2007 AMC 10A Problems/Problem 23

Problem

How many ordered pairs $(m,n)$ of positive integers, with $m \ge n$ , have the property that their squares differ by $96$ ?

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$

Solution 1

$m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3$

For every two factors $xy = 96$ , we have $m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}$ . Since $m \ge n > 0$ , $x+y \ge x-y > 0$ , from which it follows that the number of ordered pairs $(m,n)$ is given by the number of ordered pairs $(x,y): xy=96, x > y > 0$ . There are $(5+1)(1+1) = 12$ factors of $96$ , which give us six pairs $(x,y)$ . However, since $m,n$ are positive integers, we also need that $\frac{x+y}{2}, \frac{x-y}{2}$ are positive integers, so $x$ and $y$ must have the same parity. Thus we exclude the factors $(x,y) = (1,96)(3,32)$ , and we are left with four pairs $\mathrm{(B)}$ .

Solution 2

Similar to the solution above, reduce $96$ to $2^5*3^1$ . To find the number of distinct prime factors, add $1$ to both exponents and multiply, which gives us $6*2=12$ factors. Divide by $2$ since $m$ must be greater than or equal to $n$ . We don't need to worry about $m$ and $n$ being equal because $96$ is not a square number. Finally, subtract the two cases above for the same reason to get $\mathrm{(B)}$ .

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2007 AMC 10A Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

See also