TITLE: Show that any two consecutive odd integers are relatively prime
QUESTION [17 upvotes]: I've selected two integers $m=2k+1$ and $n=2k+3$ and I've tried to make a linear combination of the two such that it equals 1, but I'm sort of stuck and am not sure if this is a dead end or not. Any pointers or alternative ideas?

REPLY [3 votes]: A=2n+1=c*a , where c is a common divisor
B=2n-1=c*b
A-B=c(a-b)=2
A>B hence a>b hence a-b>0.
If a-b=1 then c=2, and then A and B (since A=ca, B=cb) are not odd, which is a contradiction. Hence a-b is not 1.
If a-b=2 then c=1.
If a-b>2 then c*(a-b)>2 which is a contradiction.
Hence c=1 is the only option for a common divisor c.