On the divisibility of the common difference of three Prime Numbers $> 3$ in an arithmetic progression

Question

Given three primes greater than three in in an arithmetic progression, how does one show that the common difference is always divisible by $3$?

Answer 1

Take that $$p_2=p_1+d$$ $$p_3=p_1+2d$$ If $$d\equiv 1\mod 3$$ then $$p_1+2d\equiv 0 \mod 3$$ (since $p>3$) If $$d\equiv 2\mod 3$$ then $$p_1+2d\equiv 0 \mod 3$$