Solve $x^5 + y^5 + z^5 = 2015$

Question

If $x, y, z$ are integer numbers, solve:

$$x^5 + y^5 + z^5 = 2015$$

A friend of mine claims there is no known solution, and, at the same time, there is no proof that there is no solution, but I do not believe him. However, I wasn't able to make much progress disproving his claim.

I tried modular arithmetics, but couldn't reach useful conclusion.

Answer 1

Since the question doesn't mention signs, then equivalently,

$$x^n+y^n = z^n+\beta$$

This is what Noam Elkies calls a Fermat near-miss and he has a table for $n\leq20$. It is interesting to ask: Let $xyz\neq0$. For a given $n$, how small can $\beta$ get?

For $n=3$:

It is well-known that $\beta =1$. (And it has an infinite number of integer solutions.)

For $n=5$:

It seems it is $\beta = 12$,

$$13^5+16^5=17^5+12$$

D. Stork's search showed there is no $\beta =2015$ with $|x|,|y|,|z|\leq200$. We can extend that with Elkies' tables (which go as high as $8\; million$). Excluding $|x|,|y|,|z|<17$, the next smallest $|\beta|$ with $\gcd(x,y,z)=1$ are,

$$\begin{aligned} &42^5 + 71^5 = 72^5 + 2951\\ &104^5 + 133^5 = 140^5-75083\\ &133^5 + 228^5 = 231^5-87890\\ &\quad\vdots\\ &707902^5 + 5645541^5 = 5645576^5+39515947850357 \end{aligned}$$

So $\beta=2015$ has missed the train, and the chances it exists is very remote.

Answer 2

These problems are usually done allowing the variables to have mixed signs, some positive, some negative or zero.

i think I will make this an answer. The similar problem for sums of three cubes has been worked on by many people; as of the linked article, the smallest number for which there are no congruence obstructions but no known expression is $$ x^3 + y^3 + z^3 = 33. $$

See THIS for the size of numbers involved. Indeed, on the seventh page, they give a list of numbers up to 1000 still in doubt, starts out 33, 42, 74, 156...

I see nothing wrong with suggesting that your problem could be in the same unsettled state, plus I do not think as many people have worked on the sum of three fifth powers.

Answer 3

A very quick exhaustive computer search (for $0 \le x, y, z \leq \lceil 2015^.2 \rceil = 5$) shows there is no solution, as does a search with $0 \le x \le 200$ and $-200 \le y, z, \le 200$.