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From Theorems 1 and 2 implies that the prime number p>2 can not be represented as a sum of two squares if it has a view of 4k+3, and can be presented, if p= 4k+1 Criterion Girard . Natural number can be represented as a sum of squares of two integers if and only if in its decomposition into simple f...

Theorem 2. Any prime number p form 4k+1 can be represented as a sum of squares of two integers. The proof consists of two lemmas: Lemma 2. For any prime p form 4k +1 there exists an integer m, that m²+1 is multiple p Lemma 3. Any prime divisors p of number m²+1, where m - integer, can be represente...

Theorem 1. Any odd number that can be represented as the sum of two squares has the form 4k+1 Proof: Between two integers whose sum of squares is odd, one number is necessarily even and the other is odd. The square of even number is divisible by 4 completely and the square of odd number of the divi...

If perfect cuboid exists, space diagonal of minimal cuboid (cuboid of minimal size) is the product of primes 4k+1. The proof of this lemma below.

Hello, guys! I opened this topic to share with you my original idea how we would continue a searching of perfect cuboid far away from 10^11 integers space. Our goal is to check all space diagonals up to 2^63 (9'223'372'036'854'775'808) If you are interested to participate, join us! We need mathemati...

Perfect Cuboid http://mathworld.wolfram.com/images/eps-gif/Brick_1000.gif A perfect cuboid is a cuboid having integer side lengths, integer face diagonals http://mathworld.wolfram.com/images/equations/PerfectCuboid/Inline3.gif (1) http://mathworld.wolfram.com/images/equations/PerfectCuboid/Inline6....