Show that if $b^4+b^3+b^2+b+1$ is a perfect square then $b=3$

Asked Nov 07 '22 at 05:24

Active Nov 07 '22 at 08:18

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Let $a,b \in \mathbb{N}$ if $a^2=b^4+b^3+b^2+b+1$ show that $b=3$

These problem appeared in my course of ring theory i try to use something like the Euler's criterion or maybe Fermat's theorem or maybe by case but i not sure to these is correct.

Any hint or help i will be very grateful

edited Nov 07 '22 at 08:18

Bill Dubuque

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asked Nov 07 '22 at 05:24

Nick_W

3

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. - found using an Approach0 search. Other duplicates also found are Integer $x$ for which $x^4+x^3+x^2+x+1$ is perfect square., A quartic diophantine equation, etc. – John Omielan Nov 07 '22 at 05:38
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This can be directly evaluated since for integer $b$ the inequality $(b^2+b/2)^2<b^4+\dots+1=(b^2+b/2+1/2)^2-(b/2-1/2)^2+1<(b^1+b/2+1)^2$ holds which indicate that $b=-1\text{or}3$ – Oolong Milktea Nov 07 '22 at 05:40
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FYI, the Approach0 search also found various AoPS threads such as Diophantine equation, Number theory, Integral solution problem, etc. – John Omielan Nov 07 '22 at 05:42

Show that if $b^4+b^3+b^2+b+1$ is a perfect square then $b=3$

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