Problem 1626

Solution:
It is obvious that $a$ and $b$ must be non-negative.
Suppose that $3^{a} + 7^{b} = n^{2}$. We can assume that $n$ is positive. We first work modulo $4$. Since $3^{a} + 7^{b} = n^{2}$, it follows that
$$n^{2} \equiv (-1)^{a} + (-1)^{b} \pmod{4}$$
Since no square can be congruent to $2$ modulo $4$, it follows that we have either (i) $a$ is odd and $b$ is even or (ii) $a$ is even and $b$ is odd.

Case (i): Let $b = 2c$. Then
$$3^{a} = (n - 7^{c})(n + 7^{c}) .$$
It cannot be the case that $3$ divides both $n - 7^{c}$ and $n + 7^{c}$. But each of these is a power of $3$. It follows that $n - 7^{c} = 1$, and therefore
$$3^{a} = 2 \cdot 7^{c} + 1 .$$
If $c = 0$, then $a = 1$, and we obtain the solution $a = 1, b = 0$. So suppose that $c \geq 1$. Then $3^{a} \equiv 1 \pmod{7}$. This is impossible, since the smallest positive value of $a$ such that $3^{a} \equiv 1 \pmod{7}$ is given by $a = 6$, and therefore all $a$ such that $3^{a} \equiv 1 \pmod{7}$ are even, contradicting the fact that $a$ is odd.

Case (ii): Let $a = 2c$. Then
$$7^{b} = (n - 3^{c})(n + 3^{c}) .$$
Thus each of $n - 3^{c}$ and $n + 3^{c}$ is a power of $7$. Since $7$ cannot divide both of these, it follows that $n - 3^{c} = 1$, and therefore
$$7^{b} = 2 \cdot 3^{c} + 1$$
Look first at the case $c = 1$. Then $b = 1$, and we obtain the solution $a = 2, b = 1$. So from now on we may assume that $c > 1$. Then $7^{b} \equiv 1 \pmod{9}$. The smallest positive integer $b$ such that $7^{b} \equiv 1 \pmod{9}$ is given by $b = 3$. It follows that $b$ must be a multiple of $3$. Let $b = 3d$. Note that $d$ is odd, so in particular $d \geq 1$.

Let $y = 7^{d}$. Then $y^{3} - 1 = 2 \cdot 3^{c}$, and therefore
$$2 \cdot 3^{c} = (y - 1)(y^{2} + y + 1) .$$
It follows that $y - 1 = 2 \cdot 3^{u}$ for some positive $u$, and that $y^{2} + y + 1 = 3^{v}$ for some $v \geq 2$. But since
$$3y = (y^{2} + y + 1) - (y - 1)^{2},$$
it follows that $3 \mid y$, which is impossible since $3 \mid (y - 1)$.