*27) mod 13 == 12 == (-1)

(as noticed by shear discovery!!!), so then...

...

(-1)^(odd power)*2 -1 ==

(-1)^49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 ==

(-1)*2 -1 = -2 -1 and (-3)^5 -5 = -248 and [(C_5)^5 -5] mod 13 == 12.

...

thus, if [(C_5)^5 -5] mod 13 == 12, and 13 mod 12 == -1, then C_5 must be prime!

...

I've listed the 4 equations for a Mersenne prime number (M_r) where r is also prime.

...

(only one of these needs to be true!)

p = 4k+1, q = 2p+3 (both prime) [(M_r)^p-p] mod q == -1, or

p = 4k+3, q = 2p+3 (again,... ), then [(M_r)^p-p] mod q == -1 (can't be +1)

p = 4k+1, q = 2p+1 (again, both prime), then [(M_r)^p-p] mod q == p

p = 4k+3, q = 2p+1 (again, both prime) [(M_r)^p-p] mod q == p+2

...

let's have another look!!!

let 2^127 -1 = 170141183460469231731687303715884105727

...

C_5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1

...

let p = 3, q = 7 such that [(C_5)^3 -3] mod 7 = N; and 2^(27) mod 7 == 1 (by mere chance!!!),

so then...

...

(1)^(left over exponent)*2 -1 ==

(1)^2*49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 == (1)*2 -1 = 2 -1

and (1)^3 -3 = 1 -3 = -2 and [(C_5)^3 -3] mod 7 == 5

...

thus, if [(C_5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C_5 must be prime!

...

C_5 is definitely prime, if my study is correct.

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

...

...

here's a better puzzle:

let 'm' be any integer of the form m = 2^k*(2n+1). if sigma(m) = 2m -1, then 'm' can only be of

the form 2^k, or... an almost-perfect number.

...

proof:

either sigma(2n +1) = 4n +1, or sigma(2^k* (2n+1)) = sigma(2^k)* sigma(2n +1) = n*2^(k +2)

+2^(k +1) -1 implies that sigma(2^k)= [n*2^(k+2) +2^(k+1) -1]/(4n+1) is defined for all 'k'.

...

if n <> 0 in the latter statement, then sigma(2^k) will be fractional; also, if k = 0, n = 0, then

sigma(1) = (0 +2 -1)/(4(0) +1)= 1/1 = 1; if k= k when n = 0, then sigma(2^k)= (2^(k +1)) -1 =

2*(2^k) -1. if you tried to argue w/the former statement that sigma(2n +1)= 4n +1, then you'd

have to answer a much bigger question about the sigma function, since sigma(4n +3)= 8n +5;

it's certainly false, since sigma(4n +3) is ALWAYS EVEN; as if you couldn't have figured this

one out... since odd-perfect numbers don't exist, it's implied that only sigma(of odd squares)

can be equal to an odd number, and odd squares aren't of the form 4n +3, so sigma(4n +3) is

ALWAYS EVEN.

*QED

...

if sigma(4n +3) is ALWAYS EVEN, then an APN must only be of the form 2^k; this is also true!

*QED

both 03/26/2012

...

...

there are infinitely many Mersenne prime numbers! I didn't create the theorem (

...

according to Dirichlet's theorem, there are infinitely many primes of the form g(n)= 4n +3 where

a= 3, d= 4, (a,d)= 1, and g(n)= a +n*d, and f(n)= n; it implies that 1/f(0) == 1/0 << [series 1/g(n)]

== [1/3 +1/7 + 1/11 + 1/19... + 1/n] which is divergent; 'n' is an element the set of Natural (con-

secutive) Numbers.

...

similarly, there will be infinitely many Mersenne primes of the form M(x) = 2*(2^x) -1 where a = -1,

d = 2, (a,d) = 1, and M(x) = a +N(x)*d such that N(x) = 2^x is 1-to-1 and onto with respect to the

set of Whole (consecutive) Numbers. thus, if and only if, 1/N(1) = 1/2 << [series 1/M(x)] == [1/3

+1/7 +1/31 + 1/127 +1/8191 + ... +1/x]. this fact can be confirmed after computing only the first

few terms.

...

therefore, there are infinitely many Mersenne primes of the form M(p) = 2^p -1.

*QE(Dirichlet's Interpretation)

01/23/2011

...

...

there are finitely many Fermat prime numbers! I didn't create the theorem (

...

unless you discount Dirichlet's theorem,

form F(x)= 2*(2^(2^x -1)) +1 where a= +1, d= 2, (a,d)= 1, and F(x)= a +G(x)*d such that G(x)=

2^(2^x -1) is 1-to-1 and onto with respect to the set of Whole (consecutive) Numbers. Thus, iff

I'm correctly understanding his argument, then 1/G(0) = 1/1 must be >> [series 1/F(x)] == [1/3

+1/5 +1/17 + 1/257 +1/65537 + ... +1/x]. this simple fact can be confirmed after computing

just the first few terms.

...

therefore, there are finitely many Fermat primes of the form F(x)= 2^(2^x) +1.

*QE(Dirichlet's Interpretation)

01/25/2011

...

...

there are infinitely many primes of the form x^2 +1! I didn't create the theorem (

...

there will be infinitely many primes of the form M(x) = x*x +1 where a = 1, d = x, (a,d) = 1, M(x) =

a +N(x)*d such that N(x) = x is 1-to-1 and onto with respect to the set of Natural (consecutive)

Numbers. Thus, if and only if, 1/N(1) = 1/2 << [series 1/M(x)] == [1/2 +1/5 +1/17 + ... +1/x]. it

can be confirmed after computing only the first few terms.

...

therefore, there are infinitely many primes of the form M(x) = x^2 +1.

*QE(Dirichlet's Interpretation)

...

...

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

let's try counting the number of "normal magic squares" for a certain size, 'n':

...

n 1 {((n -d)^2)*(2^n)}

n= 1; N_1 = ------- = ------- = 1; {not noticing the subtraction of [((1 -1)^2)*(2^1) = 0] repeats}

1 1

...

n +1 3

n= 2; N_2 = --------- = ----- = 0; (there aren't any fractional amounts of 2x2 magic squares!)

2 2

...

(n +1)*(n^2 +1) (-1)*(n^2 +1) {minus ((n -d)^2)*(2^n) repeats!}

n= 3; N_3 = --------------------------- + ----------------------- = 10 -1 -((3-2)^2)*(2^3) = 1

2*2 2*5

...

n*(n^2 +1)*(n^3 +2) n^2*(n^3 +2)

n= 4; N_4 = --------------------------------- + ------------------------ = 748 + 132 = 880;

2*3 2*4

...

...

notice how each term must be multiplied against the generator function of (n^3+2), but it must

be done in different fractions, and there will always be 2^[floor((n +/-1)/2)] terms in each sum-

mation for a different size, 'n':

...

the solution for n= 4 was found on a hunch that I could bump up the exponent to (n^3+2), and

then when 748 +132 equaled 880, I knew that I had something. so, I prepared to match Rich

Shroeppel's answer for n= 5, and the very first term was a HUGE step in the right direction. the

only problem was... that either his or my final answer was off by a little. I wrote to him in an e-

mail, and he failed to answer me back. on Sunday, I telephoned him in Arizona to tell him that

only 1 of us could be correct. he stated that we could only differ by a factor of 32; so, I didn't

believe it until I noticed that all of the 'odd formulas' needed to have a tiny amount subtracted

from it. thus, I was able to find the corrected formulas for n= 1,3,5 w/out either of us suffering!

12/21/2013**

...

for n= 5, there will be 2^[floor((5 +1)/2)] = 8 terms; look below...

...

...

n*(n^2 +1)*(n^3 +2)*(n^4 +3)*(n^5 +4) n^2*(n^3 +1)*(n^4 +2)*(n^5 +4)

n=5; N5 = --------------------------------------------------------------- + ---------------------------------------------------- +

2*2*2*3*5 2*3*3*3*5*5

...

...

(n -1)*n^3*(n^4 +1)*(n^5 +4) -2*(n^2 +1)*n^4*(n^5 +4) (n-2)*(n^4 -1)*(n^5 +4)

+ ----------------------------------------------- + ----------------------------------------- + -------------------------------------- +

2*2*3*5*5*5 2*2*3*5*5 2*2*3*3*3

...

...

(n^3 -2)*(n^5 +4) (n^2 -3)*(n^5 +4) -3*(n -4)*(n^5 +4)

+ -------------------------------- + ------------------------------- + ------------------------------- ==

3*3*7 2*3*11 3*7

-9*32

{minus ((n -d)^2)*(2^n) where n= 5, and d= 2. this subtraction is only done when 'n' is odd!}

...

...

270,352,901 +4,577,727 +652,918 -338,975 +54,236 +6109 +1043 -447 -(288) = 275,305,224.

...

these calculation take into account the number of rotations and reflections that must be subtracted.

also, there's a total of 8 entry points into a magic square, 1 at the end of each diagonal, and 1 on

both the x- and y-axis of each edge-face, so the total number of "normal magic squares" for a given

'n' must be divisible by at least 8, and must also, separately be divisible by at least one factor of

'n-d' or 5-2= 3.

...

it's only an estimate but, here's my estimate for N_6 = 17,916,790,562,863,266,656 (for n= 6) ==

...

n*(n^2+1)*(n^3+2)*(n^4+3)*(n^5+4)*(n^6+5)*(n^7+6)

= ------------------------------------------------------------------------------------ +

2*2*2*3*3*5

...

n^2*(n^3+1)*(n^4+2)*(n^5+3)*(n^6+4)*(n^7+6)

+ --------------------------------------------------------------------------- +

2*2*2*2*3*3*3*3*5

...

(n-1)*(n^3)*(n^4+1)*(n^5+2)*(n^6+3)*(n^7+6)

+ ------------------------------------------------------------------------ +

2*2*2*3*3*3*5*13

...

(n^2-1)*(n^4)*(n^5+1)*(n^6+2)*(n^7+6)

+ --------------------------------------------------------------- +

2*2*2*2*3*3*5*7

...

(n-4)*(n^3-1)*(n^5)*(n^6+1)*(n^7+6)

+ ---------------------------------------------------------- +

2*2*2*2*2*2*3*3*5

...

(n^2-2)*(n^4-1)*(n^6)*(n^7+6)

+ ------------------------------------------------ +

2*2*2*2*2*3*7

...

(n-3)*(n^3-2)*(n^5-1)*(n^7+6) (n^2-3)*(n^4-2)*(n^7+6)

+ ------------------------------------------------ + --------------------------------------- +

3*5*13 3*13

...

(n-4)*(n^3-3)*(n^7+6) (n^2-4)*(n^7+6)

+ ------------------------------------ + ---------------------------- =

2*13 2*13

...

17746708604195767404 +159000499324823156 + 10136033805551796

...

+914215937783148 +30328231541868 +855765897480 +21497607540

...

+3984694428 +119255292 +344544 == 17916790562863266656.

...

the total is divisible by 4 and 8, separately, leaving an odd number; the number is also

divisible by 37 and 97, primes from the generator function portion of (n^7+6). also, every

term is positive! hopefully, there's not a typo. this is my best estimate for the number of

normal-6x6 magic squares that best fits the profile for what the answer could be...

...

here's why a "normal magic square" must be further divisible by (n -d), and a "normal magic cube"

must be further divisible by (n -4). the formula is always (n -(L -D)) where L is the number of unique

lines that can pass diagonally or face-wise through the center, and D is the number of dimensions

for the magic shape. thus, a MSQ is divisible by (n -(4 -2))= (n -2), a MCu is divisible by (n -(7 -3))=

(n -4). that is the REAL reason why a MSQ of n= 2 and why a MCube of n= 4 are both not possible!

furthermore, if a tesseract has 16 corners and 12 parallel faces, then there won't be a magic tesseract

for n= 7, because (n -((16/2 +6/2) -4)= (n -(11 -4))= (n -7); half of the parallel faces can't be counted

in this calculation.

...

and, the count for a magic shape, once it's big enough, must be divisible by f= D*(2^(2*(D-1)))

where 'D' is the dimension of the magic object. thus, a MSQ count will be divisible by 8, a MCu

count will be divisible by 48, and a MT(magic tesseract) count will be divisible by 256, again,

once the size of 'n' becomes big enough to register these factors. nice, huh???

...

*QED

12/7/2012

...

so, 275,305,224 is divisible by 3, and 8; I corrected the calculations when 'n' is 'odd', and if you

divide by 4, the number of unique magic squares will remain w/out reflections and rotations.

...

thus, that's why for n= 4, N4= 880 is divisible by 8, and another 2. the part about being divisible

by 'n-2' is the !REAL! reason that there aren't any magic squares for n= 2. it's like n= 2 is a "point

of infinity", geometrically.

...

*QED

12/06/2012

...

...

do you see the pattern? the generator function for n=5 is (n^5 +4), but we would almost have to

know the number of normal magic squares to be counted in order to form all the numerators and

denominators for the necessary fractions!

...

for n= 6, N6= 6*(6^2+1)*(6^3+2)*(6^4+3)*(6^5+4)*(6^6+5)*(6^7+6)/2/2/3/5/6 = 1.7746*10^19

for the first term, so the estimate posted on Wikipedia is somewhat close.

...

on a more interesting note, here's the formula for the "magic" number for "magic squares" that con-

tain only "square numbers" as entries: for 'n', N4^2= [(n +1)*(n^2 -3)*(n^4 +6)]/2. notice that Euler's

4x4 example equals 8515 as promised, and now I can prove that there will NOT be a 3x3 "magic

square" of only "square numbers". oh, someone complained that magic squares of squares has 2

different sums. so, I told him that since the entries are of the form x-squared, they all associate with

the power of 2. there are 2 families of functions for magic squares of squares... Euler found the ex-

ample of [(n +1)*(n^2 -3)*(n^4 +6)]/2, & the other one is [n*(n^2 +3)*(n^4 -1)]/6 which promises a

sum of 3230.

...

...

*QED

12/20/2012

...

...

the centers of "basic magic squares" for all (odd n) will follow the pattern: {subst. M= (n -((n+1)/2))

for 'n' in the triangular formula, Tn = (n*(n+1))/2}; so, the (centerMagic{Sq}n)= {n^2 - 4*[M*(M+1)]/2}.

thus, if n= 3, (centerMSq3)= 3^2 -4*(1)= 5; n= 5, (centerMSq5)= (5^2 -4*(3))= 13, n=7, (centerMSq7)=

(7^2 -4*(6))= 25, n=9, (centerMSq9)= (9^2 -4(10))= 41, etc.

...

the centers of "basic magic cubes" for all (odd n) will follow this simpler formula:

(centerMagic{Cube}n)= [(n +1)*(n^2 -(n -1))]/2. thus, if n= 3, (centerMCu3)= (4/2)*(3^2 -3 +1)= 2*7=

14; n=5, (centerMCu5)= (6/2)*(5^2 -5 +1)= 3*21 = 63; n=7, (centerMCu7) = 4*43 = 172, etc., etc.,

but I can't prove any of them. (they're purely formulary and sound very reasonable!)

...

*QED

12/3/2012

...

...

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someone w/degrees from Harvard, Stanford, and Michigan argued that "if it could have been proven

using algebra 'why a 3x3-normal magic square of squares doesn't exist?', then it would've been done

already! let's have another look...

...

...

A B C 8 1 6

...

D E F 3 5 7

...

G H I 4 9 2

...

just click on the headings above to navigate to the several different pages of this website.

...

my efforts here are to use well-formed formulas to demonstrate the solutions to a variety of

(formerly) unsolved math problems; I didn't use a type-setting format, so you'll just have to

put your pencil to work to follow along. I've also interjected with reasons for the solutions.

...

many people have problably wondered why Leonhard Euler failed to find a closed form for

Apery's constant or for any zeta(2n+1). the explanation would have been difficult for anyone

to understand. after seeing the form for zeta(3), I can explain to anyone "why it is impossible

to find a closed form for zeta(3)", and the best part about the answer??--- membership is free!

just go to the bottom of this page to find it; it'd be like taking a refreshing drink of water...

...

someone argued that the following proof was so elementary that D. H. Lehmer would have

presented it himself! I took it as a compliment, since it meant that he/she actually loved it.

...

the third time's a charm (it took me roughly 3 drafts to prove the following conjecture):

D. H. Lehmer's Totient Problem (1932): phi(n) divides (n -1) iff 'n' is prime.

(proof, backward)

if 'n' is prime, then phi(n) = (n -1) by definition, and phi(n) divides (n -1) as desired.

(proof, forward)

let phi(n) divide (n -1) such that k*phi(n)= (n -1). assume that n= b^j*C is 'composite' where

gcd(b^j, C)= 1, 'b' is a 'prime' factor, k <> 1. if C=1 when n=b^j*C, then we'd have 'n' equal

to the power of a prime, and phi(b^j)=b^j -b^(j-1). that's not equal to n−1. so, if C<>1, then

k*phi(b^j*C)= b^j*C -1 implies k*phi(b^j)*phi(C)= b^j*C -1 implies (k*b^j -k*b^(j-1))*phi(C)=

b^j*C -1, phi(C)= [b^j*C -1]/[k*b^j -k*b^(j-1)] implies that k*b^(j-1)= 1, and an integral 'phi'

function assumes that m*[k*b^j -k*b^(j-1)] = [b^j*C -1].

...

see the argument! let W -X = Y -Z where W= m*k*b^j, X= m*k*b^(j-1), Y= b^j*C, and Z= 1.

now, W has to equal either Y or -Z, but not their difference, or -X has to do the same, and

the only possibility

*IS*... that m*k*b^(j-1) = 1, if the /*phi*/ function is to make any sense!it takes a little bit of the fun out of it... if I have to explain the limiting constraint of Z= 1

along with the fact that each of the terms of the 2 expressions contain 'm's, 'k's, and 'b's.

...

or m*k*b^(j-1) = 1, or /*phi*/ would be undefined. so, phi(C) = (bC -1) / (b -1), b= C, and

phi(C)= C +1. either /*phi*/ is undefined, or 'C' cannot have more co-prime divisors than

it has a total number of possible divisors. thus, 'n' must be prime. enjoy! this idea took me

25 minutes on the back of an old receipt for my car repairs using a black sharpie marker!

...

*QED

Bill Bouris

05/26/2013 (I created some new /*phi*/ formulas. they appear after my No OPNs proof.)

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

...

the solutions to both "Ramsey Theory" and the Erdős–Szekeres conjecture can be viewed

by clicking on the Ramsey Theory link. I'm limited to only 6 pages on FatCow. Bill Bouris.

...

there won't be a proof for the Riemann Hypothesis; please click on the Reiki link above to

review how it can ONLY be found through logical deduction; it CANNOT be solved using

algebra, since any/all analysis of the geometry will be flawed!

...

if you'd like to count the number of normal magic squares, for n= a, then go a little furtherdown this page. I hope you enjoy it!

...

...

it was only after a math coach from IMSA e-mailed me several times using the pseudonym

of Bob Smith from a yahoo e-mail account to tell me that every attempt of mine was NOT

a solution to the problem. we fought for several weeks, back and forth, until I was rude to

him for his using several spoken languages in his salutations/greetings. pisica_t_cat [at]

yahoo.com was responsible for making me look much deeper into how very delicate the

situation actually was... until I noticed the real argument! the integral of a form as alluded

to by Euler with his formula over the boundary (seen by taking the derivative of the para-

metric equivalent) was equal to the integral of the differentiated form over the region). I'm

indebted to an unknown... Bob Smith for pushing over the edge... to the inevitable proof!

12/21/2013**

...

for those of you that've visited my website, I've spent the last 10 years (almost 100 drafts)

searching for a reasonable explanation for why odd-perfect numbers don't exist; here it is...

...

(a perfect number) = (a square) *(1 +r +r^2 + ... +r^n);

...

(an even-perfect number), EP(N) = (2^(n-1))*(2^n -1), but please dismiss the first EP(N) = 2*

(1 +2) = 6 such that EP(N)'s = 2^2*(1 +2 +4)= 28, 2^4*(1 +2 +4 +8 +16)= 496, etc.

...

(an odd-perfect number), (OPN), M = Q^2 *p^q= (4z+1)^2 *(4x+1)^(4y +1) for any/all x, y, z.

giving Q^2 *(1 +r +r^2 + ... +r^n)= Q^2 *(1-r^(n+1))/(1-r), or (4x+1)^(4y+1) = (1-r^(n+1))/(1-r);

...

the first derivative is symmetric around sigma's unity at n = 0: (EP(N) = (2^0)*(2^0 -1) = 1*1,

and for x, y, & z = 0: (OPN), M = (4*0+1)^2*(4*0+1)^(4*0+1) = 1*1 such that Q^2 = (4z+1)^2.

...

(d/dr) [(1-r)* (4x+1)^(4y+1)] = [1-r^(n+1)] (d/dr); (the derivative acts as an operator applied

to a tensor whereby we discover the restriction or boundary of Euler's summation formula!)

...

-(4x+1)^(4y+1) = -(n+1)* r^n; n = 4x; r = 4x+1; -(4x+1)^(4y+1) = -(4x+1)^(4x+1) and x = y;

-4x* (4x+1)^(4x+1) = 1 - (4x+1)^(4x+1); -1 -4x* (4x+1)^(4x+1) = -(4x+1)^(4x+1) iff x = 0.

...

<><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>

mathematically speaking...

the summation of the 'form' over the boundary of a region (== Euler's formula) is equal to the

summation of the derivative of the 'form' over the region (== a perfect square). this becomes

visible only after taking the derivative of the parametric portion of Euler's formula. hence, the

latter summation is a perfect square over the entire region, and according to the generalized

Stokes Theorem, and someone's proof that OPN's can't be perfect squares, there can be "NO"

odd-perfect numbers as noted by Euler's exceptional attempt. no new information is needed.

<><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><><>

...

thus, odd-perfect numbers don't exist!

...

what's this stuff ??? just cereal,... supposed to be good for you,... you try it,... I'm not gonna try

it,... let's get Mikey,... yeah, he won't eat it,... he hates everything!... ... he likes it... hey, Mikey!

...

(OR) you can expect to comprehend the heuristic and believe that it's unlikely that they exist!

...

obviously, when someone as prolific as Euler stated that this problem was too complicated

to be easily solved, every mathematician was easily persuaded to believe it; and if you (OR)

I decided to retain Euler's already-proven formula for an odd-perfect number, it's surprising

to me that we could have been easily mislead by his prior conclusion! there are further con-

clusions from the fact that odd-perfect numbers don't exist; read on...

...

*QED

Bill Bouris

05/26/2012

...

...

when you first take a look at the "algebra" for Euler's /*phi*/ function, you'll say... "they don't

know much about it", then you'll say, "we don't know much about it", and then you'll realize that

"I don't know much at all about the /*phi*/ function", and that sets the stage for learning some-

thing about Euler's /*phi*/ function: if and only if... phi(n^2 -1) = 2*phi(n +1)*phi(n -1) can be

proved, then because phi(n^k) = n^(k -1)*phi(n), we can surely understand the following two

equations: phi(n+1) = [(n -1)*phi(n^2 -1)] / [2*phi((n -1)^2)] & phi(n -1) = [phi((n-1)^2) / (n -1)].

now, I understand them; we... and everyone can understand them! they must be written in the

subtractive format, and it must be noted that 'n' is prime. Also, I can see how the -1's could be

replaced by -k's such that we'd come to know that phi(n +k) & phi(n -k) can be meaningful, iff

gcd(n, k) =1 is closely monitored, and 'n' is prime.

...

...

C_5 must be prime! let's take a little excursion...

remember... C_5 is a 51,217,599,719,369,681,875,006,054,625,051,616,350-digit number!

...

if p= 4*k +1, and q= 2*p +3 are both prime, then if [(M_r)^p -p] mod q == N, and q mod N ==

+/-1, then (M_r), the base... is prime. also, if (M_r) mod p = 1, then choose a different 'p' or if N

is a square, then (M_r) is prime. (someone would have to prove this conjecture.)

...

let 2^127 -1 = 170141183460469231731687303715884105727

...

C_5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1

...

let p = 5, and q = 13 such that [(C_5)^5 -5] mod 13 = N, but 2^(2*27) mod 13 == 12 == (-1)

(as noticed by shear discovery!!!), so then...

...

(-1)^(odd power)*2 -1 ==

(-1)^49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 ==

(-1)*2 -1 = -2 -1 and (-3)^5 -5 = -248 and [(C_5)^5 -5] mod 13 == 12.

...

thus, if [(C_5)^5 -5] mod 13 == 12, and 13 mod 12 == -1, then C_5 must be prime!

...

I've listed the 4 equations for a Mersenne prime number (M_r) where r is also prime.

...

(only one of these needs to be true!)

p = 4k+1, q = 2p+3 (both prime) [(M_r)^p-p] mod q == -1, or

p = 4k+3, q = 2p+3 (again,... ), then [(M_r)^p-p] mod q == -1 (can't be +1)

p = 4k+1, q = 2p+1 (again, both prime), then [(M_r)^p-p] mod q == p

p = 4k+3, q = 2p+1 (again, both prime) [(M_r)^p-p] mod q == p+2

...

let's have another look!!!

let 2^127 -1 = 170141183460469231731687303715884105727

...

C_5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1

...

let p = 3, q = 7 such that [(C_5)^3 -3] mod 7 = N; and 2^(27) mod 7 == 1 (by mere chance!!!),

so then...

...

(1)^(left over exponent)*2 -1 ==

(1)^2*49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 == (1)*2 -1 = 2 -1

and (1)^3 -3 = 1 -3 = -2 and [(C_5)^3 -3] mod 7 == 5

...

thus, if [(C_5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C_5 must be prime!

...

C_5 is definitely prime, if my study is correct.

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

...

...

here's a better puzzle:

let 'm' be any integer of the form m = 2^k*(2n+1). if sigma(m) = 2m -1, then 'm' can only be of

the form 2^k, or... an almost-perfect number.

...

proof:

either sigma(2n +1) = 4n +1, or sigma(2^k* (2n+1)) = sigma(2^k)* sigma(2n +1) = n*2^(k +2)

+2^(k +1) -1 implies that sigma(2^k)= [n*2^(k+2) +2^(k+1) -1]/(4n+1) is defined for all 'k'.

...

if n <> 0 in the latter statement, then sigma(2^k) will be fractional; also, if k = 0, n = 0, then

sigma(1) = (0 +2 -1)/(4(0) +1)= 1/1 = 1; if k= k when n = 0, then sigma(2^k)= (2^(k +1)) -1 =

2*(2^k) -1. if you tried to argue w/the former statement that sigma(2n +1)= 4n +1, then you'd

have to answer a much bigger question about the sigma function, since sigma(4n +3)= 8n +5;

it's certainly false, since sigma(4n +3) is ALWAYS EVEN; as if you couldn't have figured this

one out... since odd-perfect numbers don't exist, it's implied that only sigma(of odd squares)

can be equal to an odd number, and odd squares aren't of the form 4n +3, so sigma(4n +3) is

ALWAYS EVEN.

*QED

...

if sigma(4n +3) is ALWAYS EVEN, then an APN must only be of the form 2^k; this is also true!

*QED

both 03/26/2012

...

...

there are infinitely many Mersenne prime numbers! I didn't create the theorem (

*strong*portion)......

according to Dirichlet's theorem, there are infinitely many primes of the form g(n)= 4n +3 where

a= 3, d= 4, (a,d)= 1, and g(n)= a +n*d, and f(n)= n; it implies that 1/f(0) == 1/0 << [series 1/g(n)]

== [1/3 +1/7 + 1/11 + 1/19... + 1/n] which is divergent; 'n' is an element the set of Natural (con-

secutive) Numbers.

...

similarly, there will be infinitely many Mersenne primes of the form M(x) = 2*(2^x) -1 where a = -1,

d = 2, (a,d) = 1, and M(x) = a +N(x)*d such that N(x) = 2^x is 1-to-1 and onto with respect to the

set of Whole (consecutive) Numbers. thus, if and only if, 1/N(1) = 1/2 << [series 1/M(x)] == [1/3

+1/7 +1/31 + 1/127 +1/8191 + ... +1/x]. this fact can be confirmed after computing only the first

few terms.

...

therefore, there are infinitely many Mersenne primes of the form M(p) = 2^p -1.

*QE(Dirichlet's Interpretation)

01/23/2011

...

...

there are finitely many Fermat prime numbers! I didn't create the theorem (

*strong*portion)......

unless you discount Dirichlet's theorem,

*again*, there will be finitely many Fermat primes of theform F(x)= 2*(2^(2^x -1)) +1 where a= +1, d= 2, (a,d)= 1, and F(x)= a +G(x)*d such that G(x)=

2^(2^x -1) is 1-to-1 and onto with respect to the set of Whole (consecutive) Numbers. Thus, iff

I'm correctly understanding his argument, then 1/G(0) = 1/1 must be >> [series 1/F(x)] == [1/3

+1/5 +1/17 + 1/257 +1/65537 + ... +1/x]. this simple fact can be confirmed after computing

just the first few terms.

...

therefore, there are finitely many Fermat primes of the form F(x)= 2^(2^x) +1.

*QE(Dirichlet's Interpretation)

01/25/2011

...

...

there are infinitely many primes of the form x^2 +1! I didn't create the theorem (

*strong*portion)......

there will be infinitely many primes of the form M(x) = x*x +1 where a = 1, d = x, (a,d) = 1, M(x) =

a +N(x)*d such that N(x) = x is 1-to-1 and onto with respect to the set of Natural (consecutive)

Numbers. Thus, if and only if, 1/N(1) = 1/2 << [series 1/M(x)] == [1/2 +1/5 +1/17 + ... +1/x]. it

can be confirmed after computing only the first few terms.

...

therefore, there are infinitely many primes of the form M(x) = x^2 +1.

*QE(Dirichlet's Interpretation)

...

...

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

let's try counting the number of "normal magic squares" for a certain size, 'n':

...

n 1 {((n -d)^2)*(2^n)}

n= 1; N_1 = ------- = ------- = 1; {not noticing the subtraction of [((1 -1)^2)*(2^1) = 0] repeats}

1 1

...

n +1 3

n= 2; N_2 = --------- = ----- = 0; (there aren't any fractional amounts of 2x2 magic squares!)

2 2

...

(n +1)*(n^2 +1) (-1)*(n^2 +1) {minus ((n -d)^2)*(2^n) repeats!}

n= 3; N_3 = --------------------------- + ----------------------- = 10 -1 -((3-2)^2)*(2^3) = 1

2*2 2*5

...

n*(n^2 +1)*(n^3 +2) n^2*(n^3 +2)

n= 4; N_4 = --------------------------------- + ------------------------ = 748 + 132 = 880;

2*3 2*4

...

...

notice how each term must be multiplied against the generator function of (n^3+2), but it must

be done in different fractions, and there will always be 2^[floor((n +/-1)/2)] terms in each sum-

mation for a different size, 'n':

...

the solution for n= 4 was found on a hunch that I could bump up the exponent to (n^3+2), and

then when 748 +132 equaled 880, I knew that I had something. so, I prepared to match Rich

Shroeppel's answer for n= 5, and the very first term was a HUGE step in the right direction. the

only problem was... that either his or my final answer was off by a little. I wrote to him in an e-

mail, and he failed to answer me back. on Sunday, I telephoned him in Arizona to tell him that

only 1 of us could be correct. he stated that we could only differ by a factor of 32; so, I didn't

believe it until I noticed that all of the 'odd formulas' needed to have a tiny amount subtracted

from it. thus, I was able to find the corrected formulas for n= 1,3,5 w/out either of us suffering!

12/21/2013**

...

for n= 5, there will be 2^[floor((5 +1)/2)] = 8 terms; look below...

...

...

n*(n^2 +1)*(n^3 +2)*(n^4 +3)*(n^5 +4) n^2*(n^3 +1)*(n^4 +2)*(n^5 +4)

n=5; N5 = --------------------------------------------------------------- + ---------------------------------------------------- +

2*2*2*3*5 2*3*3*3*5*5

...

...

(n -1)*n^3*(n^4 +1)*(n^5 +4) -2*(n^2 +1)*n^4*(n^5 +4) (n-2)*(n^4 -1)*(n^5 +4)

+ ----------------------------------------------- + ----------------------------------------- + -------------------------------------- +

2*2*3*5*5*5 2*2*3*5*5 2*2*3*3*3

...

...

(n^3 -2)*(n^5 +4) (n^2 -3)*(n^5 +4) -3*(n -4)*(n^5 +4)

+ -------------------------------- + ------------------------------- + ------------------------------- ==

3*3*7 2*3*11 3*7

-9*32

{minus ((n -d)^2)*(2^n) where n= 5, and d= 2. this subtraction is only done when 'n' is odd!}

...

...

270,352,901 +4,577,727 +652,918 -338,975 +54,236 +6109 +1043 -447 -(288) = 275,305,224.

...

these calculation take into account the number of rotations and reflections that must be subtracted.

also, there's a total of 8 entry points into a magic square, 1 at the end of each diagonal, and 1 on

both the x- and y-axis of each edge-face, so the total number of "normal magic squares" for a given

'n' must be divisible by at least 8, and must also, separately be divisible by at least one factor of

'n-d' or 5-2= 3.

...

it's only an estimate but, here's my estimate for N_6 = 17,916,790,562,863,266,656 (for n= 6) ==

...

n*(n^2+1)*(n^3+2)*(n^4+3)*(n^5+4)*(n^6+5)*(n^7+6)

= ------------------------------------------------------------------------------------ +

2*2*2*3*3*5

...

n^2*(n^3+1)*(n^4+2)*(n^5+3)*(n^6+4)*(n^7+6)

+ --------------------------------------------------------------------------- +

2*2*2*2*3*3*3*3*5

...

(n-1)*(n^3)*(n^4+1)*(n^5+2)*(n^6+3)*(n^7+6)

+ ------------------------------------------------------------------------ +

2*2*2*3*3*3*5*13

...

(n^2-1)*(n^4)*(n^5+1)*(n^6+2)*(n^7+6)

+ --------------------------------------------------------------- +

2*2*2*2*3*3*5*7

...

(n-4)*(n^3-1)*(n^5)*(n^6+1)*(n^7+6)

+ ---------------------------------------------------------- +

2*2*2*2*2*2*3*3*5

...

(n^2-2)*(n^4-1)*(n^6)*(n^7+6)

+ ------------------------------------------------ +

2*2*2*2*2*3*7

...

(n-3)*(n^3-2)*(n^5-1)*(n^7+6) (n^2-3)*(n^4-2)*(n^7+6)

+ ------------------------------------------------ + --------------------------------------- +

3*5*13 3*13

...

(n-4)*(n^3-3)*(n^7+6) (n^2-4)*(n^7+6)

+ ------------------------------------ + ---------------------------- =

2*13 2*13

...

17746708604195767404 +159000499324823156 + 10136033805551796

...

+914215937783148 +30328231541868 +855765897480 +21497607540

...

+3984694428 +119255292 +344544 == 17916790562863266656.

...

the total is divisible by 4 and 8, separately, leaving an odd number; the number is also

divisible by 37 and 97, primes from the generator function portion of (n^7+6). also, every

term is positive! hopefully, there's not a typo. this is my best estimate for the number of

normal-6x6 magic squares that best fits the profile for what the answer could be...

...

here's why a "normal magic square" must be further divisible by (n -d), and a "normal magic cube"

must be further divisible by (n -4). the formula is always (n -(L -D)) where L is the number of unique

lines that can pass diagonally or face-wise through the center, and D is the number of dimensions

for the magic shape. thus, a MSQ is divisible by (n -(4 -2))= (n -2), a MCu is divisible by (n -(7 -3))=

(n -4). that is the REAL reason why a MSQ of n= 2 and why a MCube of n= 4 are both not possible!

furthermore, if a tesseract has 16 corners and 12 parallel faces, then there won't be a magic tesseract

for n= 7, because (n -((16/2 +6/2) -4)= (n -(11 -4))= (n -7); half of the parallel faces can't be counted

in this calculation.

...

and, the count for a magic shape, once it's big enough, must be divisible by f= D*(2^(2*(D-1)))

where 'D' is the dimension of the magic object. thus, a MSQ count will be divisible by 8, a MCu

count will be divisible by 48, and a MT(magic tesseract) count will be divisible by 256, again,

once the size of 'n' becomes big enough to register these factors. nice, huh???

...

*QED

12/7/2012

...

so, 275,305,224 is divisible by 3, and 8; I corrected the calculations when 'n' is 'odd', and if you

divide by 4, the number of unique magic squares will remain w/out reflections and rotations.

...

thus, that's why for n= 4, N4= 880 is divisible by 8, and another 2. the part about being divisible

by 'n-2' is the !REAL! reason that there aren't any magic squares for n= 2. it's like n= 2 is a "point

of infinity", geometrically.

...

*QED

12/06/2012

...

...

do you see the pattern? the generator function for n=5 is (n^5 +4), but we would almost have to

know the number of normal magic squares to be counted in order to form all the numerators and

denominators for the necessary fractions!

...

for n= 6, N6= 6*(6^2+1)*(6^3+2)*(6^4+3)*(6^5+4)*(6^6+5)*(6^7+6)/2/2/3/5/6 = 1.7746*10^19

for the first term, so the estimate posted on Wikipedia is somewhat close.

...

on a more interesting note, here's the formula for the "magic" number for "magic squares" that con-

tain only "square numbers" as entries: for 'n', N4^2= [(n +1)*(n^2 -3)*(n^4 +6)]/2. notice that Euler's

4x4 example equals 8515 as promised, and now I can prove that there will NOT be a 3x3 "magic

square" of only "square numbers". oh, someone complained that magic squares of squares has 2

different sums. so, I told him that since the entries are of the form x-squared, they all associate with

the power of 2. there are 2 families of functions for magic squares of squares... Euler found the ex-

ample of [(n +1)*(n^2 -3)*(n^4 +6)]/2, & the other one is [n*(n^2 +3)*(n^4 -1)]/6 which promises a

sum of 3230.

...

...

*QED

12/20/2012

...

...

the centers of "basic magic squares" for all (odd n) will follow the pattern: {subst. M= (n -((n+1)/2))

for 'n' in the triangular formula, Tn = (n*(n+1))/2}; so, the (centerMagic{Sq}n)= {n^2 - 4*[M*(M+1)]/2}.

thus, if n= 3, (centerMSq3)= 3^2 -4*(1)= 5; n= 5, (centerMSq5)= (5^2 -4*(3))= 13, n=7, (centerMSq7)=

(7^2 -4*(6))= 25, n=9, (centerMSq9)= (9^2 -4(10))= 41, etc.

...

the centers of "basic magic cubes" for all (odd n) will follow this simpler formula:

(centerMagic{Cube}n)= [(n +1)*(n^2 -(n -1))]/2. thus, if n= 3, (centerMCu3)= (4/2)*(3^2 -3 +1)= 2*7=

14; n=5, (centerMCu5)= (6/2)*(5^2 -5 +1)= 3*21 = 63; n=7, (centerMCu7) = 4*43 = 172, etc., etc.,

but I can't prove any of them. (they're purely formulary and sound very reasonable!)

...

*QED

12/3/2012

...

...

{{{ leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com; leavemsg1 [at] gmail.com }}}

someone w/degrees from Harvard, Stanford, and Michigan argued that "if it could have been proven

using algebra 'why a 3x3-normal magic square of squares doesn't exist?', then it would've been done

already! let's have another look...

...

...

A B C 8 1 6

...

D E F 3 5 7

...

G H I 4 9 2

...

we can prove that '5' must be the central term. when A +I= 2*E, B +H= 2*E, C +G= 2*E, D + F= 2*E,

the magic number, M= n*(n^2 +1) / 2 = A +E +I. if we let n= 3, then 3*(3^2 +1) / 2 = 3*E, and E = 5.

now, just because A = A^2, B = B^2, etc., our logic shouldn't falter. if n= 3, then we have E^2= 5, or

E= sqrt(5). the entries must be integral. here's one of the closest solutions:

...

127^2 46^2 58^2

...

2^2 113^2 94^2

...

74^2 82^2 97^2

(as borrowed/copied from an article published by John P. Robertson.)

...

3 rows, 3 columns, and 1 diagonal have the sum= 21,609; but, unfortunately, the other diagonal has a

sum= 38,307. (I'm NOT sure why anyone would think that a solution to such an easily-defined problem

127^2 46^2 58^2

...

2^2 113^2 94^2

...

74^2 82^2 97^2

(as borrowed/copied from an article published by John P. Robertson.)

...

3 rows, 3 columns, and 1 diagonal have the sum= 21,609; but, unfortunately, the other diagonal has a

sum= 38,307. (I'm NOT sure why anyone would think that a solution to such an easily-defined problem

would necessarily be larger numbers than a modern-day computer could hold???) {corrected 7/19/14}

...

...

...

...

*QED

1/17/2013

...

...

the reason "why there won't be a 3x3-normal magic square of cubes" is the very same argument as

"of squares", since x^3 +y^3 = 2*z^3 has no integral solutions as confirmed by AlphaWolfram; just

plug it in... also, there's an overall metric which reveals why there's not going to be any 2x2-normal

magic squares, and, moreover, NO examples of a 4x4-normal magic square "of squares" larger than

that found by Euler. the formula that describes the largest absolute

magic square" is Z(n) = (n-1)^n +(n-1) +1. thus, there's one 0x0-nMSQ with up to 0 or 1 entries, one

1x1-nMSQ with entries up to 0 or 1, NULL 2x2-nMSQ's w/up to only 3 numbers that can be entered,

one 3x3-nMSQ which may be formed by 1-11 base entries, and lastly, the largest example of a magic

square of squares is Euler's example with the largest base entry not to exceed 85; his uses a 79^2 as

the largest base "squared".

*QED

5/17/2013

...

...

here's a glimpse for why there AREN'T ANY 4x4-normal magic squares of cubes. you'll notice the

famous taxicab problem when observing their diagonals. we only need to concern ourselves with

the following 2 equations which are "the sum of 2 cubes":

...

a^3 +b^3= 2*(c^3 +d^3) which doesn't have any integral solutions as verified using the Wolfram

Alpha equation-solver, and e^3 +f^3 = g^3 +h^3 along with its quadratic/eliptical restrictions that

I will describe better than Kevin A. Broughan in his paper "Characterizing the Sum of 2 Cubes". let

(the third root(n= e^3 +f^3)) <= m <= (the third root(4*n)) such that j*(e^3 +f^3) = m* (x^2 +k*x), m

<> 2 and x = e +f and j <= (the 6th root of (n = e^3 +f^3)), k <= j. thus, n = e^3 +f^3 = g^3 +h^3,

only has one solution per set S= {e, f, g, and h} due to this modular restriction.

...

here are 3 examples... 2*n= 2*1729 = 2*(1^3 +12^3) = 2*(9^3 +10^3) = 19*14*13 = (g +h)*(e +f +1)

*(e +f) where m = 19, (k= 1) <= (2= j), 2*n= 2*4104= 2*(2^3 +16^3) = 2*(9^3 +15^3)= 24*19*18 =

(g +h)*(e +f +1)*(e +f) where m = 24, (k= 1) <= (2= j), and 2*n= 2*20683= 3*(10^3 +27^3) = 2*(19^3

+24^3)= 43*39*37= (g +h) *(e +f +2)*(e +f) where m= 43, (k= 2) <= (3= j). otherwise, [a = b = c = d]

would be trivial.

...

hence, there will NOT be a 4x4-normal magic square of cubes, because the digonals demand much

more from us than just filling in the cells w/cubic quantities. the magic-square entry locations easily

help us to modularly eliminate the possibilities.

...

*QED

4/21/2013

...

there's a perfectly-good reason for why Euler couldn't find a closed form for Apery's constant, nearly

280 years ago! I skimmed through a paper on the internet written by Walther Janous; he stated that

zeta(3)= 1/(1^3) +1/(2^3) +1/(3^3) + ... = 8* Int{[ln(sin(x))*ln(cos(x))]/[sin(x)*cos(x)]} from 0 to pi()/4,

meaning that an infinite series was equal to a "definite integral or closed form" to be evaluated with

specific limits. it's a little tricky, but HIDDEN within that closed form is a very fine argument. please

don't ever go looking for a closed form for Apery's constant --- it doesn't exist, and we can prove it!

...

for u-substitution, let u = sin(x), du = cos(x) dx, (cos(x))^2 = 1 -(sin(x))^2, and du / cos(x) = dx.

...

8*Int{ [ln(sin(x))* ln(cos(x))] / [sin(x)* cos(x)] }dx evaluated from 0 to pi()/4 = 4*Int{ [ln(1 -u^2)*ln(u)]

/ [ (1 -u^2)*u ] } du evaluated from 0 to (sqrt(2))/2, respectively.

...

for v-substitution, let v = 1 -u^2, dv = -2u du, u = (1 -v)^(1/2), u^2 = 1 -v, and du = dv / (-2u).

...

4*Int{ [ln(1-u^2)*ln(u)] / [(1-u^2)*u] }du = 1*Int{ [ln(v)*ln(1 -v)] / [v*(1 -v)] }dv; the 4 is taken care of by

halfs from the substitution, and the limits go from 1/2 to 1, after removing the opposite of the integral.

...

finally, let t = ln(1 -v), dt = (-1/(1 -v))dv, 1 -e^t = v, and e^t = 1 -v, and let the wolfram alpha engine

show you using "integration by parts" that the initial closed form is actually an open form by the very

nature of its factors, etc.

...

zeta(3)= Int{ [t* ln(1-e^t)] / [(1 -e^t)] }dt = [t^2*ln(1 -e^t)] / [2*(1 -e^t)] }dt minus an integral that is more

non-conforming than the orginal "t-substituted" integral. you can forget about the limits of integration.

ALL successive integrals are absolutely intolerable and will remain unresolved!!!

...

the problem of finding a closed form for zeta(3) leads to more unresolved definite integrals, regardless

of the limits that bound them. review it using the wolfram alpha computational engine, if you like...

*QED

5/20/2014

...

1/17/2013

...

...

the reason "why there won't be a 3x3-normal magic square of cubes" is the very same argument as

"of squares", since x^3 +y^3 = 2*z^3 has no integral solutions as confirmed by AlphaWolfram; just

plug it in... also, there's an overall metric which reveals why there's not going to be any 2x2-normal

magic squares, and, moreover, NO examples of a 4x4-normal magic square "of squares" larger than

that found by Euler. the formula that describes the largest absolute

__base__that can appear in a "normalmagic square" is Z(n) = (n-1)^n +(n-1) +1. thus, there's one 0x0-nMSQ with up to 0 or 1 entries, one

1x1-nMSQ with entries up to 0 or 1, NULL 2x2-nMSQ's w/up to only 3 numbers that can be entered,

one 3x3-nMSQ which may be formed by 1-11 base entries, and lastly, the largest example of a magic

square of squares is Euler's example with the largest base entry not to exceed 85; his uses a 79^2 as

the largest base "squared".

*QED

5/17/2013

...

...

here's a glimpse for why there AREN'T ANY 4x4-normal magic squares of cubes. you'll notice the

famous taxicab problem when observing their diagonals. we only need to concern ourselves with

the following 2 equations which are "the sum of 2 cubes":

...

a^3 +b^3= 2*(c^3 +d^3) which doesn't have any integral solutions as verified using the Wolfram

Alpha equation-solver, and e^3 +f^3 = g^3 +h^3 along with its quadratic/eliptical restrictions that

I will describe better than Kevin A. Broughan in his paper "Characterizing the Sum of 2 Cubes". let

(the third root(n= e^3 +f^3)) <= m <= (the third root(4*n)) such that j*(e^3 +f^3) = m* (x^2 +k*x), m

<> 2 and x = e +f and j <= (the 6th root of (n = e^3 +f^3)), k <= j. thus, n = e^3 +f^3 = g^3 +h^3,

only has one solution per set S= {e, f, g, and h} due to this modular restriction.

...

here are 3 examples... 2*n= 2*1729 = 2*(1^3 +12^3) = 2*(9^3 +10^3) = 19*14*13 = (g +h)*(e +f +1)

*(e +f) where m = 19, (k= 1) <= (2= j), 2*n= 2*4104= 2*(2^3 +16^3) = 2*(9^3 +15^3)= 24*19*18 =

(g +h)*(e +f +1)*(e +f) where m = 24, (k= 1) <= (2= j), and 2*n= 2*20683= 3*(10^3 +27^3) = 2*(19^3

+24^3)= 43*39*37= (g +h) *(e +f +2)*(e +f) where m= 43, (k= 2) <= (3= j). otherwise, [a = b = c = d]

would be trivial.

...

hence, there will NOT be a 4x4-normal magic square of cubes, because the digonals demand much

more from us than just filling in the cells w/cubic quantities. the magic-square entry locations easily

help us to modularly eliminate the possibilities.

...

*QED

4/21/2013

...

there's a perfectly-good reason for why Euler couldn't find a closed form for Apery's constant, nearly

280 years ago! I skimmed through a paper on the internet written by Walther Janous; he stated that

zeta(3)= 1/(1^3) +1/(2^3) +1/(3^3) + ... = 8* Int{[ln(sin(x))*ln(cos(x))]/[sin(x)*cos(x)]} from 0 to pi()/4,

meaning that an infinite series was equal to a "definite integral or closed form" to be evaluated with

specific limits. it's a little tricky, but HIDDEN within that closed form is a very fine argument. please

don't ever go looking for a closed form for Apery's constant --- it doesn't exist, and we can prove it!

...

for u-substitution, let u = sin(x), du = cos(x) dx, (cos(x))^2 = 1 -(sin(x))^2, and du / cos(x) = dx.

...

8*Int{ [ln(sin(x))* ln(cos(x))] / [sin(x)* cos(x)] }dx evaluated from 0 to pi()/4 = 4*Int{ [ln(1 -u^2)*ln(u)]

/ [ (1 -u^2)*u ] } du evaluated from 0 to (sqrt(2))/2, respectively.

...

for v-substitution, let v = 1 -u^2, dv = -2u du, u = (1 -v)^(1/2), u^2 = 1 -v, and du = dv / (-2u).

...

4*Int{ [ln(1-u^2)*ln(u)] / [(1-u^2)*u] }du = 1*Int{ [ln(v)*ln(1 -v)] / [v*(1 -v)] }dv; the 4 is taken care of by

halfs from the substitution, and the limits go from 1/2 to 1, after removing the opposite of the integral.

...

finally, let t = ln(1 -v), dt = (-1/(1 -v))dv, 1 -e^t = v, and e^t = 1 -v, and let the wolfram alpha engine

show you using "integration by parts" that the initial closed form is actually an open form by the very

nature of its factors, etc.

...

zeta(3)= Int{ [t* ln(1-e^t)] / [(1 -e^t)] }dt = [t^2*ln(1 -e^t)] / [2*(1 -e^t)] }dt minus an integral that is more

non-conforming than the orginal "t-substituted" integral. you can forget about the limits of integration.

ALL successive integrals are absolutely intolerable and will remain unresolved!!!

...

the problem of finding a closed form for zeta(3) leads to more unresolved definite integrals, regardless

of the limits that bound them. review it using the wolfram alpha computational engine, if you like...

*QED

5/20/2014

...