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$x^8-y^5=z^3$

I believe it is some form of a Diophantine equation but since each variable is to a different power, I am unable to solve it.

Nick
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1 Answers1

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We can get examples using powers of $2$, and exploiting the fact that $2^n-2^{n-1}=2^{n-1}$. We just need $n$ such that $8\,|\,n$ and $15\,|\,(n-1)$.

We see that $$2^{16}-2^{15}=2^{15}\implies (2^2)^8-(2^3)^5=(2^5)^3$$

for instance.

In this way we get an infinite family of solutions, given by $n\equiv 16\pmod {120}$. Of course, these are not all the solutions. If $(a,b,c)$ is a solution then so is $(m^{15}a, m^{24}b, m^{40}c)$ for any $m\in \mathbb N$. I don't know if there are any solutions in which the three integers are relatively prime.

lulu
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    Other solutions include $(a,b,c) = (3^{4},; 3^{6},; 2 \cdot 3^{10})$ and $(a,b,c) = (2^{3}\cdot 3^{2},; 2^{5}\cdot 3^{3},; 2^{8}\cdot 3^{5})$. – Robert Israel Feb 06 '20 at 02:36