1. integers

If x and y are integers such that $\displaystyle (x-y)^2 +2y^2 = 27$ 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 $\displaystyle (x-y)^2 +2y^2 = 27$ 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:

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

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

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

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

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

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

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

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