Diophantine equation

**alexmahone** · January 14th 2011, 11:54 PM

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

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

Can someone please tell me how to proceed?

**Pranas** · January 15th 2011, 11:00 AM

Originally Posted by alexmahone

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

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

Can someone please tell me how to proceed?

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

May be rewritten as (quite unusual form)

$\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...

**orange gold** · January 15th 2011, 03:59 PM

Re-wrote it as:

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

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

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

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

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

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

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

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

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

I don't know where I'm going with this, but maybe that helped

**tonio** · January 15th 2011, 06:31 PM

Originally Posted by orange gold

Re-wrote it as:

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

Your left side here doesn't equal the OP's left side: it must be a quartic pol. in x, and you wrote a quintic.

Tonio

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

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

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

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

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

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

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

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

I don't know where I'm going with this, but maybe that helped

.

**Opalg** · January 16th 2011, 12:32 AM

Originally Posted by alexmahone

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

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

Can someone please tell me how to proceed?

Let $p(x) = x^4+2x^3+2x^2+2x+5$ . Check that

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

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

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

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

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

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

Therefore the only values of $\normalsize x$ that might give solutions to the problem are $x = 0,\ -1$ and $\pm2$ . If $\normalsize x=0$ then $p(x) = 5$ , which is not a square. But the other three values of $\normalsize x$ give the solutions to the problem, namely $(x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7)$ .

**roninpro** · January 16th 2011, 03:26 AM

Originally Posted by Opalg

Let $p(x) = x^4+2x^3+2x^2+2x+5$ . Check that

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

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

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

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

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

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

Therefore the only values of $\normalsize x$ that might give solutions to the problem are $x = 0,\ -1$ and $\pm2$ . If $\normalsize x=0$ then $p(x) = 5$ , which is not a square. But the other three values of $\normalsize x$ give the solutions to the problem, namely $(x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7)$ .

Could you elaborate a little bit more on why this method is exhaustive?

**Opalg** · January 16th 2011, 12:41 PM

Originally Posted by roninpro

Originally Posted by Opalg

Let $p(x) = x^4+2x^3+2x^2+2x+5$ . Check that

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

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

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

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

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

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

Therefore the only values of $\normalsize x$ that might give solutions to the problem are $x = 0,\ -1$ and $\pm2$ . If $\normalsize x=0$ then $p(x) = 5$ , which is not a square. But the other three values of $\normalsize x$ give the solutions to the problem, namely $(x,y) = (-2,\pm3),\ (-1,\pm2),\ (2,\pm7)$ .

Could you elaborate a little bit more on why this method is exhaustive?

The strategy is quite simple really. We want to find an integer that is a square root for $p(x)$ . My claim is that the best candidate for this square root is $z \stackrel{\mathrm{d{e}f}}= x^2+x+1$ . Equation (1) shows that $z-1$ is definitely too small to be a square root for $p(x).$ Similarly, equation (3) shows that $z+1$ is too large to be a square root for $p(x)$ except possibly when $x = -1$ or $0.$ So, for all other values of $\normalsize x$ , the square root must be bigger than $z-1$ and smaller than $z+1$ , which leaves $z$ as the only remaining possibility. But equation (2) shows that for $z$ to be the square root, it is necessary that $x^2-4=0$ , and that only happens when $x=\pm2.$

**chiph588@** · January 16th 2011, 01:06 PM

Originally Posted by Opalg

The strategy is quite simple really. We want to find an integer that is a square root for $p(x)$ . My claim is that the best candidate for this square root is $z \stackrel{\mathrm{d{e}f}}= x^2+x+1$ . Equation (1) shows that $z-1$ is definitely too small to be a square root for $p(x).$ Similarly, equation (3) shows that $z+1$ is too large to be a square root for $p(x)$ except possibly when $x = -1$ or $0.$ So, for all other values of $\normalsize x$ , the square root must be bigger than $z-1$ and smaller than $z+1$ , which leaves $z$ as the only remaining possibility. But equation (2) shows that for $z$ to be the square root, it is necessary that $x^2-4=0$ , and that only happens when $x=\pm2.$

Just to pick your brain a bit, what led you to choose $z=x^2+x+1$ in the first place?

Edit: Wait, I think I see why. $z^2=x^4+2x^3+3x^2+2x+1 \approx p(x)$

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