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This post got me curious. The Diophantine equation with $x\neq1$,

$$x^n+y^3 = z^2\tag1$$

has infinitely many coprime solutions when $n\leq5$. Are there at least a few co-prime ones when $n>5$?

$n=6:$ $\color{red}{??}$

No solutions with $x<100$ and $0<y<100000$. (I checked.)

$n=7:$

$$2^7+17^3=71^2$$ $$17^7+76271^3=21063928^2$$

$n=8:$

$$43^8+96222^3 = 30042907^2$$

Q: Does $n=6$ in fact have a coprime solution? (It seems strange that $n=7,8$ has but $n=6$ doesn't. A larger search radius might yield a result.)

P.S.: From work by Darmon and Granville, if $1/p+1/q+1/r<1$, then the equation $ax^p+by^q+cz^r=0$ has only finitely many coprime solutions. So $(1)$ has only finitely many for $n\geq7$.

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  • $\begingroup$ Considering only strictly positive integers first solutions to $x^6+y^3=z^2$ are $\left( \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 8 & 24 \\ 3 & 18 & 81 \\ 4 & 32 & 192 \\ 5 & 50 & 375 \\ 6 & 72 & 648 \\ 7 & 98 & 1029 \\ \end{array} \right)$ $\endgroup$ – Raffaele Jul 28 '17 at 11:53
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Proposition 14.6.1. In the parabolic case $1/p + 1/q + 1/r = 1$, the equation $x^p + y^q = z^r$ has no solutions in nonzero coprime integers, except that the equation $x^3 + y^6 = z^2$ has the solutions $(x, y, z) = (2, ±1, ±3)$, and the equation $x^3 + y^2 = z^6$ has the solutions $(x, y, z) = (−2, ±3, ±1)$.

From Henri Cohen Number Theory Volume II: Analytic and Modern Tools

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  • 1
    $\begingroup$ Ah, I should have known. So that is what makes $n=6$ different from the $n=7,8$ cases since $1/6+1/3+1/2=1$. $\endgroup$ – Tito Piezas III Jul 28 '17 at 7:41

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