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

Let :
         au+bv=1
         au=1-bv ............................(i)

Squaring both sides of equation (i) :
we get,

      (au)^2=(1-bv)^2 ........................(ii)
      a^2(u^2) + b(2v-bv^2)= 1
  so,
        gcd(a^2,b)=1
=>  (a^2)K1 + bK2 =1
=>  (bK2)^2=(1-a^2K1)^2
=>  b^2{(K2)^2}=1+a^4(K1)^2 -2(a^2)(K1)
=>  b^2(K2)^2 +a^2(2K1-(a^2)(K1)^2)=1

=>  gcd(a^2,b^2)=1  {Proved}


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