Show that there are infinitely many solutions $x, y, z \in\mathbb N$ of the following equation $x^2 + y^3 = z^7.$

Question

Show that there are infinitely many solutions $x, y, z \in\mathbb N$ of the following equation $x^2 + y^{3} = z^7$.

I am thinking about using proof by infinite descent, but I am not too sure how to start or how to show there are infinitely many solutions in the natural numbers.

Answer 1

For any $k\in \mathbb N$: $$ ((k^2+1)^3\cdot k)^2 + ((k^2+1)^2)^3 = (k^2+1)^6\cdot k^2 + (k^2+1)^6 =(k^2+1)^7 $$

(found by looking for a solution in the form $x=kt^3, y=t^2$)

Answer 2

For every $a \in \mathbb{N}$, $$(x,y,z)=(2213459a^{21}, 1414a^{14}, 65a^6)$$

is a solution.

Answer 3

You see that $$ (2^3)^2+(2^2)^3=2\cdot 64 = 2^7$$ is a solution. So, one idea would to think, whats about powers of $2$. Then we have $$(2^a)^2+(2^b)^2=(2^c)^7.$$ In the example we furthermore have $x^2=y^3$, so we can also try this, which leads to $$2^{7c}=2^{2a+1}=2^{3b+1}.$$ This is equivalent to $7c=2a+1=3b+1$. For $a=3d$ and $b=2d$ we get $7c=6d+1$. Hence, it suffices to show that there are infinite $c,d\in \mathbb{N}_0$ such that $7c-6d=1$. This is a typical linear Diophantine equation which can be solved by the usual procedure (or WolframAlpha) or you just see that $d=7n+1$ and $c=6n+1$ is a solution for all $n\in\mathbb{N}_0$. Back substitution gives you that $$(2^{3(7n+1)},2^{2(7n+1)},2^{6n+1})$$ is a solution for all $n\in\mathbb{N}_0$.

Answer 4

This is a generalization of the answer by Zermelo-Fraenkel. Start with any one specific solution; Z-F identifies $8^2+4^3=2^7$, but it is possible to start with any other specific solution.

For all $k\in \mathbb N$, $(8k^{21})^2+(4k^{14})^3=(2k^{6})^7$

More generally, if $x,y,z$ is a solution, then $(x\cdot k^{21})^2+(y\cdot k^{14})^3=(z\cdot k^{6})^7$

Thus, if there are any solutions (and there is at least one), there are infinity many solutions.