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I've been thinking about the integer solutions to the equation $x^3+x^2+x+1=y^2$. So far the only solutions I've found are $(0,1)$, $(1,2)$, and $(7,20)$. Do we know what the other solutions are (if any), or is this an open problem?

Joe
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OEIS
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  • One idea I have is to note that $x^3+x^2+x+1=(x+1)(x^2+1)$, and consider the prime factorisation of $x^3+x^2+x+1$ (each prime in the factorisation must be repeated). However, I am not yet sure if this yields a solution. – Joe Sep 10 '22 at 20:50
  • You have at least three more solutions since $y^2=(-y)^2$. – Andrew Chin Sep 10 '22 at 20:53
  • A quick search returns this list of results via aproach0.xyz. – Andrew Chin Sep 10 '22 at 20:55

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Of course the reflections $(0, -1), (1, -2), (7, -20)$ of the three points you identified across the $x$-axis, and inspection shows that $(-1, 0)$ is also on the curve (its unique intersection with the rational $x$-axis).

There are algorithms that compute lists of integral points on elliptic curves, and they are implemented in MAGMA and SAGE, the latter of which is open source. Executing the SAGE code

E = EllipticCurve([0,1,0,1,1])
E.integral_points(both_signs=True)

returns exactly the above seven points, namely, the $3$ you found, their mirror images, and $(-1, 0)$, so there are no further solutions. (The lattermost point, by the way, generates the torsion subgroup of the elliptic curve.)

Travis Willse
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