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

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Discussion Overview

The discussion revolves around the question of whether the expression x^3 - 8 can be proven to be prime if both x and x^2 + 8 are primes. Participants explore various mathematical approaches and reasoning related to this assertion.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that if x and x^2 + 8 are primes, then x^3 - 8 should also be prime, framing it as a simple problem solvable with high school mathematics.
  • Another participant proposes x = 3 as a potential solution, implying it might be the only case where x^3 - 8 is prime.
  • A different participant provides a factorization of x^3 - 8, indicating that it can only be prime when x = 3, and expresses confidence in their reasoning.
  • Some participants argue that for values of x greater than 3, the expression x^3 - 8 must be composite, supporting the earlier factorization argument.
  • One participant introduces a different approach involving properties of primes and modular arithmetic, suggesting that for primes other than 3, the expression x^2 + 8 would be divisible by 3, leading to a contradiction.
  • Another participant challenges the original assertion, stating that the claim cannot be proven true and pointing out that x^3 - 8 is not prime for x = 3, as it results in a composite number.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the original claim. While some support the idea that x^3 - 8 can be prime under certain conditions, others argue against it, indicating that the statement is not universally true.

Contextual Notes

There are unresolved mathematical steps and assumptions regarding the conditions under which x and x^2 + 8 are primes, as well as the implications for x^3 - 8. The discussion includes various interpretations of the factorization and properties of prime numbers.

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 [tex]x^2+8[/tex] are primes then so is [tex]x^3-8[/tex]
 
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x = 3?

cookiemonster
 
Well [itex]x^3 - 8 = (x - 2)(x^2 + 2x + 4)[/itex] meaning that this can only be a prime when [itex]x = 3[/itex].

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 [itex]x[/itex] and [itex]x^2 + 8[/itex] 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 [tex]x^2+8[/tex] are primes then so is [tex]x^3-8[/tex]

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 "[itex]x^3 - 8 = (x - 2)(x^2 + 2x + 4)[/itex]" 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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