Why does d divide gcd(a,-b) in a gcd proof?

[issacnewton]

HelloI am studying the problem given in the attachement. In the solution given, it says "Similarly \( d|\gcd(a,-b) \) ". I could not understand why this is so.thanks

 

[caffeinemachine]

HelloI am studying the problem given in the attachement. In the solution given, it says "Similarly \( d|\gcd(a,-b) \) ". I could not understand why this is so.thanks

$d|\gcd(a,-b)$ follows from the fact that $d=\gcd(a,b)$.

 

[issacnewton]

thanks...I should have realized that...