My task is to prove that gcd(n, n+1)=1 for all n>0. It is obvious that 1 is a common divisor of both n and n+1 since $$ 1|n → 1x=n $$ if x=n, and $$ 1|n+1 → 1y=n+1 $$ if y=n+1.
To prove that 1 is the greatest common divisor, I did as follows:
From the integers n and n+1 the other must be an even integer, and the other an odd integer. Since one of them must be odd, the gcd of the two can't be an even number, because the odd one doesn't have any even divisors. If the gcd of the two were an odd integer greater than 1, for example 3, then $$ 3|n → 3x=n → n=3x $$ and $$ 3|n+1 → 3y=n+1 → n=3y-1, $$ and there are no integers x and y such that both of these equations would hold at the same time. This happens with every odd integer larger than 1.
Is the part about the odd integers greater than 1 rigorous enough for a proper proof? I don't think it is; how would you prove this more rigorously? If anyone has any other improvement ideas of any kind, they're more than welcome!