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

Revision as of 12:32, 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 $xy=\{0,3,4 \}$ , 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

@@ Line 20: / Line 20: @@
 Case 4: <math>x+y=5, xy = 0</math>
-The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 5</math>, which are <math>0</math> and <math>5</math>.
+The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 5x</math>, which are <math>0</math> and <math>5</math>.
 Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>.

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

Revision as of 12:32, 11 July 2021

Problem

Solution

Solution 2