Question on finding the GCD of 4 numbers

[noxx88]

Homework Statement

a_1, a_2, b_1, b_2 are all positive integers greater than one.

Given that (a_2-1)/(a_1*a_2-1)=(b_2-1)/(b_1*b_2)

Show that GCD(a_2-1,a_1*a_2-1,b_2-1,b_1*b_2)=1

Homework Equations

(a_2-1)/(a_1*a_2-1)=(b_2-1)/(b_1*b_2)

The Attempt at a Solution

If I let the GCD=d, then d|a_2-1 and d|a_1*a_2-1, so this implies that d|a_1*a_2-a_2+1-1

-> d|a_1*a_2-a_2
-> d|a_2(a_1-1)
-> d|a_1-1 since d does not divide a_2

Also, since d|b_1*b_2, and d does not divide b_2 (due to the fact that d|b_2-1), d|b_1.

 

[kurt.physics]

Now, since d|a_1-1 and d|b_1, -> d|(a_1-1)(b_1-1)-> d|(a_1*b_1-a_1-b_1+1)-1-> d|(a_1*b_1-a_1-b_1)We also have that (a_2-1)/(a_1*a_2-1)=(b_2-1)/(b_1*b_2).-> (a_2-1)(b_1*b_2)=(b_2-1)(a_1*a_2-1)-> a_2b_1b_2-a_2b_2=b_2a_1a_2-b_2a_2-b_1a_1a_2+b_1a_2-> a_2b_2-a_2b_2=b_2a_2-b_2a_2-b_1a_1a_2+b_1a_2-> 0=b_2a_2-b_2a_2-b_1a_1a_2+b_1a_2-> 0=a_2(b_1a_1-b_1)Since a_2 is not equal to zero, this implies that b_1a_1-b_1=0-> b_1(a_1-1)=0Since b_1 is not equal to zero, this implies that a_1-1=0-> a_1=1Since a_1=1, this implies that d|a_1-1=0. This implies that the GCD=d=0, which is a contradiction. Therefore, GCD=1