Let . Are and relatively prime? How about and ?

My answer is yes, both and . But is there a way to show these are true?

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- Sep 12th 2009, 05:28 PMdori1123coprime
Let . Are and relatively prime? How about and ?

My answer is yes, both and . But is there a way to show these are true? - Sep 12th 2009, 09:46 PMsimplependulum
- Sep 13th 2009, 04:19 AMHallsofIvy
To show that (n, n+1)= 1, suppose it is not. Suppose (n, n+1)= m> 1. Then m is a common factor of n and n+1 and we must have n= mk and n+1= mp for integers k and p. Then n+ 1= mk+ 1= mp so 1= mp-mk= m(p- k). But, since the only factor of 1 is 1 itself, we must have both m and p-k equal to 1 and m= 1 contradicts m> 1.

- Sep 13th 2009, 04:45 AMPaulRS
If divides and the it divides for any - it is easy to see why, just write for some k in , and similary for y, then factor out the 's in the sum!-

For for example, if divides and then it divides

Similarly if a number divides both and then it divides their difference.