## All composites?

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• Hello all:   32(n!)**8+1 is composite for all n positive integer ??   Sincerely   Sebastian [Non-text portions of this message have been removed]
Message 1 of 6 , Jul 1, 2009
Hello all:

32(n!)**8+1 is composite for all n positive integer ??

Sincerely

Sebastian

[Non-text portions of this message have been removed]
• ... For which n 1 is 4*n!^8 + 1 prime ?? Maximilian
Message 2 of 6 , Jul 1, 2009
--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
>
> Hello all:
> ï¿½
> 32(n!)**8+1 is composite for all nï¿½positive integerï¿½??

For which n>1 is 4*n!^8 + 1 prime ??

Maximilian
• ... Sebastian Martin Ruiz wrote:   ... On the contrary, the PNT makes it probable (but not yet provable) that there is an *infinite* number of
Message 3 of 6 , Jul 1, 2009
Sebastian Martin Ruiz <s_m_ruiz@...> wrote:

> 32(n!)**8+1 is composite for all n positive integer ??

On the contrary, the PNT makes it probable (but not yet provable)
that there is an *infinite* number of primes of this form.

The probability that (2*(n!))^8+1 is prime
is (heuristically) asymptotic to exp(Euler)/(8*n).
[If I got the constant wrong, Chris will correct me.]
The integral of 1/n diverges.

Moral: just because you didn't find a small one,
doesn't mean there isn't an infinity of larger ones.

David
• correcting a typo, which does not seem to me to affect the argument: The probability that ((2*(n!))^8)/2+1 is prime is (heuristically) asymptotic to
Message 4 of 6 , Jul 1, 2009
correcting a typo, which does not seem to me to
affect the argument:

The probability that ((2*(n!))^8)/2+1 is prime
is (heuristically) asymptotic to exp(Euler)/(8*n).
[If I got the constant wrong, Chris will correct me.]
The integral of 1/n diverges.

David
• ... For none. Proof: Set x = (n!)^2 in the identity 4*x^4 + 1 = (2*x^2 + 1)^2 - (2*x)^2 David (per proxy Léon François Antoine)
Message 5 of 6 , Jul 1, 2009
"Maximilian Hasler" <maximilian.hasler@...> wrote:

> For which n>1 is 4*n!^8 + 1 prime ??

For none. Proof: Set x = (n!)^2 in the identity
4*x^4 + 1 = (2*x^2 + 1)^2 - (2*x)^2

David (per proxy Léon François Antoine)
• ... Right... of course! Bad example - that s not what I meant to say. Maybe, Is 8*n!^8+1 composite for all n 4 ? Probably not. Actually I don t mind at all.
Message 6 of 6 , Jul 1, 2009
>
> > For which n>1 is 4*n!^8 + 1 prime ??
>
> For none. Proof: Set x = (n!)^2 in the identity
> 4*x^4 + 1 = (2*x^2 + 1)^2 - (2*x)^2

Right... of course! Bad example - that's not what I meant to say.
Maybe,

Is 8*n!^8+1 composite for all n>4 ?

Probably not. Actually I don't mind at all. It was just to express the same "Moral" given elsewhere more explicitely. But I should have chosen my example more carefully...
At least this shows that the "32" is in some sense simply the 3rd possibiliy, after 2 and 8. For 2^7 there happen to be 3 small primes; for 2^9 and 2^11 only two. Thus, whenever the first prime is found for 2^5, then the same post can be made replacing 32 with 2^13.)

Maximilian
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