Question

I am trying to prove that if gcd(a,b) = 1 and gcd(c,d) = 1 then gcd(ac+bd, bc-ad) = 1

Answer 1

do you have any idea?

Answer 2

well I've gotten that if we assume the opposite and let gcd = p, then if p | ac+bd and if p | bc-ad then p | their sum so p | c(a+b) + d(b-a) and p | a(c-d) + b(c+d)
and then maybe there's some way to show a contradiction with gcd(a,b) = 1 and gcd(c,d) = 1. I don't know if this direction is the best but I am stuck

Answer 3

did you think about prime? do you know about Euler (phi) function?

Answer 4

yeah phi(a number) = how many numbers between 1 and that number are relatively prime to it. will that help here?

Answer 5

I think so, try to solve on that way. use phi function for co-prime numbers

Answer 6

could you explain what you mean by applying phi function to this problem/what is this theorem you mention?

Answer 7

sorry,

Answer 8

ok thanks, I'm really stuck on this

Answer 9

Maybe you can use a clever arguement involving complex numbers. The only reason I think complex numbers might help is because:\[(a+bi)(c-di)=(ac+bd)+(bc-ad)i\]