Show that if $p$ is a prime integer, then there do not exist nonzero $x \bmod p$ and $y \bmod p$ in $\Bbb Z/p\Bbb Z$ such that $XY = 0$ $\text{mod}$ $p$. HINT: what do you know about a prime dividing a product of two integers?
I am not seeking a full solution but rather an additional hint.
I write an answer because there is an important point here.
Firstly, in terms of integers we need to prove that if $p|xy$ then $p$ divides at least one of them. In modern mathematics, this is the general definition of a prime element in an algebraic structure which is called a ring.
However, in everyday life we are (wrongfully) taught that a prime number is a number which divides only itself and 1. In other words, $p$ is prime if the equality $p=ab$ can only happen if $a=1$ or $a=p$. In ring theory, such numbers are called irreducible. For the integer numbers, these definitions happen to coincide.
In fact, the integers constitute a U.F.D., unique factorization domain. By this I mean that every integer can be written uniquely as a product of ${irreducible}$ integers.
In every ring a prime element is irreducible, and in every U.F.D all irreducible elements are prime.
The condition you wrote is equivalent to the correct definition of a prime number, so I guess that your question is why is every irreducible integer prime?
Well, assume $p$ is irreducible and assume $xy\equiv0\pmod p$. This means there exists an integer $z$ such that $pz=xy$. Write out $x,y,z$ as a product of irreducible numbers. Since the factorization of $xy$ is unique, and since $p$ is an irreducible multiple of the left hand side it must appear in the right hand side. So $p$ occurs in at least one of the factorizations of $x,y$ and so in particular $p$ divides at least one of $x$,$y$, and hence, $p$ is prime.
Lastly, note that the fact that the integers are a U.F.D. follows from the division with remainder theorem. That is, for every positive integers $a,b$ there exist integers $q$ and $0\leq r<b$ such that $a=qb+r$. Proving that this property implies U.F.D is quite challenging. Finally, there are U.F.D.'s in which there is no division with remainder, for example, the ring of polynomial functions in two variables with real coefficients.