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

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2021 JMPSC Accuracy Problems/Problem 13

Problem

Solution