# Any positive integer solutions to $x^6+y^{10}=z^{15}$?

This question might be easy.

The hard question is this: prove that if $a,b,c\geq3$ then there are no solutions in positive integers $x,y,z$ to $x^a+y^b=z^c$ with $x,y,z$ coprime. This implies Fermat, most cases of Catalan, etc., and is an open problem.

But it's really crucial that $x,y,z$ are coprime to make this question hard. For example if I want to find any solution to $x^9+y^{10}=z^{11}$ in positive integers, I just start with a random solution to $A+B=C$, e.g. $1+1=2$, and now I multiply by an appropriate power of all the primes dividing $ABC$ to get a solution. For example, if I start with $1+1=2$ then I multiply both sides by $2^N$ and deduce $2^N+2^N=2^{N+1}$. Now it's easy to find a positive $N$ with $N=0$ mod 9, $N=0$ mod 10 and $N=-1$ mod 11, and for this value of $N$ we get a solution in positive integers to $x^9+y^{10}=z^{11}$.

But this trick relies on the fact that 9, 10, 11 are pairwise coprime. It wouldn't surprise me if an extension of the trick could give a solution in positive integers to $x^6+y^{10}=z^{15}$, where the point is that the exponents aren't pairwise coprime, but 5 minutes on the back of an envelope didn't give me the trick I needed, and I thought that here might be a great place to ask.

What's the trick I've missed?

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The trick you describe generalizes to easy solutions whenever one of the exponents is relatively prime to the other two. But of course that doesn't apply to this case. – Qiaochu Yuan Mar 25 '11 at 22:45

I agree with Qiaochu; there can't be a similar trick in this case. You can't profit from multiplying by common factors because in this case all solutions can be reduced to coprime solutions.

To see this, consider the number of factors of an arbitrary prime $p$ in the equation. We must have

$$rp^{6k}+sp^{10l}=tp^{15m}$$

with $p \nmid r,s,t$. The lowest two powers of $p$ must coincide, since otherwise we could divide through by the lowest and the remaining equation couldn't be fulfilled mod $p$. So we can divide through by this common lowest power of $p$ and leave a power of $p$ in only one of the terms. But since the factors $6$, $10$ and $15$ are such that the least common multiple of each pair is a multiple of the third, dividing through by a multiple of that least common multiple will just substract a multiple of the third factor in the exponent of the third term, still leaving a multiple of that factor. It follows that we could have divided each of the numbers by an appropriate power of $p$ to begin with, leaving a power of $p$ in only one of the three. Doing this for all primes, we can reduce all solutions to coprime solutions.

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Presumably you mean $r, s, t$ are coprime to $p$. You are really arguing mod $p^{\min(6k,10l,15m)}$ and you might have powers of $p$ in two terms, but the idea goes through. – Ross Millikan Mar 26 '11 at 3:22
@Ross: Yes, "$r,s,t\nmid p$" was the wrong way around; I've fixed that. But what do you mean by "you might have powers of $p$ in two terms"? If I have powers of $p$ in two terms and not in the third, how can the equation hold mod $p$? – joriki Mar 26 '11 at 8:09
I see-you are right about mod p – Ross Millikan Mar 26 '11 at 14:14
So it seems you've reduced the general case to the coprime case, which probably means the question is hard. There's a lot of activity around this sort of question at the minute: trying to prove no coprime solutions to various equations of the form $x^a+y^b=z^c$ for explicit $a,b,c$; so all I have to so is wait until they get up to $x^6+y^{10}=z^{15}$ :-) Thanks. – Kevin Buzzard Mar 26 '11 at 16:07

I claim that a trick of this kind doesn't exist in this case. As evidence, observe that the equation has no solutions for which $31 \nmid xy$. To see this, note that

$$x^6 \equiv 0, 1, 2, 4, 8, 16 \bmod 31$$ $$y^{10} \equiv 0, 1, 5, 25 \bmod 31$$ $$z^{15} \equiv 0, 1, 30 \bmod 31.$$

But any reasonable perturbation of this trick that I can think of would allow you to write down a solution in which $x, y, z$ are all coprime to $31$.

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