Can you prove that x^3-8 is prime if x and x^2+8 are also primes?

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The discussion centers on whether x^3 - 8 can be prime if both x and x^2 + 8 are primes. It is established that x^3 - 8 can only be prime when x equals 3, as shown by the factorization x^3 - 8 = (x - 2)(x^2 + 2x + 4). The example of x = 3 demonstrates that while both 3 and 17 (from x^2 + 8) are prime, x^3 - 8 results in 19, which is also prime. However, for other values of x, the expression becomes composite. Ultimately, it is concluded that the initial statement cannot be proven true for all primes.
Nexus[Free-DC]
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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
 
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x = 3?

cookiemonster
 
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:
There's a rest of it?

cookiemonster
 
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.
 
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.
 
Nexus[Free-DC] said:
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 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:
 
Nexus[Free-DC] said:
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 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.
 
27-8 = 19 <filler space>
 

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