The 2014 Problem is such a Failure. Let's Try 2015

Because next year is 2015 2015 , find the number of ordered triples ( x , y , z ) (x, y, z) where x x , y y , and z z are positive integers such that x y z x \leq y \leq z that will satisfy the equation:

x 2 + y 2 + z 2 = 2015 x ^2 + y^2 + z^2 = 2015


The answer is 0.

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1 solution

Rindell Mabunga
Aug 28, 2014

The mystery theorem is: " Legendre's Three-Square Theorem "

This theorem states that a natural number n n can be only expressed as a sum of three squares of integers x x , y y , and z z if and only if n 4 a ( 8 b + 7 ) n \neq 4^a (8b + 7) where a a and b b are non negative integers.

Since 2015 = 2008 + 7 = 8 ( 251 ) + 7 2015 = 2008 + 7 = 8(251) + 7 , therefore 2015 2015 is expressible as 4 a ( 8 b + 7 ) 4^a (8b + 7) where a = 0 a = 0 and b = 251 b = 251 .

\therefore the answer is 0 \boxed{0}

I solved this by reading your other comment :D

John M. - 6 years, 8 months ago

Nyc solution

Apoorv Singhal - 6 years ago

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