Let $n\geq 100$ an even number. Consider the quantities $n-91$, $n-93$ and $n-95$, one of these is a multiple of 3, and not exactly 3 cause $100-95>3$, then is a composite odd number.
Observing thet 91,93 and 95 are composite, you conclude that every $n\geq 100$ works. Now check directly the remaining numbers, and you have the solution.
Lemma 1. If $m$ is a positive integer and $\sqrt m$ is rational, then $\sqrt m$ is an integer.
Proof. Easy.
Lemma 2. If $m,n$ are positive integers and $\sqrt m+\sqrt n$ is rational, then both $\sqrt m$ and $\sqrt n$ are integers.
Proof. Say $\sqrt m+\sqrt n=x\in\Bbb Q$. Then
$$\sqrt m-\sqrt n=\frac{m-n}{x}$$
is rational and so is
$$\sqrt m=\frac{(\sqrt m+\sqrt n)+(\sqrt m-\sqrt n)}{2}\ ,$$
and likewise $\sqrt n\,$. By lemma 1, $\sqrt m$ and $\sqrt n$ are integers.
Now suppose that
$$\sqrt a+\sqrt b+\sqrt c=\sqrt s\ ,$$
where $a,b,c,s$ are positive integers and $s$ is squarefree. Squaring and rearranging,
$$2\bigl(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\bigl)=s-a-b-c\ .$$
Now add to this equation the identity $2\sqrt a\sqrt a=2a$ and factorise to obtain
$$2\sqrt{bc}+2\sqrt{as}=s+a-b-c\ .$$
By lemma 2, we see that $\sqrt{as}$ is an integer; since $s$ is squarefree, $a$ must be a square times $s$, say $a=p^2s$. Similarly $b=q^2s$ and $c=r^2s$, so
$$p\sqrt s+q\sqrt s+r\sqrt s=\sqrt s\ ,$$
but as $p+q+r>1$, this is impossible.
Best Answer
A solution using Gaussian integers, which perhaps clarifies a point.
By assumption $a = x^2 + y^2 = (x + i y) (x - i y)$.
Then $$ 2 a = (1 + i) (1 -i) (x + i y) (x - i y) = ((x-y) + i (x+y)) ((x-y) - i (x+y)), $$ so that $$ 2 a = (x-y)^2 + (x+y)^2 $$ and now use the fact that $x \ne y$.