x^2007
xy-x=y-4
x(y-1)=y-4
x=(y-4)/(y-1)=(y-1-3)/(y-1)= (y-1)/(y-1) -3/(y-1)=1-3/(y-1)
In R XR ecuatia are o infinitate solutii in (R\{1}) X (R\[1}) ale caror perechi (x;y) se afla pe graficul functiei f(x) =(x-4)/(x-1)
pt. inceput vom rezolva in N xN
y-1=3 , y=4 si x=1-1=0
deci x=0; y=4
intr-adevar 0=0+4-4
pt y-1=1, y=2, x=1-3=-2∉N
dac am rezova in Z , avem si valorile
pt y-1=1, y=2, x=1-3=-2
y-1=-1 y=0, x=1+3=4 pereche de valori care e buna si pt N
y-1=-3 y=-2 x=1+1=2
Deci in NXN, (x;y) ∈{( 4;0) si ( 0;4)}
in Z X Z (x;y) ∈{(4;0) (0;4) (-2;2) (2;-2)}
ceea ce era de asteptat, expresia fiind simetrica in x si y
