Question

Tìm `x;y in ZZ` biết `xy-x^2-y=0`

Answer 1

 $xy-x^2-y=0$

$⇔(xy-y)-x^2+1-1=0$

$⇔y.(x-1)-(x^2-1)=1$

$⇔y.(x-1)-(x-1).(x+1)=1$

$⇔(x-1).(y-x-1)=1$

$⇔$\(\left[ \begin{array}{l}(x-1).(y-x-1)=1.1\\(x-1).(y-x-1)=(-1).(-1)\end{array} \right.\) 

vì $x;y $ nguyên :

$⇒$\(\left[ \begin{array}{l}\begin{cases}x-1=1\\y-x-1=1\end{cases}\\\begin{cases}x-1=-1\\y-x-1=-1\end{cases}\end{array} \right.\) 

$⇒$\(\left[ \begin{array}{l}\begin{cases}x=2\\y-2=2\end{cases}\\\begin{cases}x=0\\y-0=0\end{cases}\end{array} \right.\) 

$⇒$\(\left[ \begin{array}{l}\begin{cases}x=2(T/M)\\y=4(T/M)\end{cases}\\\begin{cases}x=0(T/M)\\y=0(T/M)\end{cases}\end{array} \right.\)