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Nexus[Free-DC]
Apr4-04, 11:46 PM
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

cookiemonster
Apr5-04, 12:00 AM
x = 3?

cookiemonster

Zurtex
Apr5-04, 12:26 AM
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.

cookiemonster
Apr5-04, 12:42 AM
There's a rest of it?

cookiemonster

Janitor
Apr5-04, 12:50 AM
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.

Nexus[Free-DC]
Apr5-04, 01:15 AM
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.

Zurtex
Apr5-04, 02:21 AM
']Zurtex is exactly right. That was exactly the solution I had in mind. There is another one though I just realised, that doesn't involve having to factor a cubic.

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

HallsofIvy
Apr5-04, 07:17 AM
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Show that if x and x^2+8 are primes then so is x^3-8

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 x[sup]3= 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.

matt grime
Apr5-04, 07:29 AM
27-8 = 19 <filler space>