The Question:
Show that there are infinitely many positive integers that are not the sum of $15$ fourth powers.
My Attempt:
I am completely stuck.
The previous parts of this question were to show that there are infinitely many positive integers that are not the sum of:
(i) $3$ squares
(ii) $8$ sixth powers
(iii) $11$ tenth powers
For example, I did (i) by showing that all squares are congruent to either $0$, $1$, or $4 \pmod 8$ and hence all numbers of the form $7+8k$ cannot be the sum of three squares.
Similarly, I found $9+27k$ for (ii) and $12+50k$ for (iii) through exhaustive trial and error.
However, I couldn't do $15$ fourth powers, even with the help of a computer efficiently computing all the modulos for me.
Is there a better way to do this?
Try $$ 31 \cdot 16^n $$
It begins with the observation that $x^4 \equiv 0, 1 \pmod {16}$ but there is more to be done
For $31 \cdot 16^n$ with $n \geq 1,$ if any of the fourth powers are odd, there are at least sixteen odd fourth powers. On the other hand, if we have all even numbers, then dividing each by $2$ gives a representation of $31 \cdot 16^{n-1},$ so induction says sixteen are required