Numbers of form 31 x 16^n are sum of 16 4th Powers

Theorem
Let $m \in \Z$ be an integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$.

Then in order to express $m$ as the sum of fourth powers, you need $16$ of them.

Proof
We have:
 * $31 \times 16^n = \paren {2^{n + 1} }^4 + 15 \times \paren {2^n}^4$

so every integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$ can be expressed as the sum of 16 fourth powers.

Now we show that we cannot use less than $16$ fourth powers.

Observe that for an even number $2 k$:

For an odd number $2 k + 1$:

It is obvious that for $n = 0$, $31$ requires $16$ fourth powers to express:
 * $31 = 2^4 + 15 \times 1^4$

for some $n > 0$, $31 \times 16^n$ requires less than $16$ fourth powers to express.

Let $m$ be the smallest of those $n$.

Suppose $x$ of the summands are odd, where $x < 16$.

By the above, we must have:
 * $31 \times 16^m \equiv x \pmod {16}$

Since $31 \times 16^m$ is divisible by $16$, $x = 0$.

Hence each summand is even.

Dividing each summand by $2$ gives a representation of $31 \times 16^{m - 1}$ as a sum of less than $16$ fourth powers.

This contradicts the minimality condition on $m$.

Hence the result by Proof by Contradiction.