TITLE: A number theory question on $4^a+4a^2+4=b^2$
QUESTION [5 upvotes]: Find all $(a,b)$ natural number pairs that satisfy the equation
$$4^a+4a^2+4=b^2$$ 
The first observation is that $b$ must be even, and using this fact and then dividing both sides by $4$, we get 
$$4^a+4a^2+4=4c^2$$ 
$$4^{a-1}+a^2+1=c^2$$ 
At this step I'm stuck, but I thought of using the factorization. What would you suggest me here? 
Thanks!

REPLY [8 votes]: Note that $4^a$ is a perfect square, namely $(2^a)^2$. The next even perfect square is $(2^a+2)^2=4^a+4(2^a)+4$. 
For most $a$ this will be too big!  So poor $4^a+4a^2+4$ is usually stuck between two consecutive even squares, and therefore can't be a square.  
Thus our only candidates are the $a$ such that $4a^2+4\ge 4(2^a)+4$, that is, $a^2\ge 2^a$.  There aren't many.
Remark: You can use your $4^{a-1}+a^2+1$ idea first, and then the size argument.