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?

  • $\begingroup$ Look at Waring's problem in Wikipedia. $\endgroup$ – Somos Apr 28 '18 at 22:26
  • $\begingroup$ I already did, but I have yet to find anything that I could use in there? $\endgroup$ – glowstonetrees Apr 28 '18 at 22:49

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

  • $\begingroup$ But why the $31$? $\endgroup$ – glowstonetrees Apr 28 '18 at 23:41
  • $\begingroup$ I tried $31 \cdot 16=496$ and it doesn't work? $\endgroup$ – glowstonetrees Apr 28 '18 at 23:42
  • $\begingroup$ Thank you for your answer. However, I am not sure I exactly follow your argument. Would you mind to elaborate a bit please? $\endgroup$ – glowstonetrees Apr 29 '18 at 13:13

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