Question

Aflati perechile de numere naturale (x,y) care verifica relatia:
y^3 = x^3 +7x^2+4x+15

Answer 1

y^3=x^3+7x^2+4x+15
y^3 - 27=x^3+7x^2+4x -12
(y-3)(y^2+3y+9)=(x-1)(x+2)(x+6)
deoarece x,y∈N avem sistemele cu 2 ecuatii si 2 necunoscute in N:
1.
y-3=x-1
y^2+3y+9=x^2+8x+12

2.
y-3=x+2
y^2+3y+9=x^2+5x-6

3.
y-3=x+6
y^2+3y+9=x^2+x-2

1. x=y-2
y^2+3y+9=y^2-4y+4+8y-16+12
y=13
x=11

2. x=y-5
y^2+3y+9=x^2+5x-6=y^2-10y+25+5y-25-6
8y=-15 nu se admite

3. x=y-9
y^2+3y+9=x^2+x-2=y^2-18y+81+y-9-2
20y=61 nu se admite

ar mai fi de verificat variantele:

y-3=x^2+x-2
y^2+3y+9=x+6

y-3=x^2+5x-6
y^2+3y+9=x+2

y-3=x^2+8x+12
y^2+3y+9=x-1

e de vazut daca admite solutii in N , who knows?