I think the fundamentals of the theory of quadratic residues modulo a prime qualify.
It is easy to explain what residue classes modulo a prime $p$ are, and to formulate statements of this kind:
1. The product of two quadratic residues modulo $p$ is a quadratic residue.
2. The product of a quadratic residue and a quadratic nonresidue modulo $p$ is a quadratic nonresidue.
3. The product of two quadratic nonresidues modulo $p$ is a quadratic residue.
(Note that I am not counting $0$ as a quadratic residue, nor as a quadratic nonresidue.)
Now 1 and 2 are very easy to show. 3 is not. What do we do?
First, it is easy to see that every quadratic residue is the square of exactly $2$ distinct residues modulo $p$. Thus there are exactly $\frac{p-1}{2}$ quadratic residues modulo $p$. Hence, there are exactly $\left(p-1\right)-\frac{p-1}{2}=\frac{p-1}{2}$ quadratic nonresidues modulo $p$. Now let $a$ and $b$ be two quadratic nonresidues. If $ab$ is a quadratic nonresidue, then there are at least $\frac{p-1}{2}+1$ different residues $x$ modulo $p$ for which $ax$ is a quadratic nonresidue (namely, each of the $\frac{p-1}{2}$ quadratic residues qualifies as such $x$ (by statement 2), but the quadratic nonresidue $b$ also qualifies), which leads to at least $\frac{p-1}{2}+1$ different quadratic nonresidues (since distinct $x$'es lead to distinct $ax$'es), contradicting the fact that there are only $\frac{p-1}{2}$ quadratic nonresidues modulo $p$. Thus, $ab$ must be a quadratic residue, and 3 is proven.
This indirect argument is, I believe, understandable to high school students. The only two theorems we used are:
A. Every quadratic residue is the square of exactly $2$ distinct residues modulo $p$.
B. If $a$ is a nonzero residue modulo $p$, then distinct $x$'es lead to distinct $ax$'es.
Both of these theorems can be derived from the following well-known fact:
F. If a prime divides a product of two integers, then it divides one of these two integers.
Proof of A: Assume that $a^2 \equiv b^2 \equiv c^2 \mod p$ for three integers $a$, $b$, $c$ pairwise incongruent modulo $p$. Then, $a^2 \equiv b^2 \mod p$ rewrites as $p\mid \left(a+b\right)\left(a-b\right)$. Hence (by fact F), at least one of $a+b$ and $a-b$ is divisible by $p$. Since $a$ and $b$ are incongruent modulo $p$, this can only mean that $a+b$ is divisible by $p$. Similarly, $b+c$ and $c+a$ are divisible by $p$. But therefore $2a=\left(a+b\right)+\left(c+a\right)-\left(b+c\right)$ must also be divisible by $p$. Since $p$ cannot be $2$ (as there are no three integers pairwise incongruent modulo $2$), this yields that $a$ is divisible by $p$. Similarly, $b$ and $c$ are divisible by $p$, which contradicts with their being incongruent. This proves A.
The proof of B is much simpler. The fact F is also used in one possible proof of statement 2. (However we can also prove 2 using 1 by the same trick as we used to prove 3 using 2.)
We have thus used the fact F a lot of times, but other than that, we didn't apply anything nontrivial - not even the theorem that a nonzero polynomial over a field cannot have more roots than its degree (this fact is often used in university-level treatises of quadratic residues).
The hard part is to tell students what is interesting about quadratic residues. Maybe cryptography?