For which integers x,y is (4-6*sqrt(2))^2 = x+y*sqrt(2)?

[danielw]

Hi All

I have the following question.View attachment 5866

I have reviewed my notes but have not been able to crack this.

I tried two different ways, both wrong.

First:

\(\displaystyle
(4-6*\sqrt2)^2=\)

\(\displaystyle 16-24*\sqrt2-24*\sqrt2+(36*2)

= 88-218*\sqrt2\)

so, $x=88$ and

$y=218$My second method was

\(\displaystyle (4-6*\sqrt2)^2= 4^2-(6*2)=28\)

but this is not in the form they want.

I'd really appreciate some advice on how to go about solving this kind of problem.

Thanks!

Daniel

 

[MarkFL]

I think your first method is the way I would use, so that the LHS of the equation is in the form $a+b\sqrt{2}$:

\(\displaystyle (4-6\sqrt{2})^2=x+y\sqrt{2}\)

\(\displaystyle 16-48\sqrt{2}+72=x+y\sqrt{2}\)

\(\displaystyle 88-48\sqrt{2}=x+y\sqrt{2}\)

And so we see that:

\(\displaystyle (x,y)=(88,-48)\)