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Let $a$ and $b$ be two integers. Is it true that $\operatorname{lcm}(a^2, ab, b^2) = \operatorname{lcm}(a^2, b^2)$ ?

$a, b$ are integers and the result that i tried to use, in this case,is $ \operatorname{lcm}(a^2,b^2)\gcd(a^2,b^2)=a^2 b^2$ but i don't know if it's true $\operatorname{lcm}(a^2,ab,b^2)\gcd(a^2,ab,b^2)=a^3 b^3$

I did some examples apparently it's true but I can't prove it. So could give me a hint?

Bill Dubuque
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lepstein
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    Please show at least some attempts or partial results you found towards the proof. Also, are $a$ and $b$ integers? polynomials? multivariate polynomials? – darij grinberg Nov 25 '19 at 01:29
  • a, b are integers and the result that i tried to use, in this case,is $lcm(a^2,b^2)gcd(a^2,b^2)=a^2b^2$ but i don't know if it's true $lcm(a^2,ab,b^2)gcd(a^2,ab,b^2)=a^3b^3$ – lepstein Nov 25 '19 at 01:39
  • $lcm(x,y)=xy/gcd(x,y)$, let $d=gcd(a,b)$, $lcm(a^2,ab)=a(ab)/d$, $lcm(a(ab)/d,b^2)=b,lcm(a^2/d,b)=$.. It suffices to show that $gcd(a^2/gcd(a,b),b)=gcd(a,b)$ which follows from $gcd(a/d,b)=1$ – Alexey Burdin Nov 25 '19 at 01:39
  • Okay, this is better. But keep in mind that $\operatorname{lcm}\left(u,v,w\right)\gcd\left(u,v,w\right)$ is not the same as $uvw$, so you cannot apply the gcd$\cdot$lcm rule to three numbers. – darij grinberg Nov 25 '19 at 01:42
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    I don't understand @AlexeyBurdin's comment, but here is my hint: First show that $\operatorname{lcm}\left(a^2,b^2\right) = \left(\operatorname{lcm}\left(a,b\right)\right)^2$. (For this, you can use the gcd$\cdot$lcm rule.) Then, you only need to show that $\left(\operatorname{lcm}\left(a,b\right)\right)^2$ is a multiple of $ab$ is well. This has a very direct proof. – darij grinberg Nov 25 '19 at 01:44

1 Answers1

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Note $\,a^2,b^2\mid m\Rightarrow(ab)^2\!\mid m^2\Rightarrow \color{#c00}{ab}\mid m,\,$ by $\,r\in\Bbb Q,\,r^2\in\Bbb Z\Rightarrow r\in \Bbb Z,\,$ by Rational Root Test
Thus $\,\color{#0a0}{a^2,b^2}\mid m\!\iff a^2,\color{#c00}{ab},b^2\mid m^{\phantom{|^|}}\!$
so $\,\ {\rm lcm}(\color{#0a0}{a^2,b^2}) = {\rm lcm}(a^2,ab,b^2)^{\phantom{|^|}}\! $ via LCM Universal Property (or by: above $(\!\!\iff\!\!)$ shows both sides have same set $S$ of common multiples $m,\,$ so the same least common multiple $= \min S$).

Alternatively with $[x,y]:={\rm lcm}(x,y)\,$ & applying LCM distributive law to expand products

$ [a,b][aa,bb] = [aaa,aab,abb,bbb] = [a,b]^3\,$ so $\,[aa,bb] = [a,b]^2\,$ by cancelling $[a,b]\neq 0,\,$

same as in the GCD Freshman's Dream $\,(a^n,b^n) = (a,b)^n = (a^n, a^{n-1}b,\ldots, a b^{n-1}a, b^n)$

Bill Dubuque
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