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.
...
D. H. Lehmer's (1932) Totient Conjecture Stated: 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) | (n -1), exactly.
(proof, forward)
let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime; let n= bC where 'b' is one prime
factor of 'n', and 'C' is one or more other factors, combined, and k = 1. assume that k <> 1; so,
k*phi(bC) = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1] / [b*k -k]; it has
to be true that k= C= 1 in order for phi(C) to not be fractional while phi(1) = (b -1) / (b -1) = 1
is verified. hence, 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1] / [b -1]
implies that C= b as the only solution. phi(C) = [b^2 -1] / [b -1]; therefore, it's not possible to
have phi(C)= C +1, (OR) there cannot be more co-prime divisors of C than there are the total
number of possible divisors! thus, by contradiction, 'n' must be prime.
...
it took me about 20 minutes to solve Lehmer's Conjecture using a sharpie marker on the back
of a receipt for car repairs. at first glance, it looks as though the problem needs to be solved
using Fermat's little theorem; that is DEFINITELY NOT the case. enjoy!
...
*QED
05/21/2013
...
...
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!
...
look near the bottom of this page to see why the zeta(3) or... sum of 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 norm-
al magic squares, for n= a, then go directly to the bottom of this page. I hope you enjoy it!
...
The proof for Olson's conjecture has been moved to the Other Short Proofs page. 5/4/2013
...
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.
...
(d/dr) [(1-r)* (4x+1)^(4y+1)] = [1-r^(n+1)] (d/dr); (the derivative acts as an operator applied
to a tensor!)
...
-(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.
...
it's hard to explain, but when the restriction x = 0 is substituted into the original equation, it
must yield an erroneous or meaningless result, or it must form the solution to the equation. it
cannot NOT do both. it tells me that Q^2 must be the solution, and someone proved that that
isn't possible; hence, the set must be vacuous. it's like a "double-positive" result... as if some-
one answered the question by saying... "yeah, right."
...
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!
...
...
C5 must be prime! let's take a little excursion...
remember... C5 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 [(Mr)^p -p] mod q == N, and q mod N ==
+/-1, then (Mr), the base... is prime. also, if (Mr) mod p = 1, then choose a different 'p' or if N
is a square, then (Mr) is prime. (someone would have to prove this conjecture.)
...
let 2^127 -1 = 170141183460469231731687303715884105727
...
C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
...
let p = 5, and q = 13 such that [(C5)^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 [(C5)^5 -5] mod 13 == 12.
...
thus, if [(C5)^5 -5] mod 13 == 12, and 13 mod 12 == -1, then C5 must be prime!
...
I've listed the 4 equations for a Mersenne prime number (Mr) where r is also prime.
...
(only one of these needs to be true!)
p = 4k+1, q = 2p+3 (both prime) [(Mr)^p-p] mod q == -1, or
p = 4k+3, q = 2p+3 (again,... ), then [(Mr)^p-p] mod q == -1 (can't be +1)
p = 4k+1, q = 2p+1 (again, both prime), then [(Mr)^p-p] mod q == p
p = 4k+3, q = 2p+1 (again, both prime) [(Mr)^p-p] mod q == p+2
...
let's have another look!!!
let 2^127 -1 = 170141183460469231731687303715884105727
...
C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
...
let p = 3, q = 7 such that [(C5)^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 [(C5)^3 -3] mod 7 == 5
...
thus, if [(C5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C5 must be prime!
...
C5 is definitely prime, if my study is correct.
...
...
a short proof that "there exists a prime between x^2 and (x +1)^2.", or Legendre's conjecture
is actually a theorem! look below...
...
assume that... (x^2)/ln(x^2) +1 < ((x +1)^2)/ln((x +1)^2) using the Prime Number Theorem,
(x^2 +ln(x^2))/ln(x^2) < (x^2 +2x +1)/ln((x +1)^2) combining the formula w/+1 on the LHS,
x^2 +2*ln(x) < (x^2 +2x +1)*[(ln(x^2))/(ln((x +1)^2)] multiplying both sides by ln(x^2),
...
but the limit [(ln(x^2))/(ln((x +1)^2)] -> 1 almost immediately, and we have ln(x^2) < 2x +1,
and taking the derivative wrt 'x', we have... (d/dx)[ln(x^2)] < (d/dx)[2x +1], or... 2/x <2, 1<
x; so our beginning statement has to be true for x > 1, and it is...
...
x= 2; LHS= 3.885 < 4.096 = RHS, and x= 3; LHS= 5.096 < 5.770 = RHS...and so on, and
so on. we simply check between 1 and 4 to see that the primes 2 and 3 exist. I'm using
approximations of pi(x) to secure his conjecture--- it's like I'm teeing off from "being in the
rough" and "still making a superb hole-in-one!"
*QED
...
when considering Legendre's stronger conjecture, please substitute the 'sqrt(x)' in for '1' in
the previous beginning statement. then subtract x^2 to get... sqrt(x)*ln(x^2) < 2x, and w/out
loss of generality, subtract '1' from only the RHS; divide by '2' and "square" both sides to
arrive at... x*[ln(x)]^2 < [x]^2; take the derivative of both sides twice (using the chain rule),
multiply by 'x', and divide by '2'. we have... 1 +ln(x) < 2x, which is true for x > 0. {corrected
on 04/05/2012}
*QED
09/17/2011 (both)
...
...
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 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 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.
*QED(irichlet's Interpretation)
...
...
note: the sum of fractions 1/(n^4) from n= 1 to infinity is equal to [(pi^4)/90]. the summation of
1/(n^3) from n= 1 to infinity WON'T HAVE A CLOSED FORM!
(proof:)
...
infinity { 1 } 1 1 1 1 1
sum of { ---------- } = ----- + ----- + ----- + ----- + ------- + ...
for n=1 { n^3 } 1 8 27 64 125
...
which is equal to...
...
1 1 1 1
= ----- + ------ + ----- + ------- + ...
1 16 81 256
...
1 1 1
+ ------ + ----- + ------- + ...
16 81 256
...
1 1
+ ----- + ------- + ...
81 256
...
1
+ ------- + ...
256
...
turned upright, equals...
...
1
+ --- {+1 -1} +
1
...
1 1 1
+ ------ + ------ + ------ + ...
16 81 16
...
1 1 1 1 1
+ ------ + ------ + ------ + ------ + ------ + ...
81 256 625 256 81
...
1 1 1 1 1 1 1
+ ------ + ------ + ------- + ------- + ------- + ------ + ------ + ...
256 625 1296 1296 1296 625 256
...
which is equal to...
...
{2 * [(pi^4)/90] -1} + {1/16 -1/16 + 1 -1} +
...
1
+ ------ + ...
81
...
1 1
------ + ------ + ...
256 256
...
1 1 1
+ ------ + ------ + ------ + ...
625 625 625
...
etc., etc.
...
giving us...
...
infinity { 1 } [ pi^4 ] [ ] [ 1 ] [ 1 ]
sum of { -------- } = n* [ -------- ] -(n-1)* [ 1 ] -(n-2)* [-------] -(n-3)* [-------] - ...
for n=1 { n^3 } [ 90 ] [ ] [ 16 ] [ 81 ]
...
infinity { 1 } [ pi^4 ] infinity { 1 }
sum of { -------- } = n * [ ------- ] - sum of { (n-a)* ------- } + {some small error function} ...
for n=1 { n^3 } [ 90 ] for a=1 { a^4 }
...
this is a "double-telescopic" sum for a "fixed n"; you would have to sum the first ~50 terms
where n= 100 (fixed) and ranging from a=1 to 50 to find convergence; no one is going to
get famous for counting fractions, and there'll be NO CLOSED FORM, since 'n' can be fixed
in infinitely many ways.
...
the sum of 1/(n^5) from n=1 to infinity can be summed using 1/(n^6) fractions similarly.
...
*QED
10/24/2012
...
look at this spreadsheet to see the concrete results:
...
sum of cubed fractions equal to double-telescopic sum of quartic-
... powered fractions for fixed, n= 1000
1 1 1.000000000000 1000 1.082323234 1082.323233711140
1 2 0.125000000000 -999 1 1 -999.000000000000
1 3 0.037037037037 -998 1 2 -62.375000000000
1 4 0.015625000000 -997 1 3 -12.308641975309
1 5 0.008000000000 -996 1 4 -3.890625000000
1 6 0.004629629630 -995 1 5 -1.592000000000
1 7 0.002915451895 -994 1 6 -0.766975308642
1 8 0.001953125000 -993 1 7 -0.413577675968
...
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
...
1 991 0.000000001027 -10 1 990 -0.000000000010
1 992 0.000000001024 -9 1 991 -0.000000000009
1 993 0.000000001021 -8 1 992 -0.000000000008
1 994 0.000000001018 -7 1 993 -0.000000000007
1 995 0.000000001015 -6 1 994 -0.000000000006
1 996 0.000000001012 -5 1 995 -0.000000000005
1 997 0.000000001009 -4 1 996 -0.000000000004
1 998 0.000000001006 -3 1 997 -0.000000000003
1 999 0.000000001003 -2 1 998 -0.000000000002
1 1000 0.000000001000 -1 1 999 -0.000000000001
1 1001 0.000000000997 sum=1.202056736493
1 1002 0.000000000994
1 1003 0.000000000991
1 1004 0.000000000988
1 1005 0.000000000985
...
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
...
1 2995 0.000000000037
1 2996 0.000000000037
1 2997 0.000000000037
1 2998 0.000000000037
1 2999 0.000000000037
1 3000 0.000000000037
1 3001 0.000000000037
sum=1.202056847660 sum=1.202056736493
...
...
sum of quintic-powered fractions equal to double-telescopic sum of sextic-
... powered fractions for fixed, n= 1000
1 1 1.000000000000 1000 1.017343062 1017.343061984450
1 2 0.031250000000 -999 1 1 -999.000000000000
1 3 0.004115226337 -998 1 2 -15.593750000000
1 4 0.000976562500 -997 1 3 -1.367626886145
1 5 0.000320000000 -996 1 4 -0.243164062500
1 6 0.000128600823 -995 1 5 -0.063680000000
1 7 0.000059499018 -994 1 6 -0.021304869684
1 8 0.000030517578 -993 1 7 -0.008440360734
1 9 0.000016935088 -992 1 8 -0.003784179688
1 10 0.000010000000 -991 1 9 -0.001864741335
1 11 0.000006209213 -990 1 10 -0.000990000000
1 12 0.000004018776 -989 1 11 -0.000558264717
1 13 0.000002693291 -988 1 12 -0.000330879201
1 14 0.000001859344 -987 1 13 -0.000204482920
1 15 0.000001316872 -986 1 14 -0.000130950964
...
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
...
1 195 0.000000000004 -806 1 194 -0.000000000015
1 196 0.000000000003 -805 1 195 -0.000000000015
1 197 0.000000000003 -804 1 196 -0.000000000014
1 198 0.000000000003 -803 1 197 -0.000000000014
1 199 0.000000000003 -802 1 198 -0.000000000013
1 200 0.000000000003 -801 1 199 -0.000000000013
1 201 0.000000000003 -800 1 200 -0.000000000013
sum=1.036927754992 sum=1.036927755606
...
*QED
11/01/2012
...
...
let's try counting the number of "normal magic squares" for a certain size, 'n':
...
n 1 {((n -d)^2)*(2^n)}
n= 1; N1 = ------- = ------- = 1; {not noticing the minus of [((1 -1)^2)*(2^1) = 0] repeats}
1 1
...
n +1 3
n= 2; N2 = --------- = ----- = 0; (there aren't any fractional amounts of magic squares!)
2 2
...
(n +1)*(n^2 +1) (-1)*(n^2 +1) {minus ((n -d)^2)*(2^n) repeats!}
n= 3; N3 = --------------------------- + ----------------------- = 10 -1 -((3-2)^2)*(2^3) = 1; w/out removing
2*2 2*5 rotations and reflections}
{so, there's only 1 answer after removing 'no' reflections.}
...
n*(n^2 +1)*(n^3 +2) n^2*(n^3 +2)
n= 4; N4 = --------------------------------- + ------------------------ = 748 + 132 = 880; there are
2*3 2*4 220 unique after removing
... rotations and reflections.
...
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':
...
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 calculations don't take into account the number of excluded reflections, and rotations, etc.
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 N6 = 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
...
...
there's a subtle reason for "why a 3x3-normal magic square with 'square' entries doesn't exist." if
we examined a 3x3-normal magic square, we would immediately see the following, more basic
relationship:
...
A B C
...
D E F
...
G H I
...
for n = 3, we notice that B +H = 2*E, D +F = 2*E, G +C = 2*E, and A +I = 2*E. so, if we let each
entry be a 'square', then by explicit substitution, we have...
...
a^2 b^2 c^2
...
d^2 e^2 f^2
...
g^2 h^2 i^2
...
thus, if we chose any of those above equations, then we'd have... x^2= 2*z^2 -y^2. however, with
Pythagorean triples from... a^2 = b^2 -c^2, solutions can ONLY exist when a, b, and c are co-prime,
and ONLY when b^2 -c^2 can be factored properly. but, this isn't the case with 2*z^2 -y^2; it can't
be factored. hence, a 3x3-normal magic square w/square entries can't exist! (I'm not sure why any-
one would think that the solution to such an easily-defined problem would necessarily come from
larger numbers than a modern-day computer could hold???)
*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 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
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
...
just click on the headings above to navigate to the several different pages of this website.
...
D. H. Lehmer's (1932) Totient Conjecture Stated: 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) | (n -1), exactly.
(proof, forward)
let phi(n) divide (n -1), or k*phi(n) = (n -1), and 'n' isn't prime; let n= bC where 'b' is one prime
factor of 'n', and 'C' is one or more other factors, combined, and k = 1. assume that k <> 1; so,
k*phi(bC) = bC -1; k*(b -1)*phi(C) = bC -1 such that phi(C) is equal to [bC -1] / [b*k -k]; it has
to be true that k= C= 1 in order for phi(C) to not be fractional while phi(1) = (b -1) / (b -1) = 1
is verified. hence, 1*phi(bC) = bC -1, and phi(b)*phi(C) = (bC -1), and phi(C) = [bC -1] / [b -1]
implies that C= b as the only solution. phi(C) = [b^2 -1] / [b -1]; therefore, it's not possible to
have phi(C)= C +1, (OR) there cannot be more co-prime divisors of C than there are the total
number of possible divisors! thus, by contradiction, 'n' must be prime.
...
it took me about 20 minutes to solve Lehmer's Conjecture using a sharpie marker on the back
of a receipt for car repairs. at first glance, it looks as though the problem needs to be solved
using Fermat's little theorem; that is DEFINITELY NOT the case. enjoy!
...
*QED
05/21/2013
...
...
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!
...
look near the bottom of this page to see why the zeta(3) or... sum of 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 norm-
al magic squares, for n= a, then go directly to the bottom of this page. I hope you enjoy it!
...
The proof for Olson's conjecture has been moved to the Other Short Proofs page. 5/4/2013
...
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.
...
(d/dr) [(1-r)* (4x+1)^(4y+1)] = [1-r^(n+1)] (d/dr); (the derivative acts as an operator applied
to a tensor!)
...
-(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.
...
it's hard to explain, but when the restriction x = 0 is substituted into the original equation, it
must yield an erroneous or meaningless result, or it must form the solution to the equation. it
cannot NOT do both. it tells me that Q^2 must be the solution, and someone proved that that
isn't possible; hence, the set must be vacuous. it's like a "double-positive" result... as if some-
one answered the question by saying... "yeah, right."
...
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!
...
...
C5 must be prime! let's take a little excursion...
remember... C5 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 [(Mr)^p -p] mod q == N, and q mod N ==
+/-1, then (Mr), the base... is prime. also, if (Mr) mod p = 1, then choose a different 'p' or if N
is a square, then (Mr) is prime. (someone would have to prove this conjecture.)
...
let 2^127 -1 = 170141183460469231731687303715884105727
...
C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
...
let p = 5, and q = 13 such that [(C5)^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 [(C5)^5 -5] mod 13 == 12.
...
thus, if [(C5)^5 -5] mod 13 == 12, and 13 mod 12 == -1, then C5 must be prime!
...
I've listed the 4 equations for a Mersenne prime number (Mr) where r is also prime.
...
(only one of these needs to be true!)
p = 4k+1, q = 2p+3 (both prime) [(Mr)^p-p] mod q == -1, or
p = 4k+3, q = 2p+3 (again,... ), then [(Mr)^p-p] mod q == -1 (can't be +1)
p = 4k+1, q = 2p+1 (again, both prime), then [(Mr)^p-p] mod q == p
p = 4k+3, q = 2p+1 (again, both prime) [(Mr)^p-p] mod q == p+2
...
let's have another look!!!
let 2^127 -1 = 170141183460469231731687303715884105727
...
C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
...
let p = 3, q = 7 such that [(C5)^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 [(C5)^3 -3] mod 7 == 5
...
thus, if [(C5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C5 must be prime!
...
C5 is definitely prime, if my study is correct.
...
...
a short proof that "there exists a prime between x^2 and (x +1)^2.", or Legendre's conjecture
is actually a theorem! look below...
...
assume that... (x^2)/ln(x^2) +1 < ((x +1)^2)/ln((x +1)^2) using the Prime Number Theorem,
(x^2 +ln(x^2))/ln(x^2) < (x^2 +2x +1)/ln((x +1)^2) combining the formula w/+1 on the LHS,
x^2 +2*ln(x) < (x^2 +2x +1)*[(ln(x^2))/(ln((x +1)^2)] multiplying both sides by ln(x^2),
...
but the limit [(ln(x^2))/(ln((x +1)^2)] -> 1 almost immediately, and we have ln(x^2) < 2x +1,
and taking the derivative wrt 'x', we have... (d/dx)[ln(x^2)] < (d/dx)[2x +1], or... 2/x <2, 1<
x; so our beginning statement has to be true for x > 1, and it is...
...
x= 2; LHS= 3.885 < 4.096 = RHS, and x= 3; LHS= 5.096 < 5.770 = RHS...and so on, and
so on. we simply check between 1 and 4 to see that the primes 2 and 3 exist. I'm using
approximations of pi(x) to secure his conjecture--- it's like I'm teeing off from "being in the
rough" and "still making a superb hole-in-one!"
*QED
...
when considering Legendre's stronger conjecture, please substitute the 'sqrt(x)' in for '1' in
the previous beginning statement. then subtract x^2 to get... sqrt(x)*ln(x^2) < 2x, and w/out
loss of generality, subtract '1' from only the RHS; divide by '2' and "square" both sides to
arrive at... x*[ln(x)]^2 < [x]^2; take the derivative of both sides twice (using the chain rule),
multiply by 'x', and divide by '2'. we have... 1 +ln(x) < 2x, which is true for x > 0. {corrected
on 04/05/2012}
*QED
09/17/2011 (both)
...
...
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 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 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.
*QED(irichlet's Interpretation)
...
...
note: the sum of fractions 1/(n^4) from n= 1 to infinity is equal to [(pi^4)/90]. the summation of
1/(n^3) from n= 1 to infinity WON'T HAVE A CLOSED FORM!
(proof:)
...
infinity { 1 } 1 1 1 1 1
sum of { ---------- } = ----- + ----- + ----- + ----- + ------- + ...
for n=1 { n^3 } 1 8 27 64 125
...
which is equal to...
...
1 1 1 1
= ----- + ------ + ----- + ------- + ...
1 16 81 256
...
1 1 1
+ ------ + ----- + ------- + ...
16 81 256
...
1 1
+ ----- + ------- + ...
81 256
...
1
+ ------- + ...
256
...
turned upright, equals...
...
1
+ --- {+1 -1} +
1
...
1 1 1
+ ------ + ------ + ------ + ...
16 81 16
...
1 1 1 1 1
+ ------ + ------ + ------ + ------ + ------ + ...
81 256 625 256 81
...
1 1 1 1 1 1 1
+ ------ + ------ + ------- + ------- + ------- + ------ + ------ + ...
256 625 1296 1296 1296 625 256
...
which is equal to...
...
{2 * [(pi^4)/90] -1} + {1/16 -1/16 + 1 -1} +
...
1
+ ------ + ...
81
...
1 1
------ + ------ + ...
256 256
...
1 1 1
+ ------ + ------ + ------ + ...
625 625 625
...
etc., etc.
...
giving us...
...
infinity { 1 } [ pi^4 ] [ ] [ 1 ] [ 1 ]
sum of { -------- } = n* [ -------- ] -(n-1)* [ 1 ] -(n-2)* [-------] -(n-3)* [-------] - ...
for n=1 { n^3 } [ 90 ] [ ] [ 16 ] [ 81 ]
...
infinity { 1 } [ pi^4 ] infinity { 1 }
sum of { -------- } = n * [ ------- ] - sum of { (n-a)* ------- } + {some small error function} ...
for n=1 { n^3 } [ 90 ] for a=1 { a^4 }
...
this is a "double-telescopic" sum for a "fixed n"; you would have to sum the first ~50 terms
where n= 100 (fixed) and ranging from a=1 to 50 to find convergence; no one is going to
get famous for counting fractions, and there'll be NO CLOSED FORM, since 'n' can be fixed
in infinitely many ways.
...
the sum of 1/(n^5) from n=1 to infinity can be summed using 1/(n^6) fractions similarly.
...
*QED
10/24/2012
...
look at this spreadsheet to see the concrete results:
...
sum of cubed fractions equal to double-telescopic sum of quartic-
... powered fractions for fixed, n= 1000
1 1 1.000000000000 1000 1.082323234 1082.323233711140
1 2 0.125000000000 -999 1 1 -999.000000000000
1 3 0.037037037037 -998 1 2 -62.375000000000
1 4 0.015625000000 -997 1 3 -12.308641975309
1 5 0.008000000000 -996 1 4 -3.890625000000
1 6 0.004629629630 -995 1 5 -1.592000000000
1 7 0.002915451895 -994 1 6 -0.766975308642
1 8 0.001953125000 -993 1 7 -0.413577675968
...
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
...
1 991 0.000000001027 -10 1 990 -0.000000000010
1 992 0.000000001024 -9 1 991 -0.000000000009
1 993 0.000000001021 -8 1 992 -0.000000000008
1 994 0.000000001018 -7 1 993 -0.000000000007
1 995 0.000000001015 -6 1 994 -0.000000000006
1 996 0.000000001012 -5 1 995 -0.000000000005
1 997 0.000000001009 -4 1 996 -0.000000000004
1 998 0.000000001006 -3 1 997 -0.000000000003
1 999 0.000000001003 -2 1 998 -0.000000000002
1 1000 0.000000001000 -1 1 999 -0.000000000001
1 1001 0.000000000997 sum=1.202056736493
1 1002 0.000000000994
1 1003 0.000000000991
1 1004 0.000000000988
1 1005 0.000000000985
...
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
...
1 2995 0.000000000037
1 2996 0.000000000037
1 2997 0.000000000037
1 2998 0.000000000037
1 2999 0.000000000037
1 3000 0.000000000037
1 3001 0.000000000037
sum=1.202056847660 sum=1.202056736493
...
...
sum of quintic-powered fractions equal to double-telescopic sum of sextic-
... powered fractions for fixed, n= 1000
1 1 1.000000000000 1000 1.017343062 1017.343061984450
1 2 0.031250000000 -999 1 1 -999.000000000000
1 3 0.004115226337 -998 1 2 -15.593750000000
1 4 0.000976562500 -997 1 3 -1.367626886145
1 5 0.000320000000 -996 1 4 -0.243164062500
1 6 0.000128600823 -995 1 5 -0.063680000000
1 7 0.000059499018 -994 1 6 -0.021304869684
1 8 0.000030517578 -993 1 7 -0.008440360734
1 9 0.000016935088 -992 1 8 -0.003784179688
1 10 0.000010000000 -991 1 9 -0.001864741335
1 11 0.000006209213 -990 1 10 -0.000990000000
1 12 0.000004018776 -989 1 11 -0.000558264717
1 13 0.000002693291 -988 1 12 -0.000330879201
1 14 0.000001859344 -987 1 13 -0.000204482920
1 15 0.000001316872 -986 1 14 -0.000130950964
...
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
...
1 195 0.000000000004 -806 1 194 -0.000000000015
1 196 0.000000000003 -805 1 195 -0.000000000015
1 197 0.000000000003 -804 1 196 -0.000000000014
1 198 0.000000000003 -803 1 197 -0.000000000014
1 199 0.000000000003 -802 1 198 -0.000000000013
1 200 0.000000000003 -801 1 199 -0.000000000013
1 201 0.000000000003 -800 1 200 -0.000000000013
sum=1.036927754992 sum=1.036927755606
...
*QED
11/01/2012
...
...
let's try counting the number of "normal magic squares" for a certain size, 'n':
...
n 1 {((n -d)^2)*(2^n)}
n= 1; N1 = ------- = ------- = 1; {not noticing the minus of [((1 -1)^2)*(2^1) = 0] repeats}
1 1
...
n +1 3
n= 2; N2 = --------- = ----- = 0; (there aren't any fractional amounts of magic squares!)
2 2
...
(n +1)*(n^2 +1) (-1)*(n^2 +1) {minus ((n -d)^2)*(2^n) repeats!}
n= 3; N3 = --------------------------- + ----------------------- = 10 -1 -((3-2)^2)*(2^3) = 1; w/out removing
2*2 2*5 rotations and reflections}
{so, there's only 1 answer after removing 'no' reflections.}
...
n*(n^2 +1)*(n^3 +2) n^2*(n^3 +2)
n= 4; N4 = --------------------------------- + ------------------------ = 748 + 132 = 880; there are
2*3 2*4 220 unique after removing
... rotations and reflections.
...
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':
...
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 calculations don't take into account the number of excluded reflections, and rotations, etc.
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 N6 = 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
...
...
there's a subtle reason for "why a 3x3-normal magic square with 'square' entries doesn't exist." if
we examined a 3x3-normal magic square, we would immediately see the following, more basic
relationship:
...
A B C
...
D E F
...
G H I
...
for n = 3, we notice that B +H = 2*E, D +F = 2*E, G +C = 2*E, and A +I = 2*E. so, if we let each
entry be a 'square', then by explicit substitution, we have...
...
a^2 b^2 c^2
...
d^2 e^2 f^2
...
g^2 h^2 i^2
...
thus, if we chose any of those above equations, then we'd have... x^2= 2*z^2 -y^2. however, with
Pythagorean triples from... a^2 = b^2 -c^2, solutions can ONLY exist when a, b, and c are co-prime,
and ONLY when b^2 -c^2 can be factored properly. but, this isn't the case with 2*z^2 -y^2; it can't
be factored. hence, a 3x3-normal magic square w/square entries can't exist! (I'm not sure why any-
one would think that the solution to such an easily-defined problem would necessarily come from
larger numbers than a modern-day computer could hold???)
*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 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
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
...