
7 is the largest prime followed by a cubic number - bwasti
https://jott.live/texdown/cubic_prime
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montenegrohugo
Math is funny. There's an infinite amount of primes, and even though we know a
lot of them there's still more left (read:infinite). But at the same time,
with high-school level algebra we can prove that the largest prime followed by
a cubic number is 7.

The proof is really simple, but the fact is quite unintuitive (at least for
me).

~~~
tetha
That is number theory. Number theory is at the same time extremely profound
and very very blunt.

Every prime is double of an integer + 1, except for 2. This follows because
primes are integers, integers are either 2 times n or 2 times n+1 and numbers
representable as 2 times n have 2 as a prime factor, if n > 1\. Hence, except
for 2.

Overall, number theory is one example to me how very, very simple concepts are
actually hard - because you're not trained to look for them. The proof is just
... obvious once pointed out. But why would you look for it with such simple
manners?

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Tomminn
Huh, I never realized this was this simple. Fleshing out the general case the
post mentions.

Suppose p=x^n-1 is prime.

Note that:

    
    
      x^n - 1  = (x^n + x^(n-1) +... + x^2 + x) 
                -(      x^(n-1) +... + x^2 + x +1)
    
      =>x^n-1  = (x^(n-1) + x^(n-2) + ... + x + 1)(x-1)
    

So if p=x^n-1 is prime, one of the factors must be equal to 1. If the first
factor is 1, this implies x=0 and so the second factor is -1. Therefore the
second factor must be 1, and x=2.

Therefore if x^n-1 is prime, x must be 2. Damn that's cool. At first I thought
any number 2^n-1 would be prime, but then I realized that obviously the left
hand factor can have subfactors. Which is to say I remembered 15 exists.

~~~
jpegslayer
What about for n=4? I think the issue is that you're supposing that 2^n-1 is
already prime.

Primes of this form are called Mersenne primes if you want more information on
them.

~~~
Tomminn
See update, I realized my facepalm at roughly the same time as you:)

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BennyH26
The proof states: "For n to be prime either x² + x + 1 = 1 or x - 1 = 1"

Could somebody tell me why each of the factor components must be equal to one?

~~~
gizmo686
I believe there is a typo, and it should say "for p to be prime".

Since p = (x^2 + x + 1)(x - 1), and both of those factors are integers, if
both factors were not 1, then we would have demonstrated a factoring of p;
therefore p would be composite.

If you wanted to be super formal, you would have to deal with the possibility
that the factors were -1, but that is more of an uninteresting technicality.

~~~
bwasti
ah good catch on the typo, just fixed it

~~~
amluto
There’s another minor error. The proof states:

> For $ p $ to be prime either $ x^2 + x + 1 = 1 $ or $ x - 1 = 1 $.

That’s not quite true. In general, if p is prime and p=a·b for integers a and
b, then at least one of a and b is 1 _or -1_. The rest of the proof still
works since, as you say, x != 0, so x + 1 != -1.

~~~
bwasti
I'll leave that omitted to keep things concise. I think it is quite common to
assume a domain of natural numbers when talking about primes.

> then at least one of a and b is 1 or -1

I think you mean exactly one of a and b is 1 or -1

~~~
amluto
You can’t just assume a range of natural numbers in this context. You have a
prime p that is equal to f(x) for some polynomial f with integer coefficients
and some integer x. You observe that f(x)=a(x)b(x) where a and b are also
polynomials with integer coefficients. It does not follow that a and b are
non-negative. As an example, let f(x)=x^3-2x^2-3x+6. Then a(x)=x-2 and
b(x)=x^2-3. Now you could try to argue that, if f(x) is prime, then one of
a(x) and b(x) is 1, so x is 2 or 3, but f(2)=0 and f(3)=6, so f(x) can’t be
prime for integer x. But you would be wrong, and, indeed, f(1)=2, which is
prime.

You really do have to consider negative factors.

~~~
bwasti
that's a great counter example! updated the post

~~~
jwilk
You linked to the reply URL, which works only for logged-in users. Please
consider using this instead:

[https://news.ycombinator.com/item?id=17904921](https://news.ycombinator.com/item?id=17904921)

~~~
bwasti
done

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kowdermeister
What's the deal with all these $ symbols?

~~~
bwasti
Assuming you have JavaScript disabled, that is latex notation.

~~~
huhtenberg
I'm too seeing raw $ markup. Nothing's blocked, no console errors, but there's
a "425 No Reason Phrase" response for
[https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax...](https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-
MML-AM_CHTML), not sure if it's relevant. That's in recent Firefox.

More specifically, when I open the page by clicking on the HN link or copy-
pasting the URL into the bar and pressing Enter, the page renders OK. If I am
to reload the page with Ctrl-R after that, it shows up in raw $ markup and no
amount of refreshing helps. Furthermore, full reload (via Ctrl-Refresh) yields
that 425 response.

Just FYI.

~~~
bwasti
thanks for the heads up. The site is self-hosting mathjax now

