[math-fun] Number theory question
Can one show that there are infinitely many $n$ divisible by neither 41 or 42 such that n^2-n+41 (Euler's famous prime-generating polynomial) is composite? Jim Propp
Sure; all n of the form 43 i + 2 will be divisible by 43. Similar values can be found for many primes. On Sun, Jul 21, 2019 at 12:37 PM James Propp <jamespropp@gmail.com> wrote:
Can one show that there are infinitely many $n$ divisible by neither 41 or 42 such that n^2-n+41 (Euler's famous prime-generating polynomial) is composite?
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That was quick! Thanks. Jim On Sun, Jul 21, 2019 at 3:43 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Sure; all n of the form 43 i + 2 will be divisible by 43.
Similar values can be found for many primes.
On Sun, Jul 21, 2019 at 12:37 PM James Propp <jamespropp@gmail.com> wrote:
Can one show that there are infinitely many $n$ divisible by neither 41 or 42 such that n^2-n+41 (Euler's famous prime-generating polynomial) is composite?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Likewise, for all n of the form 47k+3, n(n-1)+41 is divisible by 47. Jim On Sun, Jul 21, 2019 at 3:47 PM James Propp <jamespropp@gmail.com> wrote:
That was quick! Thanks.
Jim
On Sun, Jul 21, 2019 at 3:43 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Sure; all n of the form 43 i + 2 will be divisible by 43.
Similar values can be found for many primes.
On Sun, Jul 21, 2019 at 12:37 PM James Propp <jamespropp@gmail.com> wrote:
Can one show that there are infinitely many $n$ divisible by neither 41 or 42 such that n^2-n+41 (Euler's famous prime-generating polynomial) is composite?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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By mod 47: nn-n+41 = nn-n-6 = (n-3)(n+2) has roots $\, n = 3,-2 (mod 47) On Sun, Jul 21, 2019 at 3:51 PM James Propp <jamespropp@gmail.com> wrote:
Likewise, for all n of the form 47k+3, n(n-1)+41 is divisible by 47.
Jim
On Sun, Jul 21, 2019 at 3:47 PM James Propp <jamespropp@gmail.com> wrote:
That was quick! Thanks.
Jim
On Sun, Jul 21, 2019 at 3:43 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Sure; all n of the form 43 i + 2 will be divisible by 43.
Similar values can be found for many primes.
On Sun, Jul 21, 2019 at 12:37 PM James Propp <jamespropp@gmail.com> wrote:
Can one show that there are infinitely many $n$ divisible by neither 41 or 42 such that n^2-n+41 (Euler's famous prime-generating polynomial) is composite?
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (3)
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Bill Dubuque -
James Propp -
Tomas Rokicki