Find $d=\gcd(a,b)$ if $a=p_{1}^{k_{1}}p_{2}^{k_{2}} \cdots p_{s}^{k_{s}}$ and $b=p_{1}^{l_{1}}p_{2}^{l_{2}} \cdots p_{s}^{l_{s}}$

Question

If $a=p_{1}^{k_{1}}p_{2}^{k_{2}} \cdots p_{s}^{k_{s}}$ and $b=p_{1}^{l_{1}}p_{2}^{l_{2}} \cdots p_{s}^{l_{s}}$ where $p_1,p_2,...,p_s$ are distinct primes, tell how to use this information to find $d=\gcd(a,b)$. Then argue that if $a$ and $b$ are both squares, so is $d$.

I'm not sure how to attack this problem. I started off by letting $a=2*3*5..$ and $b=2*3*5..$ but I obviously don't know what to do with the $k_s$ and the $l_s$. For example, what are they?

I think if I knew what $a$ and $b$ were I could easily prove $d$ is a square.

Answer 1

Hint $ $ If $\ p\nmid a,b\ $ then $\ (p^m a, p^n b) = p^{\min(m,n)} (a,b).\, $ Recurse similarly on $\:\!(a,b):=\gcd(a,b).$

Proof $ $ wlog $\,m = \min(m,n)\,$ so $\,(p^ma,p^nb) = p^m(\color{#c00}{a,p^{n-m}}b) = p^m(a,b)\,$ by Euclid's Lemma,

because, $ $ by $ $ hypothesis, $\,\ (a,p)=1,\ $ therefore, $\,\ (\color{#c00}{a,p^{n-m}})=1,\ $ again, $ $ by Euclid's Lemma.

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Alternatively, employ the universal definition: $\,\ \gcd(\color{#0A0}{a,b}) = \color{#C00}d \ \ $ iff $\ \ c\mid \color{#0A0}{a,b}\!\iff\! c\mid \color{#C00}d$

Notice, by unique factorization $\ \prod P_i^{J_i}\mid \prod P_i^{K_i}\!\!\iff\! J_i \le K_i.\:$ Therefore

$ \prod P_i^{J_i}\,\big|\, \color{#0A0}{\prod P_i^{K_i},\: \prod P_i^{L_i}} \!\!\iff\!\! J_i \le K_i,L_i \!\!$ $\iff\!\! J_i\le {\small \min}(K_i,L_i)\!\!\iff\!\! \prod P_i^{J_i}\,\big|\,\color{#C00}{\prod P_i^{{\small \min}(K_i,L_i)}}$

Therefore, by the stated universal definition: $\, \gcd\left(\color{#0A0}{\prod P_i^{K_i},\: \prod P_i^{L_i}}\right) =\, \color{#C00}{\prod P_i^{{\small \min}(K_i,L_i)}}$

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Finally $\, \prod P_i^{J_i}$ square $\,\Rightarrow$ $ \prod P_i^{J_i} = \left(\prod P_i^{K_i}\right)^2\! = \prod P_i^{2 K_i} $ $\Rightarrow$ $\:J_i$ even $ $ (and conversely), by unique factorization. So $\,J_i,K_i\:$ even $\,\Rightarrow$ $\,{\small\min}(J_i,K_i)\,$ even $\,\Rightarrow$ $\:\prod P_i^{{\small\min}(K_i,L_i)}$ square.

Or we can prove $\,\gcd(a^2,b^2) = \gcd(a,b)^2\,$ by easy gcd arithmetic.

Answer 2

Let $e_i=\min(k_i,l_i)$. (This is the smaller of $k_i$ and $l_i$, where if $k_i=l_i$ we have $e_i=k_i=l_i$.) Then $$d=p_1^{e_1}p_2^{e_2}\cdots p_s^{e_s}.$$

If $a$ and $b$ are squares, then the $k_i$ and $l_i$ are all even. So the $e_i$ are all even, and therefore $d$ is a perfect square.

What the symbols mean: The expression $a=p_1^{k_1}p_2^{k_2}\cdots p_s^{k_s}$ is the prime power factorization of $a$. The $p_i$ are distinct primes. It is an important basic theorem of number theory that every integer $\gt 1$ has a prime power factorization, and that, apart from the order of the terms, the factorization is unique.

Remark: There is a similar expression for the lcm of $a$ and $b$. Let $f_i=\max(k_i,l_i)$. Then the lcm is equal to
$$p_1^{f_1}p_2^{f_2}\cdots p_s^{f_s}.$$