can this expression be a perfect square?

**moriman** · October 9th 2011, 12:47 AM

Hi,

Can anyone show that $4a^4+8a^3 b-12ab^3-7b^4$ can be a perfect square if a and b are integers and a > b > 0 ?

I think that $a(a+2b)$ and $b(12a+7b)$ must simultaneously be squares but am struggling to get further

got further

can $48(a-b)+7$ be a perfect square?

**moriman** · October 9th 2011, 02:28 AM

$48(a-b)+7=x^2$

a and b are integers a > b > 0

$48a-48b+3=x^2-4$

$48a-48b+3=(x+2)(x-2)$

$3(16a-16b+1)=(x+2)(x-2)$

$so\ \ \ \ x+2=3\ \ \ \ x=1$

$or\ \ \ \ x-2=3\ \ \ \ x=5$

$if\ \ x=1\ \ then\ \ 48(a-b)=-6\ \ \ =>\ \ \ a < b$

$if\ \ x=5\ \ then\ \ 48(a-b)=18\ \ \ a-b=\dfrac{3}{8}\ \ \ \ =>\ \ \ \ a\ and\ b\ are\ not\ integers$

Is this correct

**Drexel28** · October 9th 2011, 10:38 AM

Originally Posted by moriman

$48(a-b)+7=x^2$

a and b are integers a > b > 0

$48a-48b+3=x^2-4$

$48a-48b+3=(x+2)(x-2)$

$3(16a-16b+1)=(x+2)(x-2)$

$so\ \ \ \ x+2=3\ \ \ \ x=1$

$or\ \ \ \ x-2=3\ \ \ \ x=5$

$if\ \ x=1\ \ then\ \ 48(a-b)=-6\ \ \ =>\ \ \ a < b$

$if\ \ x=5\ \ then\ \ 48(a-b)=18\ \ \ a-b=\dfrac{3}{8}\ \ \ \ =>\ \ \ \ a\ and\ b\ are\ not\ integers$

Is this correct

This looks right, but you could have just noted that if $48(a-b)+7=x^2$ then $3$ is a quadratic residue $\text{mod }4$ , which it isn't.

**moriman** · October 9th 2011, 12:10 PM

Thanks Drexel.

I'll have to take a look at quadratic residues, it's a loooooooonnnnng time since I was at school

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