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Re: All composites?

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  • David Broadhurst
    ... 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
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      --- In primenumbers@yahoogroups.com,
      "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)
    • Maximilian Hasler
      ... 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
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        --- In primenumbers@yahoogroups.com, "David Broadhurst" <d.broadhurst@...> 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

        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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