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

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

Then in order express $m$ as the sum of powers of $4$, you need $16$ of them.

Proof
Observe that for an even number $2 k$:
 * $\paren {2 k}^4 = 16 k^4 \equiv 0 \pmod {16}$

For an odd number $2 k + 1$:
 * $\paren {2 k + 1}^4 = 16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1 \equiv 8 k \paren {k + 1} + 1 \pmod {16} \equiv 1 \pmod {16}$

since $k \paren {k + 1}$ is even.

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

for some $n > 0$, $31 \times 16^n$ requires less than $16$ powers of $4$ 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 $31 \times 16^n \equiv x \pmod {16}$.

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 $16$ powers of $4$.

This contradicts the minimality condition on $m$.

Hence the result by proof by contradiction.