# Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"

## Problem

Let $x$ and $y$ be nonnegative integers such that $(x+y)^2+(xy)^2=25.$ Find the sum of all possible values of $x.$

## Solution

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Easily, we can see that $C=3,5$. $C$ must be $5$ because $795$ is divisible by $3$ and $793$ is not divisible by $3$. Therefore $\frac{795}{3}=265$. So, $3A+2B+C=6+12+5=23$

13. Case 1: $x+y=4,3$. There are no possible answer when $x+y=3$, but when $x+y=4$, $x$ can equal $3$ or $1$. Case 2: $x+y=5,0$ This works when $x=0,5$. Therefore, the answer is $9$. ~ kante314