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Determine the natural numbers $x,y,n$ matching equality $$x^n+y^n=2015.$$

I noticed for $n = 1$ the equation has solutions $(x, 2015-x), x$ integer.

For $n = 2$, given that $x$ and $y$ are different parities taking $x = 2k$ and $y=2m + 1$ we come to contradiction.

What must be done to $n\geq3$?

medicu
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4 Answers4

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Any solution with $n$ even can be used to give a solution with exponent $2$. So it suffices to study $n$ odd. Since $\log_2 2015 \approx 10.97 < 11$, we can ignore all exponents larger than $9$. This reduces it to $3, 5, 7, 9$.

  • Exponent $9$: $2015 - 2^9 = 1503$ is not a ninth power and $3^9$ is larger than $2015$, so no solutions.

  • Exponent $7$: $2015 - 2^7 = 1887$ is not a seventh power and $3^7$ is larger than $2015$, so no solutions.

  • Exponent $5$: Check directly that $x = 2, 3$ or $4$ leads to no solution, and $5^5$ is too large. No solutions.

  • Exponent $3$: I'll leave it to you to check there are no solutions. You can check for $x \le 12$.

  • Note that this answer is the opposite of clever. I'm sure there's some nice number theory based on the fact that $2015 = 5 \times 13 \times 31$ that can be used here. Probably at least for exponent $5$.... –  Dec 29 '15 at 20:10
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    One cleverness is that because $31$ is a factor and is $3 \pmod 4$, there is no solution for $n=2$ and therefore, as you say, any even $n$. I think this is a fine approach for the problem at hand. – Ross Millikan Dec 29 '15 at 20:15
  • @RossMillikan Ah, quite nice. Thanks. –  Dec 29 '15 at 20:16
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    You can use the fact that for odd $n$, $x+y$ divides $x^n+y^n$. For $n=3$ this pretty much means $x+y=13$ (since $5$ is too small and $31$ or bigger is too big), so it's enough to check $x\le6$. – Barry Cipra Dec 29 '15 at 20:24
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    @RossMillikan or just that 2015 is 3 mod 4. – djechlin Dec 29 '15 at 20:29
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Since $2015=5\cdot 13\cdot 31$, and the multiplicities of the primes congruent $3$ modulo $4$ are odd (only $p=31$ to consider here), it follows that $2015$ cannot be represented as the sum $x^2+y^2$, by Fermat's theorem. This does not only work for $2015$. Also, characterisation of two cubes is known, see here. Of course, $2015$ is still small enough to check all cases. For bigger numbers however, one would like to apply more number theory.

Edit: In case we would like to have more positive integer solutions to $x^n+y^n=m$, then $m=4097$ is the smallest number (besides $2$) which is a sum of two squares and two positive cubes. Indeed, the equation $x^n+y^n=4097$ has integer solutions for $n=1,2,3,4,6,12$: $$ 4097=1+4096=1^2+64^2=1^3+16^3=1^4+8^4=1^6+4^6=1^{12}+2^{12}. $$

Dietrich Burde
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Hint: $n<12$ because $1^{12}+2^{12}>2015$

Simply Beautiful Art
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Piggybacking off user's answer that we need only concern ourselves with a few odd exponents, with $x\lt y$ also pretty small, let's use the fact that $x+y$ divides $x^n+y^n$ for odd $n$ and $x^p\equiv x$ mod $p$ for prime $p$.

Note that $2014$ is not an $n$th power for any $n\gt1$, so we can assume $1\lt x\lt y$. Note also that $2015=5\cdot13\cdot31$. Finally, $\sqrt[3]{2015}\lt13$ means we don't have to worry about potential solutions with $x+y\ge25$, which means we need only consider solutions with $x+y=5$ or $x+y=13$.

For $n=3$, we have $x+y\equiv x^3+y^3=2015\equiv2$ mod $3$, so $x+y=5$ is the only possibility, which means $x=2,y=3$ is the only possibility, but $2^3+3^3$ is way smaller than $2015$.

For $n=5$, we have $x+y\equiv x^5+y^5=2015\equiv0$ mod $5$, so again $x+y=5$ is the only possibility, and again $2^5+3^5$ is too small.

For $n=7$, we have $x+y\equiv x^7+y^7=2015\equiv6$ mod $7$, so $x+y=13$ is the only possibility. But $7^7$ is already way too big. (Alternatively, note that $\sqrt[7]{2015}\approx2.965\lt3$ already rules out any solutions.)

For $n=9$, working mod $3$, we have $x+y\equiv x^9+y^9\equiv2$ mod $3$, so again $2^9+3^9$ is the only possibility, which is too large. (Or, again, $\sqrt[9]{2015}\lt3$ rules out any solutions.)

Barry Cipra
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