<s>[INST] How many solutions in integers to the following equation 
What is the number of positive integer solutions $(a, b)$ to $2016 + a^2 = b^2$?

We have,
$2016 = (b-a)(b+a) = 2^5 \cdot 3^2 \cdot 7$
$b - a = 2^{t_1} 3^{t_2} 7^{t_4}$ and
$a - b = 2^{t_5} 3^{t_6} 7^{t_7}$
Thus, we must have $t_1 + t_5 = 5$ and $t_2 + t_6 = 2$ and $t_4 + t_7 = 1$.
Total, $\binom{6}{1} \cdot\binom{3}{1} \cdot\binom{2}{1} = 36$. 
But the actual answer is $12$, where am I going wrong?
 [/INST] Just to close the problem out, since $t_1$ can not be either $0$ or $5$ (by parity), there are only $24$ possible choices of the $t_i$.  Inspection shows that $b$ is always positive, but that we switch the sign of $a$ when we exchange the choices $\{t_1,t_2,t_4\},\;\{t_5,t_6,t_7\}$.  Thus we can reject exactly half of the sextuples, getting the desired answer of $\fbox {12}$.
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