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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.

VividD
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  • Brute force should be possible, after all they can't go above 4. – Feyre Sep 08 '15 at 22:28
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    You should be aware that the similar problem for cubes is not generally known; there is no congruence obstruction for $x^3 + y^3 + z^3 = 33,$ but no solution is known. This allows the variables positive or negative or zero, otherwise it is a finite check to decide if there is a solution with all variables positive or, say, non-negative. – Will Jagy Sep 08 '15 at 22:29
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    It is well possible that no solution exists, and there is no way to prove this. Hilberts tenth problem was : Given a diophantic equation, can it be decided if it has a solution ? In general, this is not possible. – Peter Sep 08 '15 at 22:35
  • @Feyre These kinds of problems admit negative integers as part of the solution set, so brute force is not an option. – 2'5 9'2 Sep 09 '15 at 05:33
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    @user236182 You can play the same game modulo a power of $5$ or a prime that is $1$ mod $5$, since $5$ will divide the order of the multiplicative group, limiting potential values of $n^5$. I checked $25,11,31,41,61,71$ and there's no relief for $a=2015$. – 2'5 9'2 Sep 09 '15 at 19:55
  • If anyone cares to program searches, congruence relations can be used to speed up the search. For instance, examining the equation mod $25$ allows you to conclude one of the numbers must be $1$ mod $5$ and the other two must be $2$ mod $5$. – 2'5 9'2 Sep 09 '15 at 20:03
  • And mod $11$, one must be $0$ with the other two among ${1,3,4,5,9}$. And mod $31$, one must be $0$, with a more complicated restriction on the other two, etc. – 2'5 9'2 Sep 09 '15 at 20:09
  • @alex.jordan Yeah, I was being sloppy, though this is why I prefer it when it explicitly says with $x,y,z \in \mathbb{Z}$

    I realized it when it was too late to edit.

     – Feyre Sep 10 '15 at 15:06
  • @Feyre Hmm, I'm not sure that you took me meaning. I meant, for starters, that if you wanted to program a brute force search, you wouldn't need to run through triplets $(x,y,z)$ up to some cap on absolute value. Instead you could run through triplets $(1+5k,2+5m,2+5n)$, cutting the search over the same bound down to $\frac1{125}$ as big. And more complications from other modular relations would cut it down even more. – 2'5 9'2 Sep 10 '15 at 17:25
  • @alex.jordan I responded to your other comment, not that one. Regardless, bridge, water. – Feyre Sep 10 '15 at 18:26
  • @alex.jordan What about $101, 125, 131, 151, 181, 191$? – VividD Sep 10 '15 at 19:23
  • @VividD I didn't check. As these numbers get larger the chances this would help get smaller. The values of $n^5$ make up one fifth of the residues. Then when you start looking at sums of three of them, the possible sums start to cover a higher and higher proportion of all residues, ruling out fewer and fewer cases. – 2'5 9'2 Sep 10 '15 at 19:55
  • @alex.jordan Right, don't bother with that than. It looks it is not a promising path... Thanks anyway for your info so far... – VividD Sep 10 '15 at 19:58

3 Answers3

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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.

Will Jagy
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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.

Tito Piezas III
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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 $-200 \le y, z, \le 200$.

David G. Stork
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    that's fine, and I am not sure the person asking is aware of the significance, bu the problem changes dramatically if we allow the variables to have mixed signs; this is what is usually done. – Will Jagy Sep 08 '15 at 22:48
  • I see, you edited in a search with mixed signs. Good. – Will Jagy Sep 09 '15 at 01:10