# $x^5 + y^2 = z^3$

While waiting for my döner at lunch the other day, I noticed my order number was $343 = 7^3$ (surely not the total for that day), which reminded me of how $3^5 = 243$, so that $$7^3 = 3^5 + 100 = 3^5 + 10^2.$$ Naturally, I started wondering about nontrivial integer solutions to $$x^5 + y^2 = z^3 \tag{*}$$ ("nontrivial" meaning $xyz \ne 0$). I did not make much progress, though apparently there are infinitely many solutions: this was Problem 1 on the 1991 Canadian Mathematical Olympiad. The official solutions (at the bottom of this page) only go back to 1994. A cheap answer is given by taking $x = 2^{2k}$ and $y = 2^{5k}$ so that the l.h.s. is $2^{10k + 1}$. This is a cube iff $10k + 1 \equiv 0 \,(3)$ i.e. $k \equiv 2\,(3)$ thus giving an arithmetic progression's worth of solutions, starting with $$(x, y, z) = (16, 1024, 128)$$ corresponding to $k = 2$ and $$(x, y, z) = (1024, 33554432, 131072)$$ coming from $k = 5$.

What else is known about the equation $(*)$? In particular, are there infinitely many solutions with $x$, $y$, $z$ relatively prime? The one that caught my attention was $(x, y, z) = (3, 10, 7)$. Another one is $(-1, 3, 2)$ because $-1 + 9 = 8$. By Catalan's conjecture (now a theorem), this is the only solution with $x = \pm 1$ or $y = \pm 1$ or $z = 1$. Are there any solutions with $z = -1$? In this case, $(*)$ reduces to $x^5 + y^2 = -1$ and Mihăilescu's theorem does not apply.

Update. This question was essentially already asked here, since the equation $a^2 + b^3 = c^5$ is equivalent to $(-c)^5 + a^2 = (-b)^3$.

• we can rewrite this as $x^5+y^2\equiv 0 \pmod z$ could that help you ?
– user451844
Jul 27 '17 at 1:09
• You might look at OEIS sequences A070065, A070066 and A070067. Jul 27 '17 at 2:08
• Note that if $(x,y,z)$ is a solution, then so is $(c^6 x, c^{15} y, c^{10} z)$ for integers $c$. Jul 27 '17 at 2:41
• quora.com/… Aug 3 '17 at 11:45

Yes, there are infinitely many solutions. In fact, there are many parametrizations of the solutions.
According to a book${}^{\color{blue}{}}$ on my bookshelf,

Up to changing $y$ into $-y$, there are exactly 27 distinct parametrizations of the equations $x^5 + y^2 = z^3$.

One of the simplest paremetrization is given by following formula.

\begin{align} x =&\; 12st(81s^{10}-1584t^5s^5-256t^{10})\\ y =&\; \pm (81s^{10} + 256t^{10})\\ &\;\;\times (6561s^{20} - 6088608t^5s^{15} - 207484416t^{10}s^{10} + 19243008t^{15}s^5 + 65536t^{20})\\ z =&\; 6561s^{20}+2659392t^{5}s^{15}+10243584t^{10}s^{10} - 8404992t^{15}s^5 + 65536t^{20} \end{align}

For example, following two random choices of $s,t$ give you two sets of relative prime solutions.

• $(s,t) = (1,1) \leadsto (x,y,z) = (-21108,-65464918703,4570081)$
• $(s,t) = (1,2) \leadsto (x,y,z) = (-7506024,127602747389962225,-196120763999)$

The book I have is actually quoting result from a thesis${}^{\color{blue}{}}$ by J. Edwards. Consult that if you really want to get into the details.

References

• $\color{blue}{}$ Henri Cohen, Number Theory Number Theory Volume II: Analytic and Modern Tools,
$\S 14.5.2$ The Icosahedron Case $(2,3,5)$.

• $\color{blue}{}$ J. Edwards, Platonic solids and solutions to $x^2+y^3 = dz^r$, Thesis, Univ. Utrecht (2005).

• Nice answer, especially for the first reference. +1
– Xam
Jul 27 '17 at 3:38
• @achille: We can also use the icosahedral equation and scale it appropriately. Kindly see answer below. Jul 27 '17 at 12:00
• @achille: The parameterizations for $x^5+y^3=z^2$ depend on the icosahedron while $x^4+y^3=z^2$ depend on the tetrahedron. Does the book you cite say how many distinct parameterizations are there for the latter? This page gives $7$, but I just found an $8$th one, with no scaling involved. Jul 28 '17 at 5:53
• @TitoPiezasIII In $\S 14.4.1$ of that book, Cohen proved there are 7 parametrizations. Jul 28 '17 at 7:13
• @achillehui: I just re-checked the parameterization I found, and it turns out there is a common factor. Darn! In Edwards' work, he cites 7 parameterizations as well. Thanks, anyway. Jul 28 '17 at 7:35

There is a beautiful connection between $a^5+b^3=c^2$ and the icosahedron. Consider the unscaled icosahedral equation,

$$\color{blue}{12^3u v(u^2 + 11 u v - v^2)^5}+(u^4 - 228 u^3 v + 494 u^2 v^2 + 228 u v^3 + v^4)^3 = (u^6 + 522 u^5 v - 10005 u^4 v^2 - 10005 u^2 v^4 - 522 u v^5 + v^6)^2\tag1$$

By scaling $u=12x^5$ and $v=12y^5$ (or various combinations thereof like $u=12^2x^5$, etc), we then get a relation of form,

$$12^5a^5+b^3=c^2$$

• +1 interesting connection. it is unfortunate the mythical $j(\tau)$ is stuff way beyond me ;-p. Jul 27 '17 at 13:14