Prove that is p is a prime number greater than 3, then p^2 leaves a reamainder of 1 when it is divided by 12.

Is there a form I could use as a prime number to show this works for all odd numbers?

Printable View

- Oct 20th 2010, 10:18 AMmatgrlProof (Problem Solving)
Prove that is p is a prime number greater than 3, then p^2 leaves a reamainder of 1 when it is divided by 12.

Is there a form I could use as a prime number to show this works for all odd numbers? - Oct 20th 2010, 07:05 PMemakarov
I did not understand the question, "Is there a form I could use as a prime number". Anyway, this does not work for every odd p: take p = 9.

Here is one way to reason. Suppose p = 2k + 1; then p^2 = 4k^2 + 4k + 1. We would be done if we proved that k^2 + k is divisible by 3.

Now, k^2 + k = k(k + 1). If k is divisible by 3 or gives the remainder 2 when divided by 3, then k(k + 1) is divisible by 3. This leaves the case when k gives the remainder 1 when divided by 3. Show that in this case, 2k + 1 is not prime.