1. ## integers

If x and y are integers such that $(x-y)^2 +2y^2 = 27$ find the possible values of x.
p.s that's the complete question.

Any help would be so much appreciated!

2. Originally Posted by z1llch
If x and y are integers such that $(x-y)^2 +2y^2 = 27$ find the possible values of x.
p.s that's the complete question.

Any help would be so much appreciated!
Solve this equation for x:

$x = y-\sqrt{27-2y^2}~\vee~ x = y+\sqrt{27-2y^2}$

Since $27-2y^2 \geq 0$ the values of y are in $|y| \in \{1,2,3\}$

Plug in these values and check if x is an integer. I've got 4 solutions:

$(4, -1),\ (-4, 1),\ (6, 3),\ (-6, -3)$

3. hi earboth. thanks for helping!
i understand till the $x=y-\sqrt{27-2y^2}$
can please explain why is $27-y^2 \geq 0$
and how to get the four solutions please?

4. Originally Posted by z1llch
hi earboth. thanks for helping!
i understand till the $x=y-\sqrt{27-2y^2}$
can please explain why is $27-y^2 \geq 0$
and how to get the four solutions please?
Hi

$(x-y)^2 = 27 - 2y^2$ and $(x-y)^2 \geq 0$
Therefore you must have $27 - 2y^2 \geq 0$
Otherwise there are no integer solutions

$27 - 2y^2 \geq 0$ for |y| = 0 or 1 or 2 or 3