Question

prove that gcd (n-1, n+1)= 1 or 2 for each n>=2 and gcd (2n-1, 2n+1)= 1 for each n>=7.

Answer 1

Prove that $\gcd(n - 1, n + 1) = 1$ or 2 for each $n \geq 2$ and $\gcd(2n - 1, 2n + 1) = 1$ for each $n \geq 7$.

Solution.

1. Assume that $n - 1$ divides by $d$ and $n + 1$ divides by $d$.

Then $(n + 1) - (n - 1)$ divides by $d$ and $(n + 1) - (n - 1) = 2$ (the difference of the two).

The positive integer divisors of 2 is 1 or 2, so $\gcd(n - 1, n + 1) = 1$ or 2 for each $n \geq 2$.

2. Assume that $2n - 1$ divides by $d$ and $2n + 1$ divides by $d$.

Then $(2n + 1) - (2n - 1)$ divides by $d$ and $(2n + 1) - (2n - 1) = 2$.

The positive integer divisors of 2 is 1 or 2, but $n \geq 7$, so $\gcd(2n - 1, 2n + 1) = 1$ for each $n \geq 7$.