Question

a) x | 3y si y | (2x+3) b) x | (y+1) si y | x dau coroana pana maine

Answer 1

x | 3y => 3y=x*k ( k numar natural ) => y=x*k/3
y | (2x+3) => 2x+3=y*m ( m numar natural )

2x+3=m*x*k/3 | *3
6x+9=m*x*k
x*m*k-6x=9
x(m*k-6)=9

x apartine multimii divizorilor lui 9

x=1
y | (2*1+3),
y | 5

y=1 si y = 5

Solutii : (x,y) = (1,1) si (1,5)

x=3
y | (2*3+3)
y | 9

y=1 , y=3, y=9

Solutii (x,y)=(3,1) si (3,3) si (3,9)

x=9
y | (2*9+3)
y | 21

y=1 si y=3 si y=7 si y=21

Solutii (x,y)=(9,1) si (9,3) si (9,7) si (9,21 )

Deci are 9 solutii