# Hard polynomial equation

#### Cacapipi2

Hi everybody Here is the problem :
Find all the values of the integer x so that :
$$\displaystyle {x }^{4 }+{x }^{3 }+{x }^{2 }+x +1 =n$$
Where n is a perfect square. I really have no idea of where to begin...

#### topsquark

Forum Staff
Hi everybody Here is the problem :
Find all the values of the integer x so that :
$$\displaystyle {x }^{4 }+{x }^{3 }+{x }^{2 }+x +1 =n$$
Where n is a perfect square. I really have no idea of where to begin...
Hint: $$\displaystyle (x - 1) (x^4 + x^3 + x^2 + x + 1) = x^5 - 1$$

-Dan

#### Cacapipi2

I am sorry, still no idea

#### JeffM

It is a general truth that

$\displaystyle n \in \mathbb Z^+\ and\ a \ne b \implies \dfrac{a^n - b^n}{a - b} = \sum_{i=0}^na^{n-i}b^i \iff a^n - b^n = (a - b) * \sum_{i=0}^na^{n - i}b^i.$

If b = 1, we get $\displaystyle \dfrac{a^n - 1}{a - 1} = \sum_{i=0}^na^{n-i}.$

So what Dan was saying is that you can simplify as follows.

$x^4 + x^3 + x^2 + x + 1 = n \implies \dfrac{x^5 - 1}{x - 1} = n = m^2.$

Now what?

1 person

#### Idea

$$\displaystyle n^2=x^4+x^3+x^2+x+1$$

$$\displaystyle (8n)^2=\left(8x^2+4 x+3\right)^2+(40 x+55)$$

following the explanations given at Math Forum - Ask Dr. Math

the absolute value of $$\displaystyle 40 x+55$$ must be greater than or equal to $$\displaystyle 2\left(8x^2+4 x+3\right)-1$$

possible values for $$\displaystyle x$$ are {-1,0,1,2,3}

it is then easy to check that -1,0,3 are the only solutions

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1 person

#### Cacapipi2

Thank you very much But isn't there an easier explanation using the method given by JeffM and topsquark ?

#### topsquark

Forum Staff
Thank you very much But isn't there an easier explanation using the method given by JeffM and topsquark ?
To be perfectly honest I had a calculation method and it turns out I was wrong. I was thinking of Fermat's little theorem wrong. (Though since JeffM supported it then there might be a way.) In any event it's a heck of a lot easier to calculate this way.

-Dan

#### Cacapipi2

Well I am kinda curious I am sure there is an easier method but I definitely can't find it JeffM can you show us your method ? or give a hint ?

#### JeffM

Well I am kinda curious I am sure there is an easier method but I definitely can't find it JeffM can you show us your method ? or give a hint ?
I suspect as well that there is an easier method, but I also thought that difference of powers would be the key, perhaps with assistance from the integer root theorem. Here basically is as far as I have got along this line.

It is obvious that x = 1 is not a solution because

$1^4 + 1^3 + 1^2 + 1 + 1 = 5,\ not\ a\ perfect\ square.$

So we can assume henceforward that $x \ne 1\ and\ \dfrac{x^5 - 1}{x - 1} = n.$

So far this is easy, but now things get hairy. Here is the route I tried, now excluding x equal 1, and
assuming 1 < n = m^2 And m > 1.

$\dfrac{x^5 - 1}{x - 1} = n = m^2 \implies x^5 - m^2x + (m^2 - 1) = 0.$

The quintic equation above is sure to have at least one real solution. So it may be factorable into a linear term and a quartic term.

One possible factoring is $(x - r)(x^4 + rx^3 + r^2x^2 + px - q) \implies qr = m^2 - 1.$

$r = 0 \implies m^2 - 1 = 0 \implies m = 1.$

That works $0^4 + 0^3 + 0^2 + 0 + 1 = 1 = 1^2.$ Going forward we can assume r and so x is not zero.

$q = \dfrac{m^2 - 1}{r}.$ If r is an integer, then q is rational.

$-\ m^2x = -\ prx - qx \implies m^2 = pr + q$ because x = 0 has already been addressed.

If r is an integer, p is rational.

$m^2r = pr^2 + qr = pr^2 + + m^2 - 1 \implies m^2r - m^2 = pr^2 - 1 \implies$

$pr^2 - m^2r + m^2 - 1 = 0 \implies r = \dfrac{m^2 \pm \sqrt{m^4 - 4p(m^2 - 1)}}{2p} = \dfrac{m^2 \pm \sqrt{m^4 - 4pm^2 + 4p}}{2p}.$

The only way to make $m^4 - 4pm^2 + 4p$ a perfect square is for p = 1.

In that case $r = \dfrac{m^2 \pm \sqrt{(m^2 - 2)^2}}{2} \implies r = m^2 + 1\ or\ r = 1.$

But that answer is impossible. So this factoring can be excluded.

I suspect that if you continue along this line you get the correct answer. It's Thanksgiving weekend and I have family to attend to. Be back later.

1 person

#### DenisB

It's Thanksgiving weekend and I have family to attend to. Be back later.
They'd rather you don't show up...more turkey for them

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