Find all non-negative integers (a,b) such that $k^{2}$a+b is perfect square for all integers k

Question

The obvious solutions are ($a^{2}$,0) and (0, $b^{2}$) but I'm not sure if there are any other solutions.

Answer 1

You're asking to find all non-negative integers $(a,b)$ such that

$$f(k) = k^{2}a + b \tag{1}\label{eq1A}$$

is a perfect square for all integers $k$. You've already determined the solutions

$$(a,b) \in \{(c^2, 0), (0, d^2)\}, \;\; c, d \in \mathbb{Z} \tag{2}\label{eq2A}$$

for where $a$ and/or $b$ are $0$.

Consider that $a$ and $b$ are both positive integers. As lulu's comment indicates, using $k = 0$ in \eqref{eq1A} gives $f(0) = b$. This means $b$ must be a perfect square, i.e.,

$$b = d^2 \tag{3}\label{eq3A}$$

for some positive integer $d$. Using $k = ad$ in \eqref{eq1A}, we then have

$$f(ad) = (ad)^{2}a + d^2 = d^2(a^3 + 1) = e^2 \tag{4}\label{eq4A}$$

for some positive integer $e$. Since $\frac{e}{d} = \sqrt{a^3 + 1}$, and the only integers with rational roots are perfect squares, this means for some positive integer $f$ that

$$a^3 + 1 = f^2 \;\; \to \;\; f^2 - a^3 = 1 \tag{5}\label{eq5A}$$

However, by Catalan's conjecture (proven by Preda Mihăilescu in $2002$), and as shown in Find values of x and y when $x^2-y^3=1$, the only positive integer solution is $f = 3$ and $a = 2$. Using \eqref{eq3A} and $k = 1$ in \eqref{eq1A}, we then have

$$2 + d^2 = f^2 \;\;\to\;\; f^2 - d^2 = 2 \tag{6}\label{eq6A}$$

for some integer $f$. However, there are no perfect squares with a difference of $2$.

Thus, there are no solutions where $a$ and $b$ are both positive integers, so \eqref{eq2A} gives the only solutions.