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

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==Solution 2==
 
==Solution 2==
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Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4 \}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math>
+
Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4,5\}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math>
 
<math>\linebreak</math>
 
<math>\linebreak</math>
 
~Geometry285
 
~Geometry285

Revision as of 11:33, 11 July 2021

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

Notice that since $x$ and $y$ are both integers, $x+y$ and $xy$ are also both integers. We can then use casework to determine all possible values of $x$:

Case 1: $x+y=0,xy=5$.

The solutions for $x$ and $y$ are the roots of $x^2 + 5$, which are not real.

Case 2: $x+y=3, xy=4$.

The solutions for $x$ and $y$ are the roots of $x^2 - 3 + 4$, which are not real.

Case 3: $x+y=4,xy=3$.

The solutions for $x$ and $y$ are the roots of $x^2 - 4 + 3$, which are $1$ and $3$.

Case 4: $x+y=5, xy = 0$

The solutions for $x$ and $y$ are the roots of $x^2 - 5x$, which are $0$ and $5$.

Therefore, the answer is $1 + 3 + 0 + 5 = 9$.


~kante314

~Revised and Edited by Mathdreams

Solution 2

Note we are dealing with Pythagorean triples, so $xy=\{0,3,4,5\}$, and we have $x+y$ is a member of the set too. We see $x+y=4$ has $x=\{1,3 \}$ work, but $x+y=4$ has nothing work. If $x+y=0$, we have $x=\{0,5 \}$ work. The answer is $0+1+3+5=\boxed{9}$ $\linebreak$ ~Geometry285

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