Question

Determinati toate perechile de numere intregi (x,y) care verifica egalitatea:
$2x^{2} -5xy+ 2y^{2} =27$
Eu ,unul,inca mai caut o restrangere a binomului....

Answer 1

$Nu~era~chiar~atat~de~dificil:~ \\ \\ 2x^2-5xy+2y^2=2x^2-4xy-xy+2y^2=2x(x-2y)-y(x-2y)= \\ \\ =(2x-y)(x-2y). \\ \\ Deci~(2x-y)(x-2y)=27.~Avem~cazurile: \\ \\ \left \{ {{2x-y=-27} \atop {x-2y=-1}} \right.~;~ \left \{ {{2x-y=-9} \atop {x-2y=-3}} \right. ~;~ \left \{ {{2x-y=-3} \atop {x-2y=-9}} \right. ~ ; ~ \left \{ {{2x-y=-1} \atop {x-2y=-27}} \right. ~;$

$\left \{ {{2x-y=27} \atop {x-2y=1}} \right.~;~ \left \{ {{2x-y=9} \atop {x-2y=3}} \right. ~;~ \left \{ {{2x-y=3} \atop {x-2y=9}} \right. ~ ; ~ \left \{ {{2x-y=1} \atop {x-2y=27}} \right.~.$

$Deci...pentru~factorizare,~era~necesar~sa-l~transformi~pe~-5~in \\ \\ -4-1.$