Write an expression for the next 3 even numbers that are greater than p

Every even number is 2 away from the last.
0, 2, 4, 6, 8, 10, 12, 14...etc.

If we had an even number p, then the next three even numbers would be
p+2, p+4, and p+6.

(If we had an odd number p, then the next three even numbers would be
p+1, p+3, and p+5. I'm not sure if p is even is implied in the question. Technically the answer would be p - p mod 2 + 2, where p is an interger...that gets into more technical function stuff, though.)

We don't know whether ' P ' itself is odd or even.

-- If ' P ' is even, then the next three [larger] even numbers are

             P+2,  P+4, and  P+6 .

-- If ' P ' is odd, then the net three [larger] even numbers are

            P+1,  P+3, and  P+5.