Polynomials

It can be easily proven that if x is an integer and P(x) has integral coefficients, and $P(x) - P(y) = p$ , then $x-y | p$ . Thus, since $P(a) - P(2) = 2$ we must have $a-2 | 2$ . There are four cases: $a-2 = 2 , a-2 =1 , a-2 = -1 , a-2 = -2 \implies a = 0 ,1 , 3 .4$ Therefore, the sum is $\boxed{8}$

Since $P(2)= a$ ,

We can deduce that $P(x)= (x-2)Q(x) + a$ . Substituting $x=a$ , we get $P(a)= (a-2)Q(a) + a = a+2$

$Q(a)= \frac{2}{a-2}$

Since $P(x)$ has integer coefficients, $Q(x)$ should have integer coefficients as well, implying $Q(a)$ is an integer. This gives us $a= 0, 1, 3$ or $4$ . The required answer is therefore $\boxed{8}$ .