What does it mean for a prime to occur to an even power? The OEIS entry for A035251 (Positive numbers of the form $x^2 - 2y^2$ with integers $x$, $y$) states that

A positive number $n$ is representable in the form $x^2 - 2y^2$ iff every prime $p \equiv 3, 5 \pmod 8$ dividing $n$ occurs to an even power.

Probably because I didn't learn math in English, I am unable to understand what it means for a prime to occur to an even power. Primes can't be even (except $2$, but it does not fulfill the $p \equiv 3, 5 \pmod 8$ requirement), nor can they be an integer raised to any power. How would you explain this concept?
 A: It means that $p$ occurs in the factorisation of $n$ as $p^n$ with $n$ even. For example, $p=2$ occurs in $12=2^2\cdot 3$ with an even power, namely as $2^2$. However, $p=3$ does not occur in $12$ with an even power.
A: $$(\forall p\in\Bbb P)\quad 2\mid\max\{k\in\Bbb N_0: p^k\mid n\}$$
($\Bbb P$ being the set of primes)
A: The prime factorization of $479001600$ is $2^{10}3^5 5^2 7\cdot 11.$  The primes $2$ and $5$ occur to even powers because their exponents are even.  The primes $3$, $7,$ and $11$ occur to odd powers.  
A: Well, $2$ is even and I'm pretty sure it's prime. But then again, I'm so old that as a child I was actually taught that $1$ is prime.
Of course for a prime to be congruent to $3$ or $5$ modulo $8$, it has to be odd, which almost all odd primes are. What "even power" refers to is the parity of the exponent. So $3$ itself is a prime to an odd power ($1$), but $3 \times 3 = 3^2$ is a prime to an even power.
For example in $3^2 5^2 = 225$, we have two odd primes both of which have an exponent that is even. This number can indeed be represented as $x^2 - 2y^2$, with $x = 45$, $y = 30$ being a nontrivial solution.
Remember also that $0$ is an even number. Then, for example, we can choose to see $7$ as $3^0 5^0 7$, and we see that $5^2 - 2 \times 3^2 = 7$ indeed.
