, 2024

Solution:

We claim that $x^4 + a x^2 + b$ is splitty if and only if either $b$ or $a^2 - 4b$ is a perfect square. (The latter means that the polynomial splits into $(x^2 - r)(x^2 - s)$.)

Assuming the characterization, one can easily extract the answer. For $a = 16$ and $b = n$, one of $n$ and $64 - n$ has to be a perfect square. The solutions to this that are at most $64$ form $8$ pairs that sum to $64$ (if we include $0$), and then we additionally have $81$ and $100$. This means the sum is $64 \cdot 8 + 81 + 100 = 693$.

Now, we move on to prove the characterization.

## Necessity.

Take a prime $p$ such that neither $a^2 - 4b$ nor $b$ is a quadratic residue modulo $p$ (exists by Dirichlet + CRT + QR). Work in $\mathbb{F}_p$. Now, suppose that

$$x^4 + a x^2 + b = (x^2 + m x + n)(x^2 + s x + t)$$

Then, looking at the $x^3$-coefficient gives $m + s = 0$ or $s = -m$. Looking at the $x$-coefficient gives $m(n - t) = 0$.

- If $m = 0$, then $s = 0$, so $x^4 + a x^2 + b = (x^2 + n)(x^2 + t)$, which means $a^2 - 4b = (n + t)^2 - 4 n t = (n - t)^2$, a quadratic residue modulo $p$, contradiction.
- If $n = t$, then $b = n t$ is a square modulo $p$, a contradiction. (The major surprise of this problem is that this suffices, which will be shown below.)

## Sufficiency.

Clearly, the polynomial splits in $p = 2$ because in $\mathbb{F}_2[x]$, we have $x^4 + a x^2 + b = (x^2 + a x + b)^2$. Now, assume $p$ is odd.

If $a^2 - 4b$ is a perfect square, then $x^4 + a x^2 + b$ splits into $(x^2 - r)(x^2 - s)$ even in $\mathbb{Z}[x]$.

If $b$ is a perfect square, then let $b = k^2$. We then note that

- $x^4 + a x^2 + b$ splits in form $(x^2 - r)(x^2 - s)$ if $\left(\frac{a^2 - 4k^2}{p}\right) = 1$.
- $x^4 + a x^2 + b$ splits in form $(x^2 + r x + k)(x^2 - r x + k)$ if $a = 2k - r^2$, or $\left(\frac{2k - a}{p}\right) = 1$.
- $x^4 + a x^2 + b$ splits in form $(x^2 + r x - k)(x^2 - r x - k)$ if $a = -2k - r^2$, or $\left(\frac{-2k - a}{p}\right) = 1$.

Since $(2k - a)(-2k - a) = a^2 - 4k^2$, it follows that at least one of these must happen.