Find all integer solutions of the equation $x^2+15^a=2^b$ I found an integer solution for $$x^2+15^a=2^b$$ Which is $x=7$, when $a=1$ and $b=6$
In general: $x$ must be odd, $a$ must be odd and $b$ must be even
The problem is that I don't know wheater there are more solutions. Otherwise, how can I prove there is no other solution?
 A: I'll do the case where $x,a$ and $b$ are positive, the other one is easier but a bit intricate.
Working mod $3$ we have that $b$ must be even.
Let $b=2v$
We now have: $15^a=2^b-x^2=(2^v-x)(2^v+x)$
Clearly $x$ must be coprime to $15$, and so $(2^v-x)=3^a$ and $(2^v+x)=5^a$. or  $2^v-x=1$ and $2^v+x=15^a$.
We deal with the first case first ($(2^v-x)=3^a$ and $(2^v+x)=5^a$):
So $3^a+5^a=2(2^v)$.
So in fact finding solutions to your equation is equivalent to $3^a+5^a$ being a power of $2$. When $a$ is even it is not possible (work mod $4$). When $a$ is odd it only works with $a=1$, this is easy to observe with the lifting the exponent lemma.
Since we have $2^v-x=3$ and $2^v+x=5$ we have $x=1,v=2,a=1$ or $x=1,b=4,a=1$.
We now deal with the second case: ($2^v-x=1$ and $2^v+x=15^a$):
We get $15^a+1=2(2^v)$.
So by Catalan's theorem $a=1$, and so $v=3$. Hence we get $x=7,b=6,a=1$. And this gives all solutions with $x,a,b$ positive.
Clearly the sign of $x$ is irrelevant, and simple arguments show there are no solutions with $a$ or $b$ negative.
However, there remains one other possibility, $x=0$ (neither positive not negative).  For this case $15^a=2^b $, and as $15$ is not any rational power of $2$ there can be no nonzero solutions for $a $ and $b $.  Only $a=b=0$ works for the case $x=0$.
The complete solution set is then
$(x,a,b) \in \{(0,0,0),(1,1,4),(7,1,6)\} $
A: Another solution is $x=1,a=1,b=4$  
$b$ must be even for the right to be $1 \bmod 3$ as $x^2 \equiv 1 \pmod 3$.  Write $c=b/2$ and we are looking for $$(2^c-x)(2^c+x)=15^a$$  The two factors on the left are coprime, so we have the possibilities $$2^c-x=3^a\\2^c+x=5^a\\\text {and} \\2^c+x=15^a\\2^c-x=1$$
The solution $x=1,a=1,b=4,c=2$ is the first of these and $x=7,a=1,b=6,c=3$ is the second.  We cannot get any more from the second because we would have to have $2^{c+1}=15^a+1$ and the only perfect powers that differ by $1$ are $8,9$.  We cannot get any more from the first because we would have to have $2^{c+1}=5^a+3^a$ which can never be true $\bmod 16$ for $a \gt 1$
