at least $a\sqrt{n}$ solutions 
Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed. 

Consider a number  $0<a<2\pi$. How can we show that there is an infinite amount of integers $n$ such that 
$x^2 + y^2 + z^2=n$ 
has at least $a\sqrt{n}$ integer solutions?
I apologize for not editing in the boldened bit earlier. 

Assuming that $a$ and $b$ are fixed and that $x^2+y^2+z^2 = n$ has $\le a \sqrt{n}+b$ integer solutions for all $n$, we get that the number of integer solutions of $x^2+y^2+z^2 < n$ is bounded above by
$$\sum_{k=0}^{n-1} (a \sqrt{k}+b) = bn+ an^{3/2}\sum_{k=0}^{n-1} \sqrt{\frac{k}n} \frac1n \le bn+ a n^{3/2} \int_0^1 \sqrt{x} \, dx = bn+\frac{2an^{3/2}}{3},$$
since the second sum is a lower Riemann sum for the integral of $\sqrt{x}$ over $[0,1]$.
Now since the volume of the ball with radius $\sqrt{n}$ is $\frac{4\pi n^{3/2}}3$, and the number of integer solutions of $x^2+y^2+z^2 <n$ has the same asymptotic growth, we get that $\frac{2a}3 \ge \frac{4\pi}3$, i.e., $a \ge 2\pi$.
Lastly, if there are only finitely many $n$ such that $x^2+y^2+z^2 = n$ has $\ge a\sqrt{n}$ integer solutions, it is straightforward that there exists $b$ such that $x^2+y^2+z^2 = n$ has $\le a\sqrt{n}+b$ integer solutions for all $n$.
