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Difference between revisions of "2024 AMC 10A Problems/Problem 11"
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Note that <math>m</math> is a nonnegative integer.
 
Note that <math>m</math> is a nonnegative integer.
  
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We square both sides, rearrange, and apply the difference of squares formula: <cmath>(n+m)(n-m)=49.</cmath>
+
We square, rearrange, and apply the difference of squares formula to the given equation: <cmath>(n+m)(n-m)=49.</cmath>
 +
It is clear that <math>n+m\geq n-m,</math> so <math>(n+m,n-m)=(49,1),(7,7),(-7,-7),(-1,-49).</math> Each ordered pair <math>(n+m,n-m)</math> gives one ordered pair <math>(m,n),</math> so there are <math>\boxed{\textbf{(D)}~4}</math> such ordered pairs <math>(m,n).</math>

Revision as of 16:19, 8 November 2024

Problem

How many ordered pairs of integers $(m, n)$ satisfy $\sqrt{n^2 - 49} = m$?

$\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}$ Infinitely many

Solution

Note that $m$ is a nonnegative integer.

We square, rearrange, and apply the difference of squares formula to the given equation: \[(n+m)(n-m)=49.\] It is clear that $n+m\geq n-m,$ so $(n+m,n-m)=(49,1),(7,7),(-7,-7),(-1,-49).$ Each ordered pair $(n+m,n-m)$ gives one ordered pair $(m,n),$ so there are $\boxed{\textbf{(D)}~4}$ such ordered pairs $(m,n).$

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