Question Number 104100 by bramlex last updated on 19/Jul/20

(1)Find all natural pairs of  integers (x,y) such that x^3 −y^3 =xy+61.  (2)Find gcd(x^4 −x^3 , x^3 −x)

Answered by bemath last updated on 19/Jul/20

(2)x^4 −x^3  = x^3 (x−1)=x^2 .x(x−1)         x^3 −x = x(x^2 −1)=x(x−1)(x+1)  so gcd(x^4 −x^3 , x^3 −x)=  x(x−1) or x^2 −x ★

Answered by john santu last updated on 19/Jul/20

(1) x^3 −y^3  = (x−y)(x^2 +xy+y^2 )  ⇔(x−y)(x^2 +xy+y^2 )= xy+61  notice that x>y . therefore we  have to consider x^2 +xy+y^2 ≤  xy+61 or x^2 +y^2  ≤ 61.  because x>y , we have   61 ≥ x^2 +y^2  ≥2y^2  ⇒ y ∈{1,2,3,4,5}   { ((y=1; x^3 −x−62=0)),((y=2; x^3 −2x−69=0)),((y=3; x^3 −3x−89=0)),((y=4; x^3 −4x−125=0)),((y=5; x^3 −5x−186=0; x=6)) :}  we see the only working value  for x is when x=6 ,y=5 ; so the  only natural pairs of solution  is (x,y) = (6,5) (JS ⊛)

Commented bybramlex last updated on 19/Jul/20

thank you

Answered by floor(10²Eta[1]) last updated on 19/Jul/20

(2)gcd(x^4 −x^3 ,x^3 −x)=gcd(x^3 (x−1),(x−1)x(x+1))  =x(x−1).gcd(x^2 ,x+1)★  d=gcd(x^2 ,x+1)⇒d∣x^2 ∧d∣x+1  ⇒d∣(−1).x^2 +x(x+1)=x  ⇒gcd(x^2 ,x+1)=gcd(x,x+1)=1  gcd(x^4 −x^3 , x^3 −x)=  ★x(x−1)