skyspace

this is yet again, #problemfromthebook. I’ve taken a tiny break from the NT and am doing some combo stuff. Today pidgeon.

Theorem. There are infinitely many \(A\) so that \[f(x,y)=\left\lfloor x^{\frac{3}{2}} \right\rfloor+\left\lfloor y^{\frac{3}{2}} \right\rfloor=A\] has at least \(100\) lattice solutions \((x,y)\in \mathbb{N}^{2}\).

Proof.

Let’s say that all the \(A\)’s we’ve found thus-far are of size at most \(M-1\) for some \(M\in \mathbb{N}\).

Now, consider the values \(f(x,y)\) can take on in a \(M^2\)-side-length square starting at \((M,M)\) : clearly it can take on at most \[2(M+M^{2})^{\frac{3}{2}} \le 8 M^3\] unique values. But there are \(M^4\) lattice points in the square. So there is some value of \(f(x,y)\) which occurs at least \(\frac{M}{8}\) times in the square.