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Old 2021-08-26, 06:30   #10
bur's Avatar
Aug 2020

23·3·52 Posts

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)(p-2)+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 2021-08-26 at 06:30
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