Show that any two consecutive odd integers are relatively prime
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?
Solution 1:
If $d|2k+1$ and $d|2k+3$, then $d|2k+3-(2k+1)=2$. Thus $d=1$ or $d=2$. But $d\neq 2$, since $d$ is a divisor of an odd number. Then $d=1$. That is, the only common divisor of $2k+1$ and $2k+3$ is 1.
Solution 2:
If you really want to express $1$ as a linear combination of $2k+1$ and $2k+3$, note that $$(2k+3)(k+1)-(2k+1)(k+2)=1.$$