Re: [math-fun] Sums of four squares
Hi, Gene. Thanks for the answer. (I hope to see the proof sometime. I wonder if there's some use for quaternions here...) << The number, r4(n), of integer solutions (w,x,y,z) of w^2 + x^2 + y^2 + z^2 = n, for positive integer n, is 8 times the sum of the divisors of n which are not divisible by 4. Expressed another way, r4(n)/8 is multiplicative, and r4(p^n)/8 = 1 + p + p^2 + ... + p^n for p an odd prime, r4(2^n)/8 = 1 + 2 for n>0.
This formula, which I'm seeing for the first time, is amazingly interesting! --Dan
=asimovd@aol.com This formula, which I'm seeing for the first time, is amazingly interesting!
Note that the sum of divisors of n is >= n+1, with equality just when n is prime, leading to remarkably inefficient and inscrutable primality tests based on counting partitions into four squares.
Robert G. Wilson V somehow determined that the prime number in the subject divides Googolplex+1. I verified this ... but I don't know what cleverness he used to find it. --Ed Pegg Jr, www.mathpuzzle.com
Ed Pegg, math funers:
Robert G. Wilson V somehow determined that the prime number in the subject divides Googolplex+1.
I verified this ... but I don't know what cleverness he used to find it.
I suspect he used legendary cleverness... or should I say "Legendre." Numbers of the form a^n+b^n have two kinds of factors. - The algebraic factors are a^m+b^m, where M is N divided by an odd number. - The primitive factors are those primes which divide a^n+b^n, but not any algebraic factor. Legendre showed that they are all of the form k*n+1, for some K. In our case, a=10, b=1, and n=google: For each odd factor of google (each small power of 5), the cofactor C is 2^100*5^d for some D. 10^C+1 is an algebraic factor of 10^google+1, and its primitive prime factors are k*(2^100*5^d)+1. Each such primitive factor also factors 10^google+1. One can check values of D from 0 upwards, and values of K from 1 upwards. Perhaps that's what Dr. Wilson did. It happens that d=6, k=16 yields Robert's prime factor. -- Don Reble djr@nk.ca
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Don Reble -
ed pegg -
Marc LeBrun