Prime+prime+prime......

$-2q^{2}=1 - p^{2}$

$-2q^{2}= (1 - p)(1 + p)$

The prime factorization of $-2q^{2}$ is $(-1)(2)(q)(q)$

Since $1 + p$ must be positive we solve the following cases:

$1 + p = 2 \rightarrow 1 - p = -q^{2}$ Solving we get $p=1$ which is not a prime

$1 + p = 2q \rightarrow 1 - p = -q$ Solving we get $\boxed{q=2 \rightarrow p=3}$

$1 + p = 2q^{2} \rightarrow 1 - p = -1$ Doesn't have postitive solutions.

So the sum is $3 + 2 = \boxed5$

By parity, for any prime p > 2, p^2 - 1 is always even (this is modulo 2). But p^2 is always congruent to 1 modulo 4, this implies that p^2 -1 is divisible by 4. And 2q^2 must be divisible by 4 and so q must be even, but no other primes q satisfy this, except 2. Hence, the only primes p & q, that satisfies this is p = 3 and q = 2. Hence, the answer.

If $p\neq 3$ and $q\neq 3$ then $p^2\equiv q^2\equiv 1\pmod{3}$ for all primes. But then $p^2-2q^2\equiv -1≢ 1\pmod{3}$ . So $p=3$ or $q=3$ . Substituting successively gives a solution $(p,q)=(3,2)$ . The sum is $\fbox5$ .

Express the equation to $(p-q)(p+q)=1+q^2$ , then the parity of each tells us that one must be even. In another word, either p or q must be $2$ since $2$ is an only even prime.

here, for any prime p > 2, p^2 - 1 is always even (this is modulo 2). But p^2 is always congruent to 1 modulo 4, this implies that p^2 -1 is divisible by 4. And 2q^2 must be divisible by 4 and so q must be even, but no other primes q satisfy this, except 2. Hence, the only primes p & q, that satisfies this is p = 3 and q = 2.

-2q^2 = 1 - p^2

2q^2 = p^2 - 1

2q^2 = (p+1)(p-1)

RHS:

(p+1) - (p-1) = 2, difference of the two numbers must be two. The next clue is the product of the two numbers must be even by looking at LHS. This implies that the two numbers are 2 and 4, because 2*4 = 8 which is even and 4-2 = 2.

Hence, p+1 = 4 => p = 3 / p-1 = 2 => p = 3

q^2 = 4 => q = 2

Hence, (p,q) is satisfied by (3,2). Sum of the numbers = 5, which solves the problem.

for any prime p > 2, p^2 - 1 is always even (this is modulo 2). But p^2 is always congruent to 1 modulo 4, this implies that p^2 -1 is divisible by 4. And 2q^2 must be divisible by 4 and so q must be even, but no other primes q satisfy this, except 2. Hence, the only primes p & q, that satisfies this is p = 3 and q = 2.

close square number i think only 2 and 3 :p