Multiply any two of these numbers together and add 1 and you'll always get a perfect square:

1+ 1 × 3 = 4
1 + 1 × 8 = 9
1 + 3 × 8 = 25

a. Add to the above set a 4th integer, thus enabling 3 additional equalities.
b. Can you show why no 5th member maintaining the above feature exists?