Trost's Discriminant Trick

T. Nagell in [Norsk Mat. Forenings Skrifter. 1:4 (1921)] shows that, for an odd prime $q$, the equation $$ x^2-y^q=1 \qquad (*) $$ has a solution in integers $x>1$, $y>1$, then $y$ is even and $q\mid x$. In his proof of the latter divisibility he uses a similar trick as follows.

Assuming $q\nmid x$ write ($*$) as $$ x^2=(y+1)\cdot\frac{y^q+1}{y+1} $$ where the factors on the right are coprime (they could only have common multiple $q$). Therefore, $$ y+1=u^2, \quad \frac{y^q+1}{y+1}=v^2, \quad x=uv, \qquad (u,v)=1, \quad \text{$u,v$ are odd}. $$ Using these findings we can state the original equation in the form $x^2-(u^2-1)^q=1$, or $$ X^2-dZ^2=1 \quad\text{where $d=u^2-1$}. $$ The latter equation has integral solution $$ X=uv, \quad Z=(u^2-1)^{(q-1)/2}, $$ while its general solution (a classical result for this particular Pell's equation) is taken the form $(u+\sqrt{u^2-1})^n$. It remains to use the binomial theorem in $$ X+Z\sqrt{u^2-1} =(u+\sqrt{u^2-1})^n $$ (for certain $n\ge1$) and simple estimates to conclude that this is not possible.

To stress the use of similar trick: instead of showing insolvability of $x^2-(u^2-1)^q=1$, we assume that a solution exists and then use $d=u^2-1$ to produce a solution $X,Z$ of $X^2-dZ^2=1$; finally, the pair $X,Z$ cannot solve the resulting Pell's equation. (Of course, it is hard to claim that this is exactly Trost's trick, as here is a dummy variable but no discriminants, except the one for Pell's equation. Trost's trick is less trickier to my taste. $\ddot\smile$)

Note that Nagell's result was crucial for showing that ($*$) does not have integral solutions $x>1$, $y>1$ for a fixed prime $q>3$. This was shown in a very elegant way, using the Euclidean algorithm and quadratic residues by Ko Chao [Sci. Sinica 14 (1965) 457--460], and later reproduced in Mordell's Diophantine equations. The ideas of this proof are in the heart of Mihailescu's ultimate solution of Catalan's conjecture. A much simpler proof of Ko Chao's result, based on a completely different (nice!) trick, was given later by E.Z. Chein [Proc. Amer. Math. Soc. 56 (1976) 83--84].

Cassels reviews Trost's paper (MR0288077 (44 #5275)) and notes that 1) the equivalence of the unsolvability of $x^4-2y^4=z^2$ and $x^4+y^4=z^2$ is well-known (he cites Hardy and Wright, Theorem 226), and 2) "There is a second application of the same trick which shows that there are no non-trivial rational points on a certain elliptic curve if there are none on its jacobian."

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 $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$