1973 Canadian MO Problems/Problem 3

Problem

Prove that if $p$ and $p+2$ are prime integers greater than $3$, then $6$ is a factor of $p+1$.

Solution

Prime numbers greater than $3$ are odd. Thus, if $p$ and $p+2$ are prime integers greater than $3$, then they are odd, and $p+1$ is a multiple of $2$. Also, consider each group of three consecutive integers. One has remainder $1$ after division upon $3$, one has remainder $2$, and one has remainder $0$. If $p$ and $p+2$ are prime integers greater than $3$, then they cannot be divisible by $3$. Thus, $p+1$ must leave remainder $0$ after division by three, and so is a multiple of $3$. Finally, if $p+1$ is a multiple of $2$ and $3$, then it is a multiple of $2\times3 = 6$.


See also

1973 Canadian MO (Problems)
Preceded by
Problem 2
1 2 3 4 5 Followed by
Problem 4