1. Apr 4, 2004

### Nexus[Free-DC]

As a little diversion I thought i'd post this question which I call a forehead-slapper because that's what you'll likely do when you see the answer. You won't need more than high school maths to solve it.

Show that if x and $$x^2+8$$ are primes then so is $$x^3-8$$

2. Apr 5, 2004

x = 3?

3. Apr 5, 2004

### Zurtex

Well $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ meaning that this can only be a prime when $x = 3$.

Edit: I think I have proved the rest of it I'll let others have a go.

Last edited: Apr 5, 2004
4. Apr 5, 2004

There's a rest of it?

5. Apr 5, 2004

### Janitor

Given the Zurtex factorization, it seems pretty obvious that for x=4, 5, 6, ..., x^3-8 has to be composite, so I'm with Cookiemonster.

6. Apr 5, 2004

### Nexus[Free-DC]

Zurtex is exactly right. That was exactly the solution I had in mind. There is another one though I just realized, that doesn't involve having to factor a cubic.

Note that any prime p except 3 is equal to 3k+1 or 3k-1 for some integer k. Then p^2 + 8 = 9k^2 ± 6k + 9, which is divisible by 3. 3^2+8=17, which is prime and also 3^3-8=19 is prime.

7. Apr 5, 2004

### Zurtex

Yes, the rest of it was obviously to prove that $x$ and $x^2 + 8$ could never both be prime. My proof was a little bit more complex that as it is early in the morning and I can't think simple maths yet

8. Apr 5, 2004

### HallsofIvy

The point of cookiemonster's post was that x= 3 is a prime number and that x2[/sup+8= 9+8= 17 is a prime number but x3= 27-8= 21= 3*7 is NOT.

You can't prove your statement: it's not true.

What Zurtex showed with "$x^3 - 8 = (x - 2)(x^2 + 2x + 4)$" was that x3- 8 cannot be prime unless x= 3. That is essentially a converse of your original statement.

9. Apr 5, 2004

### matt grime

27-8 = 19 <filler space>