hello I am looking for an efficient way, hopefully a formula or a somewhat tight upper bound, for the number of integer solutions to the following
let $k$ be a fixed integer and $\lambda \ge 1$ and $n>m>0$. I wish to know how many choices there are for triples, $(\lambda,n,m)$ that satisfy:
$k \ge \lambda(n^2+m^2)$
Thanks in advance
If $k$ is large and $\lambda=1$ it is approximately $\frac {\pi k}8$ If $\lambda=2$ it is $\frac {\pi k}{8 \cdot 2}$ If we cut this off at $\lambda=k$ we get a total of $\frac {\pi k}8 H_k$, where $H_k$ is the $k^{\text{th}}$ Harmonic number. This is approximately $\frac {\pi k}8 (\log k + \gamma)$, where $\gamma$ is Euler's constant, about $0.577$