*First of all we need an agreement that all integers are of the following formats: 6n, 6N + 1, 6N + 2, 6N + 3, 6N + 4, 6N + 5, where N is an integer. If we go to 6N + 6 this can be written 6(N+1) which is just another version of 6N. I claim 6N + 1 and 6N + 5 are the only possible primes (having exactly 2 divisors, 1 and itself). 6N is obviously divisible by 6. 6N + 2 is the sum of 2 even numbers thus even and divisible by 2, as is 6N + 4. 6N + 3 is the sum of two numbers divisible by 3 thus divisible by three itself. Therefore we only need consider 6N + 1 and 6N + 5.*

**Format Method.**Now onto our problem:

Square them:

36N^2 + 12N + 1 or 36N^2 + 60N + 25

Add 17

36N^2 + 12N + 18 or 36N^2 + 60N + 42

With a minimum of algebra we get:

12(3N^2 + N + 1) + 6 or 12(3N^2 + 5N + 3) + 6

Dividing by 12 we can see that both these numbers leave a remainder of 6.

**. Let P be our prime number bigger than 3**

*Consecutive Integer Method*Square it: P^2

Add 17: P^2 + 17

Rearranging algebraically we get: (P^2 - 1) + 18

Now factor. (P - 1)(P + 1) + 18

Here's where consecutive integers comes in.

P - 1, P, P + 1 are consecutive integers (for example if P is 11 the we get 10, 11, 12)

Therefore either P + 1 or P - 1 is divisible by 3

And because P itself is odd (all prime numbers bigger than 3 are odd) then both P - 1 and P + 1 are even and (P - 1)(P + 1) is divisible by 4. Thus (P - 1)(P + 1) is divisible by 12.

Sooooooo (P - 1)(P +1) + 18 when divided by 12 leaves a remainder of 6 (we need only consider what happen when we divide 18 by 12).

Its better to learn the different types of numbers by making them in an n series.For example odd numbers are written as 2n+1,even numbers are written as 2n.These numbers are used to solve some lengthy questions.

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