2007 Indonesia MO Problems/Problem 8

Problem

Let $m$ and $n$ be two positive integers. If there are infinitely many integers $k$ such that $k^2+2kn+m^2$ is a perfect square, prove that $m=n$ .

Solution 1 (credit to crazyfehmy)

Note that we can complete the square to get $k^2 + 2kn + n^2 - n^2 + m^2$ , which equals $(k+n)^2 + m^2 - n^2$ .

Assume that $m > n$ . Since $m, n$ are positive, we know that $m^2 - n^2 > 0$ . In order to prove that $(k+n)^2 + m^2 - n^2$ is not a perfect square, we can show that there are values of $k$ where $(k+n)^2 < (k+n)^2 + m^2 - n^2 < (k+n+1)^2$ .

Since $m^2 - n^2 > 0$ , we know that $(k+n)^2 < (k+n)^2 + m^2 - n^2$ . In the case where $(k+n)^2 + m^2 - n^2 < (k+n+1)^2$ , we can expand and simplify to get $\begin{align*} k^2 + 2kn + n^2 + m^2 - n^2 &< k^2 + 2kn + n^2 + 2k + 2n + 1 \\ m^2 - n^2 &< 2k + 2n + 1 \\ \frac{m^2 - n^2 - 2n - 1}{2} &< k. \end{align*}$ All steps are reversible, so there are values of $k$ where $(k+n)^2 < (k+n)^2 + m^2 - n^2 < (k+n+1)^2$ , so there are no values of $m, n$ where $m > n$ that results in infinite number of integers $k$ that satisfy the original conditions.

Now assume that $m < n$ . Since $m, n$ are positive, we know that $m^2 - n^2 < 0$ . In order to prove that $(k+n)^2 + m^2 - n^2$ is not a perfect square, we can show that there are values of $k$ where $(k+n-1)^2 < (k+n)^2 + m^2 - n^2 < (k+n)^2$ .

Since $m^2 - n^2 < 0$ , we know that $(k+n)^2 + m^2 - n^2 < (k+n)^2$ . In the case where $(k+n-1)^2 < (k+n)^2 + m^2 - n^2$ , we can expand and simplify to get $\begin{align*} k^2 + 2kn + n^2 - 2k - 2n + 1 &< k^2 + 2kn + n^2 + m^2 - n^2 \\ -2k - 2n + 1 &< m^2 - n^2 \\ k &> -\frac{m^2 - n^2 + 2n - 1}{2}. \end{align*}$ All steps are reversible, so there are values of $k$ where $(k+n-1)^2 < (k+n)^2 + m^2 - n^2$ , so there are no values of $m, n$ where $m < n$ that results in infinite number of integers $k$ that satisfy the original conditions.

Now we need to prove that if $m = n$ , there are an infinite number of integers $k$ that satisfy the original conditions. By the Substitution Property, we find that $k^2 + 2kn + m^2 = k^2 + 2kn + n^2$ . The expression can be factored into $(k+n)^2$ . Since the expression is a perfect square, for all integer values of $n, k$ , there are an infinite number of integers $k$ that satisfies the original conditions when $m = n$ .

Solution 2 (credit to dskull16)

We begin by completing the square to get $k^2 + 2kn + n^2 - n^2 + m^2$ , which equals $(k+n)^2 + m^2 - n^2$ .

Then we have that $(k+n)^2 + m^2 - n^2 = a^2$ for some natural number a. This then gives us $m^2 - n^2 = a^2 - (k+n)^2$ which we can write like $(m+n)(m-n) = (a+k+n)(a-k-n)$ by difference of two squares.

Now we remark that the left hand side is a constant since we prematurely chose $m$ and $n$ . Acknowledging the fact that this equation is comprised entirely of integers, we see that $(a+k-n)$ and $(a-k-n)$ need both be factors of the left hand side of which there are finitely many. This means that there are finitely many solutions for $a$ .

If however the left hand side were 0, implying that either $m = n$ or $m = -n$ , we would be able to find infinitely many integers such that $a+k = n$ . Since $m$ and $n$ are positive integers, this means that $m=n$ as required.

2007 Indonesia MO (Problems)
Preceded by Problem 7	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Last Problem
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2007 Indonesia MO Problems/Problem 8

Contents

Problem

Solution 1 (credit to crazyfehmy)

Solution 2 (credit to dskull16)

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