Are there infinitely many integers that are not the sum of at most $7$ seventh powers of integers?

Question

So far I have been looking for a possible divisor such that the sum of $7$ seventh powers will never leave a certain remainder but there seems not to be such a divisor below $400$ and the search is too computationally heavy to continue.

Answer 1

If you mean non-negative seventh powers, the answer is yes, infinitely many misses. Indeed, the volume of the region $x_i \geq 0$ and $\sum_{i=1}^7 x_i^7 \leq 1$ is a constant $C$ smaller than $1.$ the volume of $$\sum_{i=1}^7 x_i^7 \leq N$$ is roughly $CN.$ This volume is a good approximation of the (maximum) count of numbers represented as the sum of seven non-negative seventh powers.

Here $$ C = \Gamma \left( \frac{8}{7} \right)^7 \approx 0.6267634 $$

which means that, in the long run, at least 37% of numbers fail to have a representation.

This is usually written $G(7) \geq 8,$ see https://en.wikipedia.org/wiki/Waring%27s_problem#Lower_bounds_for_G(k)

Note $\Gamma(1+z) = z \, \Gamma(z)$

Answer 2

This is a case of Waring's problem: https://en.wikipedia.org/wiki/Waring%27s_problem

It can be proven (Ben Green) that it is not possible to solve this using modular arithmetic - for $7$ seventh powers a combinatorial solution is applicable.

Answer 3

The number of ways to make a number less than $N$ as the sum of seven $7$th powers is at most the number of ways to choose seven (not necessarily distinct) integers not exceeding $\sqrt[7]{N}$. But $$\binom{\lfloor\sqrt[7]{N}\rfloor+6}{7}\sim\frac N{7!}.$$ So, from $N$ numbers, with $N$ large, just $1$ in $5040$ or even less will be the sum of seven $7$th powers. This proves what you wanted. $\blacksquare$