x^2-6xy+10y^2=4
find all the integer pairs of x and y which satisfiy the equation.
Thanks!
$\displaystyle x^2 - 6xy + 10y^2 - 4 = 0$
$\displaystyle a = 1$ , $\displaystyle b = -6y$ , $\displaystyle c = 10y^2 - 4$
the quadratic formula yields ...
$\displaystyle x = 3y \pm \sqrt{4 - y^2}$
since x is an integer value, the value under the radical must be a positive, perfect square value ... finish it.