21. How many pairs $(x, y)$ of positive integers satisfy the equation $4^{x}=y^{2}+15$ ?
A 0
B 1
C 2
D 4
$\mathrm{E}$ an infinite number
SOLUTION
C

The equation $4^{x}=y^{2}+15$ may be rearranged as $4^{x}-y^{2}=15$. Now $4^{x}=\left(2^{2}\right)^{x}=\left(2^{x}\right)^{2}$. Hence $4^{x}-y^{2}$ may be factorized, using the standard factorization of the difference of two squares. This enables us to rewrite the equation as
$$
\left(2^{x}-y\right)\left(2^{x}+y\right)=15 .
$$

It follows that, for $(x, y)$ to be a pair of positive integers that are solutions of the original equation, $2^{x}-y$ and $2^{x}+y$ must be positive integers whose product is 15 , and with $2^{x}-y<2^{x}+y$.
The only possibilities are therefore that either
$$
2^{x}-y=1 \quad \text { and } \quad 2^{x}+y=15,
$$
or
$$
2^{x}-y=3 \quad \text { and } \quad 2^{x}+y=5 \text {. }
$$

In the first case $2^{x}=8$ and $y=7$, giving $x=3$ and $y=7$.
In the second case $2^{x}=4$ and $y=1$, giving $x=2$ and $y=1$.
Therefore there are just two pairs of positive integers that satisfy the equation $4^{x}=y^{2}+15$, namely, $(3,7)$ and $(2,1)$.