Rigid pentagons and rational solutions of $s^4+s^3+s^2+s+1=y^2$

Question

Gerard 't Hooft, Nobel Prize in Physics laureate, wrote three articles on what he called "Meccano math" (1, 2, 3) – rigid constructions following rules quite similar to my earlier question on doubling the cube with unit sticks, but with the following generalisations:

• Sticks can be of any rational length (the formulation in 't Hooft's papers uses idealised Meccano strips of integral length, but they can be trivially scaled)
• Hinges can lie anywhere on a stick, not just at the ends, as long as they are at rational distances from the ends
• For rigid polygons, the polygon's sides can be extended

One of the given constructions is a rigid pentagon with just two extra sticks. However, it does not look very nice because it requires long extensions of two sides.

So I decided to make it less intrusive (in the sense of "less occupied space outside the pentagon") as follows. Let $r,t,s$ be the lengths of three consecutive sides of a quadrilateral, with $108^\circ=\frac{3\pi}5$ angles between them:

Then it is easy to show that the fourth side length $u$ is $$\sqrt{\left((r+s)\cos\frac{2\pi}5+t\right)^2+\left((r-s)\sin\frac{2\pi}5\right)^2}$$ We want all four side lengths to be rational (but they can be negative). If $u$ is rational, so is $u^2$, so the expression inside the square root must also be rational. Expanding it gives $$r^2+s^2+t^2-\frac{rs+rt+st}2+\frac{\sqrt5}2(rt+st-rs)$$ and for this to be rational we must have $rt+st-rs=0$ or $t=\frac{rs}{r+s}$. Making this substitution gives $$u=\sqrt{\frac{r^4+r^3s+r^2s^2+rs^3+s^4}{r^2+2rs+s^2}}$$ Clearly we can scale any solution $(r,s,t,u)$ by any rational number, so we set $r=1$ arbitrarily: $$u=\sqrt{\frac{s^4+s^3+s^2+s+1}{s^2+2s+1}}=\frac{\sqrt{s^4+s^3+s^2+s+1}}{|s+1|}$$ Thus, up to scale, all rational solutions correspond one-to-one with solutions of $$s^4+s^3+s^2+s+1=y^2\qquad s,y\in\mathbb Q,s\not\in\{0,-1\}\tag1$$ The same equation has been posed on this site before, but only with integers, and I could not find any good reference in this answer. By Faltings's theorem there are only finitely many solutions, but have I found all of them?

Is it true that $(1)$ has a solution only if $s$ or $1/s$ is in $\left\{3,\frac{808}{627},-\frac{11}8,-\frac{123}{35}\right\}$? References would be much appreciated.

The solution with $s=-\frac{11}8$ in particular gives a much less intrusive rigid pentagon. (All black sticks below, sides of the pentagon, are of unit length.)

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Edit: The sequence of $s$ values is now in the OEIS! Numerators at A339325, denominators at A339326.

Answer 1

Not true I'm afraid. There are, in fact, an infinite number of rational solutions.

The curve is a quartic with a rational point $(0,1)$, and is thus birationally equivalent to an elliptic curve, which has genus $1$. Faltings' Theorem only applies if the genus is strictly greater than $1$.

The equivalent elliptic curve is $v^2=u^3-5u^2+5u$ with $s=(2v-u)/(4u-5)$. The point $(0,0)$ is the only finite torsion point and we can take $(1,1)$ as a generator.

The rational solutions you give come from small multiples of the generator. Larger examples are $-20965/43993$ and $-761577/1404304$, but you can get larger and larger solutions.