Proving that if $a,b > 1$, then $5^a - 3^b=16$ has only one solution with $a=2$ and $b=2$ This may be one of those problems that is easy to state but very hard to prove.  I don't know.
I have tried to show that there is only one solution but I have not made much progress.
Here's what I have:
If there exists $a,b > 2$, then:
$$5^2(5^{a-2} - 1) = 3^2(3^{b-2}-1)$$
$a-2 = u(3-1)(3) = 6u$ where $u \ge 1$
$b-2 = v(5-1)(5) = 20v$ where $v \ge 1$
So, as I understand it, the problem comes down to showing that for all $u,v$:
$$5^2(2)(1 + 5 + 5^2 + \cdots + 5^{6u-1}) \ne 3^2(1 + 3 + 3^2 + \cdots + 3^{20v-1})$$
I can't figure out any step after this.  Is my thinking wrong?  Is there $a,b > 2$ where $5^a - 3^b = 16$
Thanks very much,
-Larry
 A: Consider everything modulo $3$. Then,
$$(-1)^a \equiv 1 \pmod 3$$
From this, $a$ must be even, i.e. $a = 2m$ for some integer $m \ge 1$.
Now, consider everything modulo $5$. Then,
$$-(-2)^b \equiv 1 \pmod 5$$
From this (why?), $b$ must even, i.e. $b = 2n$ for some $n \ge 1$. 
So we rewrite:
$$5^{2m} - 3^{2n} = 16$$
$$(5^m - 3^{n})(5^m + 3^{n}) = 2^4$$
Now, let $5^m - 3^{n} = 2^x$ and $5^m + 3^{n} = 2^{x + k}$ for some integers $x,k \ge 0$ s.t. $2x + k = 4$. Then, adding both up,
$$2\cdot5^m = 2^x + 2^{x + k} = 2^x(1 + 2^k)$$
$$5^m = 2^{x - 1}(1 + 2^k)$$
But if $x - 1 > 0$, then both sides will be even, which is false since $5^m$ is odd. So $x - 1 = 0 \implies x = 1 \implies k = 4 - 2 = 2$. Therefore, we have
$$5^m - 3^{n} = 2$$
$$5^m + 3^{n} = 8$$
The rest is trivial (Hint: sum up both, or subtract one from another).
A: Observing the equation modulo $5$ and $3$, you know $a$, $b$ to be even, say $a = 2m$, $b = 2n$. Then $5^{2m} - 3^{2n} = (5^m - 3^n)(5^m + 3^n)$. Both factors have to be positive integers if they multiply up to $16$.
