If $\gcd(a,b)=1$ prove that $(a^2,b^2)=1$ or $2$.

Question

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 $2$

Answer 1

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 $$(u^2-2u^2vb)a^2+(v^2-2uv^2a)b^2=1 $$