Prove that $(a,b)^2=(a^2,b^2,ab)$

Question

I am trying to prove that $(a,b)^2=(a^2,b^2,ab)$ and was told that this follows from some very basic $\gcd$ laws. What am I not seeing?

Answer 1

Hint: for nonnegative integers $m,n$, can you see that $$ 2\min\{m,n\}=\min\{2m,2n,m+n\}? $$ (Just consider $m\leq n$; $n\leq m$ follows by symmetry.) Now apply prime factorizations.

Answer 2

Let $(a,b) = c$, then $a = cp, b = cq$ with $(p,q) = 1$. Thus: $(a^2,b^2,ab) = (c^2p^2, c^2q^2,c^2pq) = c^2(p^2,q^2,pq)$. We need to prove: $(p^2,q^2,pq) = 1$. We have: $(p^2,q^2,pq) = d \Rightarrow d \mid p^2, d\mid q^2 \Rightarrow d \mid (p^2,q^2) = 1$ since $(p,q) = 1 \Rightarrow d = 1$. Thus the problem is solved.

Note: They mentioned to you that use the basic facts about GCD. I think they wanted to talk about the property that: $ (p,q) = 1 \iff (p^2,q^2) = 1$.

Answer 3

Let $$ c=\gcd(a,b),\quad a=a_1c,\quad b=b_1c, $$ with $\gcd(a_1,b_1)=1$. Then $$ \gcd(a^2,ab,b^2)=\gcd(a_1^2c^2,b_1^2c^2,a_1b_1c^2)=c^2\gcd(a_1^2,b_1^2,a_1b_1). $$ Since $\gcd(a_1,b_1)=1$, it follows that $\gcd(a_1^2,b_1^2)=1$, and therefore

$$ \gcd(a^2,ab,b^2)=c^2=[\gcd(a,b)]^2. $$

Answer 4

Let $d=(a,b)$. $\;\;$Then $d\big| a$ and $d\big| b\implies d^2\big|a^2, d^2\big|b^2,$ and $d^2\big|ab$; $\;$so $d^2\le(a^2,ab,b^2)$.

Since $d=ma+nb$ for some $m,n\in\mathbb{Z}$, $\;\;d^2=m^2a^2+2mnab+n^2b^2$ $\;$so $(a^2,ab,b^2)\big|d^2$.

Therefore $d^2=(a,b)^2=(a^2,ab,b^2)$.