Do you know when it is a perfect square?

Notice that this forces $a,b\geq 0$ . By looking through $\mod 4$ , we can find that exactly one of a, b is an even integer. Assume $3^a+7^b=n^2$ . We can divide this into 2 cases:

1. $2|a, a=2x$

$7^b=n^2-9^x=(n-3^x)(n+3^x)$

Then, n is an average of 2 powers of 7, which makes $n=3^x+1$ , since not both $n-3^x$ and $n+3^x$ is divisible by 7, and $n+3^x>n-3^x$ . However, $\log_7 (2\times 3^x+1) \in N$ leads to $x=1$ , and we have $(2,1)$ as the solution.

1. $2|b, b=2y$

$3^a=n^2-49^y=(n-7^y)(n+7^y)$

with the same reasoning as the first case, $n=7^y+1$ . Also, this leads to $y=0$ , hence we have $(1,0)$ as the solution.

In total, there are $\boxed{2}$ solutions.

3^1 + 7^0 = 2^2

3^2 + 7^1 = 4^2

a^n + b^n = c^n of two terms on L.H.S. cannot have answer of n > 2 {Fermat's Last Theorem}.

Since a^2 + b^2 = c^2 does not consist of {3, 7, c}, (a, b) of (1, 0) and (2, 1) are only possibilities.

Answer: 2