# Trost's Discriminant Trick

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:

1. Was Trost's trick noticed by anyone (before or after Trost)?
2. Are there any other cute applications?
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That's a neat trick. Here's another way of looking at it. I'll use the identity $a^4 + b^4 = (a^4 - b^2)^2 - a^4(a^4 - 2b^2 - 1)$. Hence if $a^4 - 2b^2 = 1$, it follows that $a^4 + b^4$ is a square. – Ravi Boppana Aug 26 '10 at 19:48
Even simpler, if $a^4 - 2b^2 = 1$, then $a^4 + b^4 = (2b^2 + 1) + b^4 = (b^2 + 1)^2$. – Ravi Boppana Aug 26 '10 at 22:55

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].

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Hi Wadim, welcome back! – Victor Protsak Oct 14 '10 at 15:50

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."

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