Diophantine equation $a^{2} = b^{4} + b^{3} + b^{2} + b + 1$

Asked Nov 18 '22 at 18:45

Active Nov 18 '22 at 18:45

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As title sugests, I want to solve this equation $a^{2} = b^{4} + b^{3} + b^{2} + b + 1$ for integers $a,b$.

Actually, the exercise just asks to show that $b=3$ and this is in the section of euclidian rings so I think that I should use some euclidian domain related with the integers such as $\mathbb{Z}[\sqrt{-5}]$ and use the usual complex norm. But I cannot figure out which norm will give me something like $a^{2} = b^{4} + b^{3} + b^{2} + b + 1$

asked Nov 18 '22 at 18:45

Yeipi

2

Does this answer your question? for which values of $n$, $1+n+n^{2}+n^{3}+n^{4}$ is a perfect square? - found using an Approach0 search. There's also other duplicate posts here, e.g., Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square., as well as multiple AoPS threads, e.g., ... – John Omielan Nov 18 '22 at 18:53
Or https://math.stackexchange.com/questions/510292/integer-x-for-which-x4x3x2x1-is-perfect-square – Parcly Taxel Nov 18 '22 at 18:54
(cont.) Dipohantine equation, equation, etc. – John Omielan Nov 18 '22 at 18:55
Does this answer your question? Determining all the positive integers $n$ such that $n^4+n^3+n^2+n+1$ is a perfect square. – Clyde Kertzer Nov 18 '22 at 23:05

Diophantine equation $a^{2} = b^{4} + b^{3} + b^{2} + b + 1$

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