What is the greatest common divisor of positive integers m

Hi,

(1) m-n and n are co-prime
what ever common factors m and n have , m-n, m+n, m and n will also have same factors..

Reason - say m and n have x in common.... so \(m = xa\) and \(n = xb.\)...
so \(m-n = xa-xb = x(a-b)..................m+n = x(a+b)................m=xa\) .............n=xb......mn = xa*xb...........
BUT not\(\frac{m}{n}\)
thus all five of them have same common factors..

so IF m-n and n are co-primes or both do not have any factor in common, so m and n will also be co-prime..
Suff

(2) m and n are consecutive integers
consecutive integers do not have any factor other than 1 in common.....
Suff

D

Here's another approach to solving statement one, though I am not totally sure if this is the correct approach.

We know that co-primes do not share any common factors and from a pattern point of view if two integers are separated by \(1\) unit (consecutive integers) then we can be sure that the two integers are co-primes E.g. \(k\) and \(k + 1\) were \(k\) is a positive integer

Now \(n\) and \(n + 1\) will be co-primes for sure and we are told that \(m - n\) and \(n\) are co-primes, so we could write \(n + 1 = m - n\) which upon simplification yields \(m = 2n + 1\)

When,
\(n = 1\), \(m = 3\) and GCD \(= 1\)
\(n = 2\), \(m = 5\) and GCD \(= 1\)
\(n = 3\), \(m = 7\) and GCD \(= 1\)
\(n = 4\), \(m = 9\) and GCD \(= 1\)
\(n = 15\), \(m = 31\) and GCD \(= 1\)
\(n = 20\), \(m = 41\) and GCD \(= 1\)
and so on...

Again I am not sure if this is the right approach. I just shared what's running in my mind!