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