# 4 squares almost in an arithmetic progression

Does there exist infinitely many coprime pairs of integers x,d such that x, x+d, x+2d, x+4d are all square numbers?

One example would be 49,169,289,529. This is the only example I have found so far and I believe it may be the only one.

The answer is yes. Here is another example, $$x=2021231^2, \qquad d=82153503191760.$$ Then $$x+d=9286489^2, \qquad x+2d=12976609^2, \qquad x+4d=18240049^2.$$
Basically, write $x=X^2$, $x+d=Y^2$, $x+2d=Z^2$, and $x+4d=W^2$. Eliminating $x$ and $d$ reduces to the two equations: $$X^2-2Y^2+Z^2=0, \qquad 3X^2-4Y^2+W^2=0.$$ This pair of equations defines a smooth curve $C$ of genus one in $\mathbb{P}^3$, which we can put into Weierstrass form by sending the point $(1,1,-1,1)$ (say) to the point at infinity. The elliptic curve you obtain is $$y^2 = x^3 - x^2 - 9x + 9$$ with Cremona reference 192A, and Mordell--Weil rank $1$. The point $(5,8)$ has infinite order, and corresponds to your solution (up to changes in sign). Taking multiples of this point and mapping back to $C$ gives infinitely many coprime pairs $x$, $d$ such that $x$, $x+d$, $x+2d$, $x+4d$ are squares.
• I guess the letters in the two equations for $C$ should be capital. And there may be a typo in the first equation ? – Chris Wuthrich Nov 21 '13 at 15:15
• I still think you meant $X^2$ instead of $Y^2$ in the first equation ... (sorry to be picky). – Chris Wuthrich Nov 21 '13 at 16:16