If $x+y=1$ and $x-y=3$, what is the value of $2^{x^{2}-y^{2}}$ ?
(A) 4
(B) 8
(C) 2
(D) 16
(E) 32
Solution 1

Since $x+y=1$ and $x-y=3$, then $x^{2}-y^{2}=(x-y)(x+y)=3$, and so $2^{x^{2}-y^{2}}=2^{3}=8$.

## Solution 2

Since $x+y=1$ and $x-y=3$, then adding these two equations, we obtain $2 x=4$ or $x=2$. Substituting this value for $x$ back into the first equation we see that $y=1$.

Therefore, $2^{x^{2}-y^{2}}=2^{2^{2}-1^{2}}=2^{3}=8$.