2014 UMO Problems/Problem 2

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.

2014 UMO (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2014 UMO Problems/Problem 2

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