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 . the explanation would have been difficult for anyone

to understand. after seeing the form for , I can explain to anyone “why it is impossible to

find a closed form for “, 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): iff ‘n’ is prime.

(proof, backward)

if ‘n’ is prime, then by definition, and as desired.

(proof, forward)

let such that . assume that is ‘composite’

where , ‘b’ is a ‘prime’ factor, . if when , then we’d

have ‘n’ equal to the power of a prime, and . that’s not equal to .

so, if , then implies

implies , implies

that , and an integral function assumes that

. see the argument! let where ,

, , and . 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 ,

if the 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 along with the fact that each of the terms of the 2

expressions contain ‘m’s, ‘k’s, and ‘b’s. or , or would be undefined.

so, , , and . either 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 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 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 or… 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 , 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) *; (an even-perfect number),

EP(N) = , but please dismiss the first EP(N) = such that

EP(N)’s = , etc.

(an odd-perfect number), (OPN), for x, y, z.

giving , or ;

the first derivative is symmetric around sigma’s unity at : (EP(N) =

, and for x, y, & z = 0: (OPN), such that

; (the derivative acts as an

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

mation formula!)

and ;

.

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

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 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 function”, and that sets the stage for learning something

about Euler’s function: if and only if which’s true,

then because , we can surely understand the following two equations:

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 and can be meaningful,

iff is closely monitored, and ‘n’ is prime.

must be prime! let’s take a little excursion…

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

if , and are both prime, then if ,

and , then , the base… is prime. also, if , then

choose a different ‘p’ or if N is a square, then is prime. (someone would have to prove

this conjecture.)

let

let , and , but

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

and and .

thus, if , and , then must be prime!

I’ve listed the 4 equations for a Mersenne prime number where is also prime.

(only one of these needs to be true!)

(both prime) , or

(again,… ), then (can’t be +1)

(again, both prime), then

(again, both prime)

let’s have another look!!!

let

let such that ; and (by mere

chance!!!), so then…

and and

thus, if , or , then must be prime!

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 . if , then ‘m’ can

only be of the form , or… an almost-perfect number.

proof:

either , or

implies that is defined for all ‘k’. if

in the latter statement, then will be fractional; also, if , then

; if when , then .

if you tried to argue w/the former statement that , then you’d have

to answer a much bigger question about the sigma function, since ;

it’s certainly false, since 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 (of odd squares)

can be equal to an odd number, and odd squares aren’t of the form , so

is ALWAYS EVEN.

*QED

if is ALWAYS EVEN, then an APN must only be of the form ; 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

where , and , and ; it implies that

which is divergent; ‘n’ is an element the set of

Natural (consecutive) Numbers. similarly, there will be infinitely many Mersenne primes of the

form where , and

such that is 1-to-1 and onto with respect to the set of Whole (consecutive) Numbers.

thus, if and only if, . this

fact can be confirmed after computing only the first few terms. therefore, there are infinitely many

Mersenne primes of the form .

*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 where , and

such that is 1-to-1 and onto with respect to the set of Whole (consecutive)

Numbers. thus, iff I’m correctly understanding his argument, then

. this simple fact can be confirmed after computing

just the first few terms.

therefore, there are finitely many Fermat primes of the form .

*QE(Dirichlet’s Interpretation)

01/25/2011

there are infinitely many primes of the form ! I didn’t create the theorem (strong portion)…

there will be infinitely many primes of the form where

such that is 1-to-1 and onto with respect to the set of

Natural (consecutive) Numbers. thus, if and only if,

. it can be confirmed after computing only the first few terms.

therefore, there are infinitely many primes of the form .

*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 repeats}

; {not noticing the subtraction of repeats}

; (there aren’t any fractional amounts of 2×2 magic squares!)

{minus repeats!}

;

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 terms in each summation for a different size, ‘n':

the solution for was found on a hunch that I could bump up the exponent to , and

then when equaled , I knew that I had something. so, I prepared to match Rich

Shroeppel’s answer for , 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 w/out either of us suffering!

12/21/2013**

for , there will be terms; look below…

{minus where , and . this subtraction is only done when ‘n’ is odd!}

.

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

‘‘ or .

it’s only an estimate but, here’s my estimate for

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

…

…

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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

…