can this expression be a perfect square?

Hi,

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

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

got further

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

Re: can this expression be a perfect square?

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

a and b are integers a > b > 0

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

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

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

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

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

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

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

Is this correct

Re: can this expression be a perfect square?

Quote:

Originally Posted by

**moriman** $\displaystyle 48(a-b)+7=x^2$

a and b are integers a > b > 0

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

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

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

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

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

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

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

Is this correct

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

Re: can this expression be a perfect square?

Thanks Drexel.

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