To do this proof, we need to show that for any prime number 'a', ((a*a) + 26)/3 is an integer. We do this by the modulus of each number.

26 is congruent to 2 mod 3. Which means if 26 is divided by 3, the remainder is 2.

It then suffices to show that a*a is congruent to 1 mod 3. Which means if a*a is divided by 3, the remainder is 1.

...... blah blah blah

-Mathematics Graduate Student

University of Missouri

Here is another way to look at this proof:

The proof requires bases. I will be working in base three (number set 0, 1, 2)

Here is a base 3 multiplication table (leaving out 0) 1*1=1, 1*2=2, 2*2=11.

So, any number divisible by 3 will end in 0 (base 3).

A prime number is not divisible by 3, so it will end in either 1 or 2.

When I square that prime number, it will end in 1, always. (see multiplication table).

So, I just need to add 2, or 3n+2 to the square to make it divisible by 3.

I can use 3n + 2 in the place of "26" to make this puzzle work. So that the addend is not obvious, n is even.

Therefore, the problem works for a*a+2 as well as a*a+38