This is a long explanation. I tried to be thorough. Let me know if you find any flaws.

Consider the set of perfect squares:

{ 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, ... }

0+1 = 1

1+3 = 4

4+5 = 9

9+7 = 16

In short let B0 represent the first element in the sequence 0,

let B1 represent the second element, and Bn represent the (n-1)th element in the sequence.

n^2 = Bn = B(n-1) + 2n -1 (This can be proven by induction)

Using this rule for the sequence of perfect squares we can now see that a pattern develops with regard to the remainders after division by 6. (I must credit WritersBlock here. I've never heard of the rule about sqares of primes expressed as 6n+1. I never would have looked at the remainders as a point of interest had I not seen the previous post.)

0/6 = 0 r 0-----1/6 = 0 r 1-----4/6 = 0 r 4-----9/6 = 1 r 3-----16/6 = 2 r 4------25/6 = 4 r 1

36/6 = 6 r 0---49/6 = 8 r 1----64/6 = 10 r 4---81/6 = 13 r 3---100/6 = 16 r 4--121/6 = 20 r 1

...

This pattern is guaranteed to continue because we're adding consecutive odd numbers and every six consecutive odd numbers will add a multiple of six. So, whatever the remainder is for any given number, it will be the same for the sixth number after it.

Proof: Let k represent any odd number.

add k and the 5 consecutive odd numbers after k.

k + k+2 + k+4 + k+6 + k+8 + k+10 =

6k + 30 = 6(k+5)

Using the pattern of remainders {0, 1, 4, 3, 4, 1}, only the second and fifth columns could possibly be the sqares of prime numbers.

If remainder is 0 then A^2 is divisible by 6 which means that A is not prime.

-----(if A is prime then the prime factors of A^2 is A*A. 6 can't be a factor)

If remainder is 4 then A^2 is divisible by 2 which means that A is 2 or "not Prime".

-----(if A is prime then the prime factors of A^2 is A*A. 2 can't be a factor unless A=2)

If remainder is 3 then A^2 is divisible by 3 which means that A is 3 or "not Prime".

-----(if A is prime then the prime factors of A^2 is A*A. 3 can't be a factor unless A=3)

Therefore, if prime numbers exist (and we know they do) then their squares can be expressed as some multiple of six plus 1. Or, A^2 = 6x + 1

A^2 = 6x + 1

A^2 + 26 = 6x + 1 + 26

= 6x + 27

= 3(2x + 9) Which is obviously divisible by 3 and therefore not prime.