Is there for all $N$ an integer that is the sum of, say, twenty $N$-th powers in two different ways?

Question

I believe the following to be true, but I have been unable to find any reference concerning the question.

Statement : for every $k\geq 1$, there exists $N = N(k)$ such that there is no non-trivial solution in non-negative integers to the equation $$x_1^N + \cdots +x_k^N = y_1^N + \cdots + y_k^N.$$

That is, there is at most one (up to order) representation of any natural number as a sum of $k$ $N$-th powers, for some sufficiently large $N$ with respect to $k$. (I also believe that it should be true for all sufficiently large $N$).

I know (from the Wikipedia page on generalized taxicab numbers) that, for $k=2$, it is unknown whether $N=5$ works, but I don't know if it is unknown whether any $N$ works.

Is the statement true, and if yes, how to prove it/where can I find a proof ?

Answer 1

Out of curiosity, where does this question come from? I remember hearing about it very recently...

Anyways, for sufficiently large $M$, we can always represent $M$ as a sum of $k$ $N$th powers.

Assume all the $x_i$ lie between $1$ and $m$. Then there are $\binom{m}{k}$ different sums of $\sum x_i^N$. But there at most $km^N$ sums, meaning that $$\binom{m}{k}< km^N$$ which doesn't hold for large $m$ - so there are at least two sums which are the same.

Answer 2

"OP" inquired about equal sums of around $20$ (nth) power's

in two ways. Well for degree two, Albert Beiler in his book

"Recreations in the theory of numbers" page $159$ has given

a solution for $28$ squares in two ways. The solution is

mentioned below.

[2,18,22,37,38,39,41,43,49,67,72,80,85,103,116,154,164,

175,178,192,200,207,215,222,230,247,422,593] equal to

[13,16,17,23,30,43,47,84,92,93,119,120,142,163,165,167,

177,183,188,199,215,219,261,270,280,363,372,382]

The sum of above is $1030225$ which is a square of $(1015)^2$

From above solution, many more similar solution's can also be

arrived at.