Powerful Squares

Obviously, one such pair is $(4, 2)$ since $5^4 - 7^2 = 25^2 - 7^2 = 24^2$ . But, there may be more. Let's let $(m, n)$ be an arbitrary pair of positive integers such that

$5^m - 7^n = k^2,$

for some nonnegative integer $k$ . First, we take $\mod 5$ of both sides of the equation to get $2^n \equiv k^2 \pmod{5}$ . Checking powers of 2, we find that $n$ has to be even, since otherwise $2^n$ is congruent to either $2$ or $3$ , but no perfect square is congruent to either of those $\mod 5$ . Similarly, if we take $\mod 7$ of the equation, we get $(-2)^m \equiv k^2 \pmod{7}$ , which implies that $m$ is also even.

Let $m = 2a$ and $n = 2b$ for some nonnegative integers $a, b$ . Now our equation becomes

$5^{2a} - 7^{2b} = k^2.$

Subtracting $5^{2a}$ from both sides and factoring, we get

$-7^{2b} = (k + 5^a)(k - 5^a).$

This equation implies that both $|k + 5^a|$ and $|k - 5^a|$ are powers of $7$ . Also, exactly one of $k + 5^a$ and $k - 5^a$ is negative. Suppose that neither $|k + 5^a|$ nor $|k - 5^a|$ are equal to $1$ . Then, the difference between $k + 5^a$ and $k - 5^a$ is also a multiple of $7$ . However, $(k + 5^a) - (k - 5^a) = 2(5^a)$ , which is never a multiple of $7$ . Thus, one of $|k + 5^a|$ and $|k - 5^a|$ is equal to $1$ . Since $k - 5^a < k + 5^a$ , we let $|k - 5^a| = 1$ , or $k = 5^a - 1$ since $k - 5^a$ must be negative.

Now, we can substitute this expression for $k$ into our equation to get

$7^{2b} = 2(5^a) - 1.$

Rearranging gives

$2(5^a) = 7^{2b} + 1.$

If we take $\mod 5$ of both sides of this equation, we get $(-1)^b \equiv -1 \pmod{5}$ , which means $b$ is odd. Let $b = 2c + 1$ . Since $b$ is odd, we can factor the expression $7^{2b} + 1$ :

$\begin{aligned} 7^{2b} + 1 &= (7^2)^b + 1 \\ &= (7^2)^{2c + 1} + 1 \\ &= (7^2 + 1)(7^{2c} - 7^{2c - 1} + \cdots - 7 + 1) \\ &= (50)(7^{2c} - 7^{2c - 1} + \cdots - 7 + 1). \end{aligned}$

From this factorization, we immediately get $a \geq 2$ . Suppose $a > 2$ , and let $d = a - 2$ . Dividing both sides of our equation by $50$ , we get

$5^d = 7^{2c} - 7^{2c - 1} + \cdots - 7 + 1.$

If we analyze powers of $7 \mod 5$ , we find

$\begin{aligned} 7^{4x} &\equiv 1 \pmod{5} \\ -7^{4x + 1} &\equiv -2 \pmod{5} \\ 7^{4x + 2} &\equiv -1 \pmod{5} \\ -7^{4x + 3} &\equiv 2 \pmod{5}, \end{aligned}$

for all nonnegative integers $x$ . Notice that $-7^{4x + 3} + 7^{4x + 2} - 7^{4x + 1} + 7^{4x} \equiv 0 \pmod{5}.$

Using these equivalences, we get either

$7^{2c} - 7^{2c - 1} + \cdots - 7 + 1 \equiv 1 + 2 - 1 - 2 + 1 \cdots -2 + 1 \equiv 1 \pmod{5}$

or $7^{2c} - 7^{2c - 1} + \cdots - 7 + 1 \equiv -1 - 2 + 1 + 2 - 1 - 2 + 1 \cdots - 2 + 1 \equiv -2 \pmod{5}.$

Thus, $7^{2c} - 7^{2c - 1} + \cdots - 7 + 1$ can never be a multiple of $5$ , let alone a power of $5$ .

This means that $c = 0$ , or $b = 2(0) + 1 = 1$ and $n = 2(1) = 2$ . Also, $d = 0$ , $a = 0 + 2 = 2$ , and $m = 2(2) = 4$ . Therefore, the only pair of positive integers $(m, n)$ that satisfies $5^m - 7^n = k^2$ is $(4, 2)$ , and the maximum value of $m + n$ is $2 + 4 = \boxed{6}$ .