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
Best Answer
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 $a_1b_1c_1$