# Please help to solve this number theory question based on gcd and lcm.

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

## 4 Answers

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 $$a_1b_1c_1$$

• I think it should be lowest power of exponent of p in gcd$(b_3,c_3)$ – Aryan 24k Jun 13 at 15:18
• @Aryan24k, No, think of the definition of GCD – lab bhattacharjee Jun 13 at 18:37

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).

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 $$(b_3,c_3)= a_2$$

• What does EXE mean? – Théophile Jun 13 at 13:05

$$\DeclareMathOperator{\lcm}{lcm}$$ 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 $$l_2=\lcm(a_1,c_1)$$: $$\gcd(b_3,c_3)=\gcd(\gcd(l_1,c_1), \gcd (l_2,b_1))=\gcd(\gcd(l_1,b_1), \gcd (l_2,c_1))=\gcd(b_1,c_1)=a_2$$

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