Find all integral solutions of the equation $\displaystyle x^4+2x^3+2x^2+2x+5=y^2$.

My attempt: The equation can be rewritten as $\displaystyle (x+1)^2(x^2+1)=(y+2)(y-2)$.

Can someone please tell me how to proceed?

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- Jan 14th 2011, 11:54 PMalexmahoneDiophantine equation
Find all integral solutions of the equation $\displaystyle x^4+2x^3+2x^2+2x+5=y^2$.

My attempt: The equation can be rewritten as $\displaystyle (x+1)^2(x^2+1)=(y+2)(y-2)$.

Can someone please tell me how to proceed? - Jan 15th 2011, 11:00 AMPranas
I hate to be writing the very first comment in this thread without any actual help within, but it has been resting for a while now...

I noticed that

$\displaystyle \displaystyle \[{x^4} + 2{x^3} + 2{x^2} + 2x + 5 = {y^2}\]$

May be rewritten as (quite unusual form)

$\displaystyle \displaystyle \[x\left( {x\left( {x\left( {x + 2} \right) + 2} \right) + 2} \right) + 5 = {y^2}\]$

I don't have a clue whether there can be a slightest use of this though, sorry... - Jan 15th 2011, 03:59 PMorange gold
Re-wrote it as:

$\displaystyle

(x+1)(x+1)(x-1)(x-1)(0.5x)=(y+2)(y-2)

...$

$\displaystyle

(x^2-1)^2(0.5x)=(y^2-4)

...$

$\displaystyle

(x^2-1)^2=(y^2-4)/(0.5x)

...$

$\displaystyle

x((x^2-1)^2)=(y^2-4)/(0.5)

...$

$\displaystyle

(x)(x^2-1)(x^2-1)=(2)(y+2)(y-2)

...$

$\displaystyle

(x)(x-1)(x+1)(x-1)(x+1)=(2)(y+2)(y-2)

...$

$\displaystyle

(x)(x^2+2x+1)(x^2-2x-1)=(2)(y+2)(y-2)

...$

$\displaystyle

(xxxxx-4xxx-4xx-x)=(2y+4)(2y-4)

...$

$\displaystyle

xxxxx-4xxx-4xx-x=4yy-16

$

I don't know where I'm going with this, but maybe that helped :D - Jan 15th 2011, 06:31 PMtonio
- Jan 16th 2011, 12:32 AMOpalg
Let $\displaystyle p(x) = x^4+2x^3+2x^2+2x+5$. Check that

$\displaystyle (1)\qquad (x^2+x)^2 = p(x) - (x^2+2x+5),$

$\displaystyle (2)\qquad (x^2+x+1)^2 = p(x) - (x^2-4),$

$\displaystyle (3)\qquad (x^2+x+2)^2 = p(x) + (3x^2+2x-1).$

Then $\displaystyle x^2+2x+5 = (x+1)^2+4>0$, so it follows from (1) that $\displaystyle x^2+x$ is always too small to be a square root for $\displaystyle p(x).$

Next, $\displaystyle 3x^2+2x-1 = (3x-1)(x+1)$, and the only integers for which this is not positive are $\displaystyle x=0$ and $\displaystyle x=-1$. For all other integers, $\displaystyle 3x^2+2x-1$ is positive, and it follows from (3) that $\displaystyle x^2+x+2$ is too large to be a square root of $\displaystyle p(x).$

Thus, unless $\displaystyle x = 0$ or $\displaystyle -1$, the only possible candidate for an integral square root of $\displaystyle p(x)$ is $\displaystyle y = x^2+x+1$, and it follows from (2) that this solution will only work if $\displaystyle x^2-4=0.$

Therefore the only values of $\displaystyle \normalsize x$ that might give solutions to the problem are $\displaystyle x = 0,\ -1$ and $\displaystyle \pm2$. If $\displaystyle \normalsize x=0$ then $\displaystyle p(x) = 5$, which is not a square. But the other three values of $\displaystyle \normalsize x$ give the solutions to the problem, namely $\displaystyle (x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7)$. - Jan 16th 2011, 03:26 AMroninpro
- Jan 16th 2011, 12:41 PMOpalg
The strategy is quite simple really. We want to find an integer that is a square root for $\displaystyle p(x)$. My claim is that the best candidate for this square root is $\displaystyle z \stackrel{\mathrm{d{e}f}}= x^2+x+1$. Equation (1) shows that $\displaystyle z-1$ is definitely too small to be a square root for $\displaystyle p(x).$ Similarly, equation (3) shows that $\displaystyle z+1$ is too large to be a square root for $\displaystyle p(x)$ except possibly when $\displaystyle x = -1$ or $\displaystyle 0.$ So, for all other values of $\displaystyle \normalsize x$, the square root must be bigger than $\displaystyle z-1$ and smaller than $\displaystyle z+1$, which leaves $\displaystyle z$ as the only remaining possibility. But equation (2) shows that for $\displaystyle z$ to be the square root, it is necessary that $\displaystyle x^2-4=0$, and that only happens when $\displaystyle x=\pm2.$

- Jan 16th 2011, 01:06 PMchiph588@