Exponential Diophantine perfect square problem I need to find all positive integer solutions $(a,b,c)$ to $1+4^a+4^b=c^2$. I am certain that the only solutions are $(t,2t-1, 2^{2t-1}+1) $ and $ (2t-1, t, 2^{2t-1}+1) $ ,  $t\in \mathbb{N}$ but I am having some trouble confirming this. Could someone help me please? Thank you.
 A: By symmetry it is enough to deal with the case $1 \le a \le b$. 
Rewrite the equation as
$$4^a(1+4^{b-a})=c^2-1=(c-1)(c+1).$$
Suppose first  that $b=a$. Then 
$4^a(1+4^{a-b})=2^{2a+1}$.  But $(c-1)(c+1)$ is a power of $2$ only if $c=3$. Thus $a=b=1$.  From now on we can assume that $b>a$.
Since $a \ge 1$, $c$ must be odd, say $c=2d+1$.
Substitute and divide by $4$.
We get
$$4^{a-1}(1+4^{b-a})=d(d+1).$$
We cannot have $a=1$, since if $a=1$ the left side is odd and the right side is even.  Thus $a>1$.
Since $b>a$, $1+4^{b-a}$ is odd.  But one of $d$ and $d+1$ is even. Whichever one this is, it must be divisible by $4^{a-1}$.  We first examine the possibility that $d$ or $d+1$ is exactly $4^{a-1}$.
Then either (i) $4^{a-1}=d$ and $4^{b-a}+1=d+1$, or (ii) $4^{a-1}=d+1$ and $4^{b-a}+1=d$.  
Possibility (ii) is easy to dismiss: two non-trivial powers of $4$ cannot differ by $2$.  
If possibility (i) holds, we have $b-a=a-1$, so $b=2a-1$.  It is easy to verify that this always does give a solution of the original system, namely your solutions with $t>1$. 
Finishing the argument:
We found that one of $d$ or $d+1$ is divisible by $4^{a-1}$, and  dealt fully with the case of equality. So now we ask: could we have $d$ or $d+1$ equal to $4^{a-1}k$, where $k>1$?  
If that is the case, then since $k \ge 3$ we have $4^{b-a}+1\ge 9\cdot 4^{a-1}$. It follows that $2b-2a>2a-2+3$, and hence $b\ge  2a+1$.
But this is impossible.  For note that since $1+4^a+4^b$ is a perfect square, we must have 
$$4^b+4^a+1 \ge (2^b+1)^2 =4^b+2^{b+1}+1$$ 
However, if $b \ge 2a+1$ then $4^{a}<2^{b+1}$, contradicting the above inequality.  
A: According to Richard Guy (Unsolved Problems In Number Theory, Problem D10), Mignotte observed that the equation $n^2=2^a+2^b+1$ is completely solved by combining the results of Le Mao-Hua on $x^2=4q^n+4q^m+1$ with those of Tzanakis and Wolfskill. Le Mao-Hua wrote dozens of papers on Diophantine equations. I think the one Guy means is The Diophantine equation $x^2=4q^n+4q^m+1$, Proc Amer Math Soc 107 (1989) 599-604, MR 90b:11024. The T and W paper is The Diophantine equation $x^2=4q^{a/2}+4q+1$, J Number Theory 26 (1987) 96-116, MR 88g:11009. 
EDIT: I think there is an elementary argument that works. I'll sketch it, in part. 
As in Andre's (currently deleted) answer, we may assume $a\lt b$. I'll do the case where $b$ is odd, and leave you the case where $b$ is even. So, $b=2t-1$. 
Show there aren't any squares between $4^b=(2^{2t-1})^2$ and $4^b+4^t+1$. Deduce that $a\ge t$. 
From $4^b+4^a=(c+1)(c-1)$, deduce that $4^{a-(1/2)}$ divides $c+1$ or $c-1$, so $c\ge4^{a-(1/2)}-1$. Square both sides to get $4^b+4^a+1\ge4^{2a-1}-4^a+1$. Play around with the inequalities and deduce $a\le t$. 
If you can fill in all the blanks, you're done. 
