= Solution
Let the common difference of $r^2,s^2,1,t^2$ be $d$. Then
$$
r^2=2s^2-1,
\qquad t^2=2-s^2.
$$
For $x=-2s^2$ and $y=2rst$,
$$
x(x+1)(x+4)
=(-2s^2)(-r^2)(2t^2)
=4r^2s^2t^2=y^2,
$$
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 $s\ne0$ is $-2$, arising from $P_2$ or $-P_2$. Thus $s^2=1$, and the four-term <arithmetic progression> has common difference zero. Consequently
$$
\boxed{\text{there is no nonconstant four-term arithmetic progression of rational squares.}}
$$
Clearing denominators gives Euler's corresponding result for integer squares.
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