Hello I have got the answer for question ie. if $\gcd(a,b)=1$ prove $(a+b,a-b)= 1$ or $2$ But I want to find answer for, if $\gcd(a,b)=1$ prove $(a^2,b^2)=1$ or $2$
Asked
Active
Viewed 272 times
0
-
4if $(a,b)=1$, then $(a^2,b^2)=1$ – Thorgott Oct 10 '17 at 11:10
-
1if GCD(a,b) = 1 the two number are not both even. How is it possible that $a^2$ and $b^2$ are both even? – Raffaele Oct 10 '17 at 11:58
1 Answers
3
If $ua+vb=1$, then $$1^2=(ua+vb)^2=u^2a^2+2uvab+v^2b^2 $$ and $$ub\cdot a^2+va\cdot b^2=ab,$$ hence $$(u^2-2u^2vb)a^2+(v^2-2uv^2a)b^2=1 $$
Hagen von Eitzen
- 374,180
-
Sir this wrong solve by considering d ie assuming a gcd d such that dla – Sanket Oct 10 '17 at 13:00
-
@Sanket The assumption used here is that there exist integers $u,v$ with $ua+vb=1$, which is equivalent to the OP assumption that $\gcd(a,b)=1$. – Hagen von Eitzen Oct 11 '17 at 09:12