Odd-Perfect Numbers Don’t Exist

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’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) \vert (n -1) iff ‘n’ is prime.

(proof, backward)

if ‘n’ is prime, then \phi(n) = (n -1) by definition, and \phi(n) \vert (n -1) as desired.

(proof, forward)

let \phi(n) \vert (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 \ne 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 \ne 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)= \frac{[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) = \frac{(bC -1)}{(b -1)}b= C, and \phi(C)= C +1either \phi() is undefined, or ‘C’ can’t 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.)

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the solutions to both “Ramsey Theory” and the Erdos–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!

 

look near the bottom of this page to see why the \zeta(3) or… \sum_{k=1}^\infty \frac{1}{n^3} fractions won’t have

a closed form; you’ll enjoy the answer! also, if you’d like to count the number of normal

magic squares, for n= a, then go a little further down 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 x, y, z.

giving Q^2 *(1 +r +r^2 + ... +r^n)= Q^2 *\frac {(1-r^{(n+1)})}{(1-r)}, or (4x+1)^{(4y+1)} = \frac {(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. \frac{d}{dr} [(1-r)* (4x+1)^{(4y+1)}] = [1-r^{(n+1)}] \frac{d}{dr}; (the derivative acts as an

operator applied to a tensor whereby we discover the restriction or boundary of Euler’s sum-

mation 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!

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

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!

 

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 something

about Euler’s \phi() function: if and only if \phi(n^2 -1) = 2*\phi(n +1)*\phi(n -1) which’s true,

then because \phi(n^k) = n^{(k -1)}*\phi(n), we can surely understand the following two equations:

    \[\phi(n+1) = \frac {[(n -1)*\phi(n^2 -1)]}{[2*\phi((n -1)^2)]}, \hspace{25pt} \phi(n -1) = \frac {\phi((n-1)^2)}{(n -1)}.\]

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) and \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 \equiv N,

and q \mod N \equiv +/-1, then (M_r), the base… is prime. also, if (M_r) \mod p \equiv 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 \vert [(C_5)^5 -5] \mod 13 \equiv N, but 2^{(2*27)} \mod 13 \equiv 12 \equiv (-1)

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

(-1)^{(odd power)}*2 -1 \equiv

(-1)^{49*19*43*73*127*337*5419*92737*649657*77158673929}*2^1 -1 \equiv

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

thus, if [(C_5)^5 -5] \mod 13 \equiv 12, and 13 \mod 12 \equiv -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 \equiv -1, or

p = 4k+3, q = 2p+3 (again,… ), then [(M_r)^p-p] \mod q \equiv -1 (can’t be +1)

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

p = 4k+3, q = 2p+1 (again, both prime) [(M_r)^p-p] \mod q \equiv 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 \equiv N; and 2^{(27)} \mod 7 \equiv 1 (by mere

chance!!!), so then…

(1)^{(left over exponent)}*2 -1 \equiv

(1)^{2*49*19*43*73*127*337*5419*92737*649657*77158673929}*2^1 -1 \equiv (1)*2 -1

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

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

C_5 is definitely prime, if my study is correct.

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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)= \frac {[n*2^{(k+2)} +2^{(k+1)} -1]}{(4n+1)} is defined for all ‘k’. if n \ne 0

in the latter statement, then \sigma(2^k) will be fractional; also, if k = 0, n = 0, then \sigma(1) =

\frac {(0 +2 -1)}{(4(0) +1)}= \frac {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 \frac {1}{f(0)} \equiv

\frac {1}{0} \ll \sum_{n=1}^\infty \frac {1}{g(n)} \equiv [\frac {1}{3} +\frac {1}{7} + \frac {1}{11} + \frac {1}{19}... + \frac {1}{n}] which is divergent; ‘n’ is an element the set of

Natural (consecutive) 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, \frac{1}{N(1)} = \frac {1}{2} \ll [\sum_{x=1}^\infty \frac {1}{M(x)}] \equiv [\frac {1}{3} +\frac {1}{7} +\frac {1}{31} + \frac {1}{127} +\frac {1}{8191} + ... +\frac {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 the

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 \frac {1}{G(0)} = \frac {1}{1} \gg [\sum_{x=1}^\infty \frac {1}{F(x)}]

\equiv [\frac {1}{3}+\frac {1}{5} +\frac {1}{7} + \frac {1}{257} +\frac {1}{65537} + ... + \frac {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, \frac{1}{N(1)} = \frac{1}{2} \ll [\sum_{x=1}^\infty \frac{1}{M(x)}] \equiv [\frac{1}{2} +\frac{1}{5} +\frac{1}{17}

+ ... + \frac {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)

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let’s try counting the number of “normal magic squares” for a certain size, ‘n':

{not noticing the subtraction of {((n -d)^2)*(2^n)} repeats}

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

\Large n= 2; N_2 = \frac {n+1}{2} = \frac{3}{2} = 0; (there aren’t any fractional amounts of 2×2 magic squares!)

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

\Large n= 3; N_3 = \frac {(n+1)*(n^2+1)}{2*2} + \frac {(-1)*(n^2+1)}{2*5} = 10 -1 -((3-2)^2)*(2^3) = 1

\Large n= 4; N_4 = \frac{n*(n^2 +1)*(n^3 +2)}{2*3} + \frac {n^2*(n^3 +2)}{2*4} = 748 + 132 = 880;

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(\frac {(n \pm 1)}{2})] terms in each summation 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(\frac{(5 +1)}{2})] = 8 terms; look below…

\Large n=5; N5 = \frac {n*(n^2 +1)*(n^3 +2)*(n^4 +3)*(n^5 +4)}{2*2*2*3*5} + \frac {n^2*(n^3 +1)*(n^4 +2)*(n^5 +4)}{2*3*3*3*5*5}

\Large + \frac {(n -1)*n^3*(n^4 +1)*(n^5 +4)}{2*2*3*5*5*5} + \frac {-2*(n^2 +1)*n^4*(n^5 +4)}{2*2*3*5*5} + \frac {(n-2)*(n^4 -1)*(n^5 +4)}{</span>2*3*3*3} +

\Large + \frac {(n^3 -2)*(n^5 +4)}{3*3*7} + \frac {(n^2 -3)*(n^5 +4)}{2*3*11} + \frac {-3*(n -4)*(n^5 +4)}{3*7} -9*32 \equiv

{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) \equiv

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-6×6 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
4×4 example equals 8515 as promised, and now I can prove that there will NOT be a 3×3 “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 3×3-normal magic square of squares doesn’t exist?’, then it would’ve been done
already! let’s have another look…

generically…
A B C 8 1 6

D E F 3 5 7; there’s an even # of even parity, and an odd # of odd parity!

G H I 4 9 2

assume that we have a 3×3-normal magic square of squares, forget that the 1 through 9 example is
unique; it doesn’t matter. now, imagine that we are comparing any two unrelated rows or columns. I
only have to prove that an even/odd “parity” of squared entries will differ. then, a 3×3-normal magic
square of squares isn’t possible.

(the sum of 3 “squares” equal to the sum of 3 unrelated “squares”, any 1 term isn’t the same as a term
from another row/column. the conclusion is… if the “parity” differs, then the 3×3-normal magic square
of squares isn’t possible.)

here’s the algebraic connection which confirms the difference in “parity” among cells in a 3×3 matrix:

(x*y)^2 +(x*(y+1))^2 +((x+1)*(y+2))^2
= ((x+1)*y)^2 +((x+1)*(y+1))^2 +((x+2)*(y+2))^2

has two solutions: x= -1, y= -1 (OR) x= -1, y = 3; so, either the “squared” entries are all “odd” & equal to
each other or their “parity” differs… i.e. there can’t be an all even or all odd matrix. here’s the closest so-
lution exhibiting a matrix that has differences in “parity”:

127^2 46^2 58^2

2^2 113^2 94^2; there’s still an even # of even parity and an odd # of odd parity entries!

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. finally, since “parity” must differ, I would use this argument to seal the deal… example…
(2a+1)^2 +(2b)^2 +(2c)^2 <> (2a+1)^2 +(2e+1)^2 +(2i+1)^2, and it’s done using algebra! hence, our
unique, 3×3-normal magic square of squares can’t exist! (I am 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 1/25/2014***}

if you don’t believe that a difference in “parity” will exist, then have a look at this identity: t^2 +u^2 +v^2
= w^2 where t= x, u = x+1, v= x*(x+1), and w= x^2 +x +1. the difference in “parity” exists even when the
sum of 3 squares is equal to a square.

*QED
1/17/2013


the reason “why there won’t be a 3×3-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 2×2-normal
magic squares, and, moreover, NO examples of a 4×4-normal magic square “of squares” larger than
that found by Euler. the formula that describes the largest absolute base that can appear in a “normal
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
1×1-nMSQ with entries up to 0 or 1, NULL 2×2-nMSQ’s w/up to only 3 numbers that can be entered,
one 3×3-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 4×4-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 4×4-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 Walther Janous, and 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 an 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