 Solution

Let the common difference of $r^2,s^2,1,t^2$ be $d$. ThenFor $x=-2s^2$ and $y=2rst$,so $(-2s^2,2rst)\in E(\mathbb{Q})$.

Part (c) says every rational point is one of the eight torsion points from part (b). Among their $x$-coordinates, the only negative value of the form $-2s^2$ with $s\ne 0$ is $-2$, arising from $P_2$ or $-P_2$. Thus $s^2=1$, and the four-term arithmetic progression has common difference zero. ConsequentlyClearing denominators gives Euler's corresponding result for integer squares.