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Let $a_1,\ b_1,\ c_1$ be natural numbers.

Let $$\gcd (b_1,c_1)=a_2,\ \gcd(a_1,c_1)=b_2,\ \gcd(a_1,b_1)=c_2$$

$${\rm lcm}(b_2,c_2)=a_3,\ {\rm lcm}(a_2,c_2)=b_3,\ {\rm lcm}(a_2,b_2)=c_3$$

Prove that $$\gcd(b_3,c_3)=a_2$$

I have tried this question a lot but I am stuck at this point.

We have to do something with the power of exponents of primes in $a_1,b_1,c_1$ but can't crack the problem and I am unable to assume the prime factorisation of $a_1,b_1,c_1$.

So please help me to do further and solve the problem.

Thanks

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Let the highest exponent of prime $p$ that divides $a_1,b_1,c_1$ respectively be

$$A_1,B_1,C_1$$

WLOG $A_1\ge B_1\ge C_1$

So, the highest exponent of $p$ divides $a_2,b_2,c_2,a_3,b_3,c_3$ will respectively be $$C_1,C_1,B_1,B_2,A_2,A_2$$

The highest exponent of $p$ in gcd$(b_3,c_3)=A_2$ which is the same in $c_3$

Now this will hold true for any prime that divides $a_1b_1c_1$

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  • $\begingroup$ I think it should be lowest power of exponent of p in gcd$(b_3,c_3)$ $\endgroup$ – Aryan 24k Jun 13 at 15:18
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    $\begingroup$ @Aryan24k, No, think of the definition of GCD $\endgroup$ – lab bhattacharjee Jun 13 at 18:37
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Since both $b_3$ and $c_3$ are multiples of $a_2$ by definition, we know that $$a_2 \mid \gcd(b_3,c_3).$$

On the other hand, $b_1$ is (again by definition) a multiple of both $\gcd(b_1,c_1)=a_2$ and $\gcd(a_1,b_1)=c_2$, so therefore $b_1$ is a multiple of lcm$(a_2,c_2) = b_3$. Similarly, $c_1$ is a multiple of both $\gcd(b_1,c_1)=a_2$ and $\gcd(a_1,c_1)=b_2$, so therefore $c_1$ is a multiple of lcm$(a_2,b_2) = c_3$. Since $b_3\mid b_1$ and $c_3\mid c_1$, it follows that $$\gcd(b_3,c_3) \mid \gcd(b_1,c_1) = a_2.$$

These mutual divisibilities show that $\gcd(b_3,c_3) = a_2$ (since $a_2$ is assumed to be positive).

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Notation : gcd $\mapsto (\ ),\ {\rm lcm}\ \mapsto ((\ ))$

(1) If $(a_1,b_1,c_1)=T$, then $$ (c_2/T,b_2/T)=1 \ \ast$$

Proof : If not, then for $l>1$ $$ Tl|c_2=(a_1,b_1) ,\ Tl | b_2=(a_1,c_1) $$

Hence $Tl|(a_1,b_1,c_1)$ which is a contradiction

(2) Hence $$b_3=((a_2,c_2))=a_2\frac{c_2}{T},\ c_3= ((a_2,b_2))=a_2\frac{b_2}{T}$$by $\ast$ Hence $ (b_3,c_3)= a_2 $

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  • $\begingroup$ What does EXE mean? $\endgroup$ – Théophile Jun 13 at 13:05
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$\DeclareMathOperator{\lcm}{lcm}$ It can be shown using the below properties. $$\gcd (\lcm(x,y),z)=\lcm (\gcd (x, z), \gcd (y, z))$$ $$\gcd (\lcm (x,y),x)=x$$ $$\gcd (\gcd (x, y), \gcd (z, t)) = \gcd (\gcd (x, z), \gcd (y, t))$$

They are either easy to prove or can be found here. So, I just use them. $$\gcd (\lcm(a_1,b_1), c_1)=\lcm(\gcd(a_1,c_1),\gcd(b_1,c_1))=\lcm(b_2, a_2)=c_3$$ Similarly, $$\gcd(\lcm(a_1, c_1), b_1)=b_3.$$ Defining $l_1=\lcm(a_1, b_1)$ and $l_2=\lcm(a_1,c_1)$: $$\gcd(b_3,c_3)=\gcd(\gcd(l_1,c_1), \gcd (l_2,b_1))=\gcd(\gcd(l_1,b_1), \gcd (l_2,c_1))=\gcd(b_1,c_1)=a_2$$

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  • $\begingroup$ The first identity is not valid (take $x=y=2$ and $z=4$ for example). $\endgroup$ – Greg Martin Jun 13 at 7:21
  • $\begingroup$ @GregMartin, I think it's now OK. $\endgroup$ – Robert Jun 13 at 7:42

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