# Find all integer solutions of $1+x+x^2+x^3=y^2$

I need some help on solving this problem:

Find all integer solutions for this following equation:

$1+x+x^2+x^3=y^2$

My attempt:

Clearly $y^2 = (1+x)(1+x^2)$, assuming the GCD[$(1+x), (1+x^2)] = d$, then if $d>1$, $d$ has to be power of 2. This implies that I can assume: $1+x=2^s*a^2, 1+x^2=2^t*b^2$. If $t=0$ then it is easy to finish. Considering $t>0$, we can get $t=1$ (simple steps only), so I come up with a “Pell-related” equation .. Then I get sticking there. It has a solution $x=7$, so I guess it’s not easy to find the rest.

#### Solutions Collecting From Web of "Find all integer solutions of $1+x+x^2+x^3=y^2$"
There is quite a bit about this equation in Dickson’s History of the Theory of Numbers. Volume 1, page 56, it says Gerono, Nouv Ann Math (2) 16 (1877) 230-234 proved the only solutions are $$(x,y)=(-1,0),\quad(0,\pm1),\quad(1,\pm2),\quad(7,\pm20)$$ On page 57, Dickson references a proof by Genocchi, Nouv Ann Math (3) 2 (1883) 306-310. Lucas, Nouv Corresp Math 2 (1876) 87-88, had noted that the problem is equivalent to solving $1+x=2u^2$, $1+x^2=2v^2$, and then letting $y=2uv$. Dickson then discusses that system in Volume 2, pages 487-488. Several references are given there.
The solution is given in Ribenboim’s book on Catalan’s conjecture, where all Diophantine equations $$y^2=1+x+x^2+\cdots +x^k$$ are studied.
For $k=3$, only $x=1$ and $x=7$ are possible.
This problem, and its “inverse” (q.v. Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$) were posed by Fermat in 1657 (see, for example, Mahoney pg. 337). Both lead to Pell equations (q.v. Solutions to $p+1=2n^2$ and $p^2+1=2m^2$ in Natural numbers.), which was exactly what Fermat was trying to get his contemporaries to study with him.