TITLE: Trost's Discriminant Trick
QUESTION [31 upvotes]: The following little trick was introduced by E. Trost 
(Eine Bemerkung zur Diophantischen Analysis, 
 Elem. Math. 26 (1971), 60-61). For showing that a diophantine equation
such as $x^4 - 2y^2 = 1$ has only the trivial solution, assume that 
$a^4 - 2b^2 = 1$; then the quadratic equation $a^4 t^2 - 2b^2 t - 1 = 0$
has the rational solution $t = 1$, hence its discriminant $4b^4 + 4a^4$ 
must be a square. Thus $a = 0$ or $b = 0$ by Fermat's Last Theorem for 
exponent $4$.
Trost gave a few other nice applications of this trick, but I have not seen
it in any textbook on number theory. My questions:

Was Trost's trick noticed by anyone (before or after Trost)?
Are there any other cute applications?

REPLY [5 votes]: 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 resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.
The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went 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, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$