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I know that both of the following formulas equal to $\text{lcm}(a,b,c)$, but how can we show that

$\dfrac{abc \cdot \gcd(a,b,c)}{\gcd(a,b) \cdot \gcd(a,c) \cdot \gcd(b,c)} = \dfrac{abc }{\gcd(ab,ac,bc)}$ ?

Especially Lime
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John Watts
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  • Can you take for granted that $a,b=\text{lcm}(a,b),\gcd(a,b)$ ? –  Mar 24 '21 at 10:18
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    By the way, you should simplify as $\gcd(a,b,c)\cdot\gcd(ab,ac,bc) = \gcd(a,b) \cdot \gcd(a,c) \cdot \gcd(b,c)$. –  Mar 24 '21 at 10:21
  • @Righter Doesn't the setup in your comment only apply when $\gcd(a,b,c)=1$ ? – coffeemath Mar 24 '21 at 10:31
  • Hint: Take $a=dpqx$, $b=dqry$, $c=drpz$, such that $d, p, q, r, x, y, z$ are all coprime. – Righter Mar 24 '21 at 11:39
  • @coffeemath Right, I made the change – Righter Mar 24 '21 at 11:40
  • @Righter In case one takes $d=1$ in your (new) version, it can be said only that $a,b,c$ are pairwise coprime, as are $y,z,x$ and also the three pairs $(p,y),(q,z),(r,x)$ are each coprime. It can still be that the other possible pairs like $(p,z)$ may not be coprime. – coffeemath Mar 24 '21 at 12:07
  • @coffeemath Hmm.... is there any correct substitution of this type then? – Righter Mar 24 '21 at 13:24
  • @Righter I should have said $p,q,r$ are pairwise coprime in my comment. Other than that, one only keeps the coprime assumptions of my previous comment. [I was definitely in error in saying in previous comment that $a,b,c$ are pairwise coprime.] Note that $d,$ the gcd of the triple $a,b,c$, need not have any coprime relation with any of the other six parameters $p,q,r,x,y,z.$ [This takes some argument as to why this can be done for unique values of the parameters for any positive $a,b,c.$] – coffeemath Mar 24 '21 at 14:21

2 Answers2

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Hint:

WLOG the highest exponents of prime $p$ that divides $a,b,c$ be $A,B,C$

WLOG $A\ge B\ge C\ge0$

So, the highest exponents of prime $p$ that divides the numerator of the LHS will be $$A+B+C+C$$

For the denominator of the left hand side the highest exponent $=B+C+C$

For the denominator of the right hand side the highest exponent $$=B+C$$

The above will hold true for any prime that divides $abc$

lab bhattacharjee
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As here, cancelling $\,abc\,$ then clearing denominators shows it is equivalent to

$$(a,b,c)(ab,bc,ca) = (a,b)(b,c)(c,a)\qquad$$

This equality is easy by gcd "polynomial" arithmetic: expanding the products on both sides (apply gcd associative, commutative, distributive laws), shows both $= (aab,aac,abb,abc,acc,bbc,bcc)$.

Note: as shown here, it is equivalent to gcd distributes over lcm (or vice-versa).

Bill Dubuque
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