Robert Gerbicz wrote: Let $ n$ a positive integer and $ x=10*{n}^{15},\; y=3*{n}^{6},\; z=7*{n}^{10}$ then $ x,y,z$ is a solution of the equation.

Peter wrote: Show that the equation $ x^2 + y^5 = z^3$ has infinitely many solutions in integers $ x, y, z$ for which $ xyz \neq 0$. Other solution: Solution 1 From $ 2^n + 2^n = 2^{n + 1}$ Chose $ z = 2^{\frac {n + 1}{3}}$ $ y = 2^{\frac {n}{5}}$ $ x = 2^{\frac {n}{2}}$ Chose $ 10|n,3|n + 1,\Rightarrow n\equiv 20 (\mod 30)$ Then it has solution. In the same way we can prove that if $ \gcd(a,b) = 1$ then the equation: $ x^a + y^b = z^c$ has integer solution and $ xyz > 0$ Solution 2 (From Zetax) We use the lemma : Every integer can represent as sum of a cube and a perfect square. The equation has equivalent $ ( - y)^5 = x^2 + ( - z)^3$ (1) Chose $ |y|$ not is a cube or a perfect square. We prove that if $ x,y$ satisfy (1) then it is different from 0. Case 1$ x = 0$ then $ y^5 = z^3\Rightarrow y = m^3$ (tradition!) Case 2 $ z = 0$ then $ ( - y)^5 = z^2,\Rightarrow y = - m^2$ (tradition!) So it has solution satisfy $ xyz$ is different from 0.

Another parametric solution: take $z=(k^{24}-1)^8k^8$ $ y=(k^{24}-1)^5$ $x=(k^{24}-1)^{12}$ then $z^3-y^5 = (k^{24}-1)^{24}k^{24}-(k^{24}-1)^{25}$ $=(k^{24}-1)^{24}(k^{24}-(k^{24}-1))$ $=(k^{24}-1)^{24}$ $=((k^{24}-1)^{12})^2 = x^2$ this doesn't required $10^2+3^5=7^3$

$x=2^{5(3k-1)}$ $y=2^{2(3k-1)}$ $z=2^{10k-3}$