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How can I find all triples $(m,n,k)$ of non-negative integers such that $5^n+7^m=k^3$?

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A quick search suggests that you already have. – Ross Millikan Jun 17 '11 at 16:37
Do a websearch for "exponential diophantine equations" and use what you find to learn the methods to use. – Gerry Myerson Jun 18 '11 at 0:30
up vote 1 down vote accepted

First, note that $n$ must be even, $m$, odd. So we have $$x^2+7y^2=k^3{\rm\qquad with\qquad}x=5^r,y=7^s$$ So $$(x+y\rho)(x-y\rho)=k^3{\rm\qquad where\qquad}\rho=\sqrt{-7}$$ Are the integers in ${\bf Q}(\rho)$ a unique factorization domain? I don't know. It's easy enough to look up. Let's suppose it's the case. Anything that divides both $x+y\rho$ and $x-y\rho$ must divide their sum and difference, hence must divide 2. In this ring, $$2={1+\rho\over2}{1-\rho\over2}$$ so I may be sweeping something under the rug when I deduce from the second display that $x+y\rho$ is a cube. We have two cases: $$x+y\rho=(a+b\rho)^3{\rm\qquad or\qquad}x+y\rho=\left({a+b\rho\over2}\right)^3$$ In the second case, we may assume $a$ and $b$ are both odd. Let's look at the first case.

Multiplying out and comparing imaginary parts we get $y=3a^2b-7b^3$. So $b$ divides $y$, and $y=7^s$, so $b=7^t$ for some $t$, so mod 3 the equation becomes $1=-1$, contradiction. So, on to the second case.

Multiplying out and comparing imaginary parts we get $8y=3a^2b-7b^3$. Now $y=7^s$, $b$ is odd and divides $8y$, so $b=7^t$ for some $t$, so $8\times7^s=3a^27^t-7^{3t+1}$, so $s=t$, so $8=3a^2-7^{2s+1}$, so $3a^2\equiv1\pmod7$. But $3$ is a quadratic non-residue mod $7$, so this is impossible.

You may want to go over the argument to see just exactly where I lost the solution $5^0+7^1=2^3$.

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The ring of integers is a UFD, but $x + y \rho$ isn't necessarily a cube; as you have seen it can instead be $\frac{1 + \rho}{1 - \rho}$ times a cube (which is divisible by $(1 - \rho)$). – Qiaochu Yuan Jun 18 '11 at 7:19
@Qiaochu, sure enough. You want to have a go at patching the argument? – Gerry Myerson Jun 18 '11 at 12:22

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