Could a mod change the thread titel to something like "Help with exercise questions from Elementary Number Theory"? So I don't have to open a new threads several times. Thanks.
Here's the next one where I'm struggling:
Let p and p^2+8 be prime. Prove that p^3+4 is also prime.
As an initial remark: This is true for p=3 and then I couldn't find any other examples. Which is due to:
(3k+1)^2+8 = 9k^2+6k+9 = 3(3k^2+2k+3)
(3k+2)^2+8 = 9k^2+12k+12 = 3(3k^2+4k+4)
So only if p=3 we can have both p and p^2+8 be prime. A bit weird way to phrase the question in that case, but the proof should still be possible, I guess?
I thought about algebraic factors of p^3+4 or p^2+8, came as far as (p+2)(p2)+12 = p^2+8.
Other than that, no idea. Or is it really just about showing that it only holds for p=3 and since 3^3+4=31 is prime, that's the proof?
Last fiddled with by bur on 20210826 at 06:30
