Question

Let a,b∈Z. Then show that gcd(a,b)=1⟹gcd(a+b,a−b)=1 or 2. How could I do this?

Answer 1

\(a\) and \(b\) cannot be both even because \[\gcd(a,b)=1\]
let \[d=\gcd(a+b,a-b)\]
\(d\) divides \((a+b)\pm (a-b)\), that is, \(d|2a\) and \(d|2b\)

if exactly one of a and b is odd, then (a+b) and (a-b) are both odd so d is odd. therefore \(\gcd(d,2)=1\) and d divides both a and b

this means that d divides as and bt, and d divides (as + bt)

because gcd(a,b) =1 means there exists s and t such that as + bt = 1

then d divides (as + bt) = 1. Thus, d = 1.

Answer 2

now, if both a and b are odd, a+b and a-b are both even, hence d is even, say d=2e

2e=d divides 2a implies e divides a
2e=d divides 2b implies e divides b

so e divides a, b, and as + bt = 1 (again, gcd(a,b) = 1)

therefore e = 1 and d = 2e = 2

Answer 3

thanks a lot

Answer 4

prove lcm(na,nb)=n lcm(a,b)

Answer 5

hint: \[\Large \text{lcm} (a,b)=\frac{ab}{\gcd(a,b)}\]

Answer 6

i know this hint, and i already prove gcd(na,nb)=n gcd(a,b). but i don't know how can i start to prove lcm(na,nb)=n lcm(a,b)