if p is prime and (a,b)=p, then (a^2,b^2)=?
Dear Deepu,
$\displaystyle (a,b)=p$
Take $\displaystyle (a^2,b^2)=k$
Now since, $\displaystyle k\mid{a^2}$ and $\displaystyle k\mid{b^2}$
Therefore, $\displaystyle k\mid{a}$ and $\displaystyle k\mid{b}$
Then since the gcd of "a" and "b" is "p" ; $\displaystyle k\leq{p}$ ---------------A
Also since $\displaystyle p=(a,b)\Rightarrow{p\mid{a}}$ and $\displaystyle p\mid{b}\Rightarrow{p\mid{a^2}}$ and $\displaystyle p\mid{b^2}$
Then since the gcd of $\displaystyle a^2$ and $\displaystyle b^2$ is "k" ; $\displaystyle p\leq{k}$ ----------------B
From A and B ; p=k
Therefore $\displaystyle (a^2,b^2)=p$
Hope this helps.
This implication is incorrect.
Consider, for example, $\displaystyle a=6,b=9 \Rightarrow a^2=36, ~b^2=81 \Rightarrow 9 | 36, ~ 9 | 81$ but 9 obviously does not divide 6.
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Let $\displaystyle (a,b)=p \Rightarrow p|a, ~ p|b \Rightarrow p^2|a^2,~p^2|b^2$
So, we now know that $\displaystyle p^2$ is a common divisor of $\displaystyle a^2,~b^2$. I'll leave it to you to show that it is the greatest common divisor.
Suppose $\displaystyle a=p_1^{\alpha_1} p_2^{\alpha_2} \cdot\cdot\cdot p_n^{\alpha_n} $ and
$\displaystyle b=p_1^{\beta_1} p_2^{\beta_2} \cdot\cdot\cdot p_n^{\beta_n} $ where $\displaystyle \alpha_i,\beta_j \geq 0 $.
Set $\displaystyle k=(a,b) $. Then $\displaystyle k=p_1^{\min{\alpha_1,\beta_1}} p_2^{\min{\alpha_2,\beta_2}} \cdot\cdot\cdot p_n^{\min{\alpha_n,\beta_n}} $.
Now $\displaystyle a^2=p_1^{2\alpha_1} p_2^{2\alpha_2} \cdot\cdot\cdot p_n^{2\alpha_n} $ and $\displaystyle b^2=p_1^{2\beta_1} p_2^{2\beta_2} \cdot\cdot\cdot p_n^{2\beta_n} $.
So $\displaystyle (a^2,b^2) = p_1^{\min{2\alpha_1,2\beta_1}} p_2^{\min{2\alpha_2,2\beta_2}} \cdot\cdot\cdot p_n^{\min{2\alpha_n,2\beta_n}} = p_1^{2\min{\alpha_1,\beta_1}} p_2^{2\min{\alpha_2,\beta_2}} \cdot\cdot\cdot p_n^{2\min{\alpha_n,\beta_n}} = k^2 $.
Hence $\displaystyle (a^2,b^2) = (a,b)^2 $.