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

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