Let $N\in\mathbb{N}$ and $a,b\in\mathbb{N}$ be such that $a+b\in(N/2,2N)$ (then of course $\max\{a,b\}\simeq N$). I'm interested in getting an upper bound (in terms of $N$) for the number of positive integer solutions $(m,n)\in\mathbb{N}^2$ to the system of inequalities: \begin{align*} \begin{cases} |m^2+n^2-a^2-b^2|<N^\alpha,\\ |m^3+n^3-a^3-b^3|<N^\alpha, \end{cases} \end{align*} where $\alpha\in(1,2)$ is a fixed constant.
The best I can do is to observe that this system has at most as many solutions as the single inequality \begin{align*} |m^3+n^3-a^3-b^3|<N^\alpha \end{align*} and then use a fact that for a fixed $k\in (-N^\alpha, N^\alpha)\cap\mathbb{Z}$ the equation \begin{align*} m^3+n^3=a^3+b^3+k \end{align*} has at most $(a^3+b^3+k)^\epsilon\simeq N^{3\epsilon}$ solutions, where $\epsilon>0$ is an arbitrary small constant. Since there are $N^\alpha$ choices of $k$ it gives the upper bound $N^{\alpha+\epsilon}$, for arbitrarily small $\epsilon>0$. Can one do better?
