3. Given that $100^{2}+1^{2}=65^{2}+76^{2}=p q$ for some primes $p$ and $q$. Find $p+q$.
Answer: 210
Solution: Recall the identity $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=(a d+b c)^{2}+(a c-b d)^{2}=(a c+b d)^{2}+(a d-b c)^{2}$. This, and the form of the equation, plus matching parities (note that $a d+b c$ and $a d-b c$ must share the same parity) suggests, as one possible setup:
$$
\begin{array}{l}
a d+b c=100 \\
a c-b d=1 \\
a d-b c=76 \\
a c+b d=65
\end{array}
$$

Hence, we have $b d=32, a c=33, a d=88, b c=12$. Wanting integers, we note that both $b c$ and $b d$ are divisible by 4 . Letting $b=4$ instantly works: we have $c=3, a=11$, and $d=8$. Plugging back into the leftmost side of the original identity, we have $\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=137 \cdot 73$. Both 137 and 73 , from the given, must be prime, and thus $p$ and $q$ are those two integers in some order.