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Find the number of positive integers $x$ for which $x^4+x^3+x^2+x+1$ is a perfect square.

My attempts:

Let $x^4+x^3+x^2+x+1=k^2\implies (x+1)^2(x^2-x+1)=(k-x)(k+x)$

I analysed this a bit found $x=0$ as one which satisfy all condition, how do I find other, please help, try to continue this further, if any other elegant method then add that too.

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  • $\begingroup$ I found $3$ also satisfy, by hit and trials. $\endgroup$
    – mathlover
    Jan 28, 2017 at 13:43
  • $\begingroup$ This is a very old problem from AMM. $\endgroup$
    – Michael Rozenberg
    Jan 28, 2017 at 14:06

1 Answer 1

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Note that the problem is equivalent to finding integer solutions to $$4y^2=4x^4+4x^3+4x^2+4x+4$$ Now proceed to note that if $x>3$, we can find $$(2x^2+x)^2=4x^4+4x^3+x^2 < 4x^4+4x^3+4x^2+4x+4=4y^2$$ And $$4y^2 < 4x^4+4x^3+5x^2+2x+1=(2x^2+x+1)^2 $$ Since $4x^4+4x^3+4x^2+4x+4$ is stuck between squares of two consecutive numbers, it cannot be a square itself, which is a contradiction.

Thus we have that if $x$ is a positive integer it must be less than, or equal to, $3$. Trial and error gives us $x=3, y=11 $ is a valid solution.

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  • $\begingroup$ you assumed $x>3$ but why? where you used is fact to establish stucking of $4y^2$ between two perfect square numbers? please explain. how it will affect if $x>3$ is not true. $\endgroup$
    – mathlover
    Jan 28, 2017 at 13:55
  • $\begingroup$ Well, if $x<3$. then the second inequality, $$4y^2 < 4x^4+4x^3+5x^2+2x+1=(2x^2+x+1)^2 $$ does not hold $\endgroup$
    – S.C.B.
    Jan 28, 2017 at 13:57
  • $\begingroup$ @mathlover The bound does not hold when $x<3$. $\endgroup$
    – S.C.B.
    Jan 28, 2017 at 13:57
  • $\begingroup$ (+1), will you please tell is there any standard results regarding this problem, where did it strikes to multiply this by $4$ and proceed. $\endgroup$
    – mathlover
    Jan 28, 2017 at 14:07
  • $\begingroup$ @mathlover A more general result here. I multiplie by $4$ to get a square. $\endgroup$
    – S.C.B.
    Jan 28, 2017 at 14:20

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