As I was browsing through the problems & solutions department of *the American Mathematical Monthly* the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, [Ernst Trost][1] resorted to this trick and, what is more, added a reference to the aforementioned paper of his for further applications of it. 

The problem asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution goes as follows:

> We consider the more general eq.

>$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

>where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: [K. Conrad][2], ["Proofs by descent"][3], p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$


  [1]: https://www.genealogy.math.ndsu.nodak.edu/id.php?id=184011
  [2]: http://mathoverflow.net/users/3272/kconrad
  [3]: http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/descent.pdf