## Re: All composites?

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• ... 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 1 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 2 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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