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

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==Solution==
 
==Solution==
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asdf
 
  
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Case 1: <math>x+y=4,3</math>.
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Notice that since <math>x</math> and <math>y</math> are both integers, <math>x+y</math> and <math>xy</math> are also both integers. We can then use casework to determine all possible values of <math>x</math>:
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There are no possible answer when <math>x+y=3</math>, but when <math>x+y=4</math>, <math>x</math> can equal <math>3</math> or <math>1</math>.
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Case 2: <math>x+y=5,0</math>
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Case 1: <math>x+y=0,xy=5</math>.
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This works when <math>x=0,5</math>.
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Therefore, the answer is <math>9</math>. ~ kante314
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 + 5</math>, which are not real.
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Case 2: <math>x+y=3, xy=4</math>.
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 3 + 4</math>, which are not real.
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Case 3: <math>x+y=4,xy=3</math>.
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The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 4 + 3</math>, which are <math>1</math> and <math>3</math>.  
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Case 4: <math>x+y=5, xy = 0</math>
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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>.  
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Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>.  
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~kante314
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~Revised and Edited by Mathdreams

Revision as of 10:46, 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 $x^2 - 5$, which are $0$ and $5$.

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


~kante314

~Revised and Edited by Mathdreams

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