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Can someone give me a hint on this problem please? Show that the polynomial $p(x)=x^2-x+41$ takes prime values for x in the set (0,1,2,...,40) Thanks in advance!

Lucy Sun
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    Just compute the values. – lulu Sep 23 '17 at 19:15
  • There is no cool theoretic way to show that "$p(n)$ is composite $\implies n>40$", as far as I know. Just check the values. – Arthur Sep 23 '17 at 19:18
  • @Arthur you can at least show all values the polynomial take on are odd etc. just with simple parity arguments. –  Sep 23 '17 at 19:19
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    @RoddyMacPhee Checking that $p(1),p(2),\cdots,p(40)$ are all prime, or checking for each prime below $\sqrt{p(40)}$ that they don't divide any $p(n)$. It's still checking, one by one, either way. – Arthur Sep 23 '17 at 19:21
  • @Arthur my point is if even if more than 1 of them is even we could disprove the claim it's a heuristic argument to use the parity argument. also just checking 2 and 3 will show if they are in the proper classes mod 6. –  Sep 23 '17 at 19:25
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    A short proof given in 1913 by Rabinowitz. I gave a proof in simple terms (no fields, just binary quadratic forms) at http://math.stackexchange.com/questions/289338/is-the-notorious-n2-n-41-prime-generator-the-last-of-its-type/289357#289357 – Will Jagy Sep 23 '17 at 19:27
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    @WillJagy That's extremely interesting! Thank you for sharing that! – Andrew Tawfeek Sep 23 '17 at 19:30
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    This is a fairly immediate consequence of the fact that the ring of integers of $,\Bbb Q(\sqrt{-163}),$ isa UFD. For proofs see the links I gave in this answer. – Bill Dubuque Sep 23 '17 at 19:33
  • take $$x=41$$ and the number is not prime – Dr. Sonnhard Graubner Sep 23 '17 at 19:35
  • @Dr.S Trivially. But nobody claimed it was prime at $,x=41,$ so what is your point? – Bill Dubuque Sep 23 '17 at 19:37
  • ok sorry i have misread this – Dr. Sonnhard Graubner Sep 23 '17 at 19:39

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One relatively low-level approach involves using quadratic residues. First render $n^2-n+41$ as an odd number. So if $p|n^2-n+41$ for any prime $p$ then $p$ must be odd.

We then test

$4(n^2-n+41)=(2n-1)^2+163\equiv 0\bmod p$

which will be contradicted whenever $-163$ is a nonquadratic residue $\bmod p$. If the contradiction holds for all odd $p<41$ then we establish primality for all polynomial values below $41^2$ (thus all nonnegative $n$ below $41$). There are eleven prime factor candidates to test, but a nonquadratic residue for each one applies simultaneously to all forty-one values of interest.

Oscar Lanzi
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