Here is the question:(adsbygoogle = window.adsbygoogle || []).push({});

-------------

Show that a perfect square is never of the form 3k+2 for any k. Conclude that if p is prime and [itex]p \geq 5[/itex], then [itex]p^2 + 2[/itex] is always composite.

-------------

We are at the very start of the book, so there is not very much available to use. We have defined a|b, primes, floor and ceiling (which I don't think will play a role here), and have the division theorem.

For the first part I think it is going to be proved by contradiction, so I have been trying to get some kind of parity going with both sides. I tried using [itex]n^2 - 1 = 3k+1[/itex] and then breaking into evens and odds, but that did not get anywhere. I was also trying to get a contradiction with [itex]3k+2 \mid n[/itex], but again was able to get nothing. Any ideas?

For the second part I was using the result of the first. Since a perfect square is not of the form 3k+2 it must be of the form 3k+1 of 3k, for the former it is easy to see that [itex]p^2+2[/itex] is composite. But for the latter I am stuck again. Any ideas here?

Thanks!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Number Thry: Divisibilty, perfect square.

**Physics Forums | Science Articles, Homework Help, Discussion**