Show that for $a,b,n>1$, $$a^n-b^n\not\mid a^n+b^n $$
By the way of contradiction, Assume $a^n-b^n\mid a^n+b^n $, Hence $$(a^n-b^n, a^n+b^n)= a^n-b^n \\ =(2a^n, a^n+b^n)=(2b^n, a^n+b^n)$$ Well, what’s next?
Show that for $a,b,n>1$, $$a^n-b^n\not\mid a^n+b^n $$
By the way of contradiction, Assume $a^n-b^n\mid a^n+b^n $, Hence $$(a^n-b^n, a^n+b^n)= a^n-b^n \\ =(2a^n, a^n+b^n)=(2b^n, a^n+b^n)$$ Well, what’s next?
Copyright © 2021 JogjaFile Inc.