NT Factoring Question

Thunder365

Determine all positive integral solutions to $x$ for which $x^4 + x^3 + x^2 + x + 1$ yields a perfect square.

This seems really easy, but I cant seem to find the proper factors. I factored it a bit to

$(x + 1)^2(x^2 + 1) - x(x^2 + x + 1)$ and then I dont know what to do.

I also need help solving these general types of problems, so if your solution could be more general rather than specific I would really appreciate it Very Happy

**Posted:** Thu Nov 05, 2009 3:59 pm

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t0rajir0u

Factoring isn't really the way. Bound it between two polynomials which are perfect squares.

**Posted:** Thu Nov 05, 2009 5:13 pm

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Ihatepie

maybe?

**Posted:** Thu Nov 05, 2009 5:29 pm

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t0rajir0u

That doesn't quite work. You want the bounds to be different by $1$ ; that's how you can conclude that there are no squares between them.

**Posted:** Thu Nov 05, 2009 5:31 pm

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Thunder365

Im a complete beginner. What exactly is bounding and how do you do it?

**Posted:** Thu Nov 05, 2009 7:04 pm

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**Posted:** Thu Nov 05, 2009 7:40 pm

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randomguy64

You can't prove it is never a perfect square because x=3 makes the polynomial have a value of 121, obviously a perfect square...

**Posted:** Thu Nov 05, 2009 7:55 pm

t0rajir0u

The bounds only work for all but finitely many $x$ .

**Posted:** Thu Nov 05, 2009 8:37 pm

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srinath.r

Here is a complete solution .
I knew this problem before and the mathlinks user Agr_94_math ,had said his beautiful solution to me .
Here it goes
For convenience sake ,let $x = n$ and $y = m$ ,so we have to solve for $m \in \mathbb{N}$ for which
$m^{4} + m^{3} + m^{2} + m + 1$ is a perfect square .
Consider the expansion of $(m^{2} + \frac {m}{2})^{2} = m^{4} + m^{3} + \frac {m^{2}}{4} = m^{4} + m^{3} + m^{2} + m + 1 - (\frac {3m^{2}}{4} + m + 1)$
Since the $\text{L.H.S}$ is positive and $m$ is natural ,we have
$m^{4} + m^{3} + m^{2} + m + 1 = n^{2} > (m^{2} + \frac {m}{2})^{2}$ ,since $m$ is natural ,
$n > m^{2} + \frac {m}{2}$
If $m$ is even ,
$n \ge m^{2} + \frac {m}{2} + 1$ ,
Squaring both sides we get
$n^{2} \ge n^{2} + \frac {5m^{2}}{4}$
$\implies 0 \ge \frac {5m^{2}}{4}$ ,so no solutions in naturals .
If $m$ is odd ,
we have $n \ge m^{2} + \frac {m}{2} + \frac {1}{2}$ ,
again repeating the same step we dont get any solutions in naturals .

**Posted:** Thu Nov 05, 2009 10:00 pm

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randomguy64

I'll finish your solution.

In the even case, 5m^2/4=0 has no solutions in naturals, but $m^2 /4-m/2 -3/4$ has one solution, namely m=3, and hence n=11.

That is the only solution.

**Posted:** Fri Nov 06, 2009 3:35 pm

Agr_94_Math · Yang-Mills Theory Offline Joined: 17 Feb 2008 Posts: 716

Thank you Srinath for posting the solution.

**Posted:** Fri Nov 06, 2009 9:34 pm