'Surd Year

Since $2016 = (8 - \sqrt p)(x + y\sqrt p) = (8x - py) + (8y - x)\sqrt p$ , we see that $\begin{cases} 8x - py = 2016 \\ 8y - x = 0\end{cases}$ so that $x = 8y$ and $(64 - p) y = 2016.$ We must therefore find all primes $p$ less than 64 such that $64 - p$ is a divisor of $2016=2^5\cdot 3^2\cdot 7$ .

First we rule out $p = 2$ because $62 \not| 2016$ .

All other primes are odd, so we look for an odd divisor of 2016, i.e. a divisor of $3^2\cdot 7$ . The equation, then, reduces to $64 - p = 1,3,7,9,21,63\ \ \therefore\ \ p = 63, 61, 57, 55, 43, 1.$ It is easy to check that only $p = 61$ or $43$ are prime numbers. The answer is therefore $61 + 43 = \boxed{104}.$

Since $p$ prime, then $\sqrt{p}$ will never be an integer. Thus, $x+y\sqrt{p}$ must be a multiple of the conjugate of $(8-\sqrt{p})$ . So let $x+y\sqrt{p}=8a+a\sqrt{p}=a(8+\sqrt{p})$

Then the equation becomes $a(8-\sqrt{p})(8+\sqrt{p})=a(64-p)=2016$

This is a basic Diophantine equation and the solution can be easily done.