# 2014 UMO Problems/Problem 2

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## Problem

(a) Find all positive integers $x$ and $y$ that satisfy $$x^2+y^2 = 2014,$$ or prove that there are no solutions.

(b) Find all positive integers $x$ and $y$ that satisfy $$x^2 + y^2 = 3222014,$$ or prove that there are no solutions.

## Solution

(a) We see that we can rewrite $x^2 + y^2 = 2014$ as $x^2 + y^2 \equiv 6 \bmod{8}$. Since $x^2$ and $y^2$ are perfect squares, their modulo can only be ${0,1,4}$. Since none of those two combinations make $6$, there are no solutions to $x^2 + y^2 = 2014$ such that $x,y \in \mathbb Z$.

(b) Similarly, we can rewrite $x^2 + y^2 = 3222014$ as $x^2 + y^2 \equiv 6 \bmod{8}$ and therefore it also does not have integer solutions.