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I am trying to come up with conjectures related to prime numbers and came up with this question. The last question I asked was wrong. So for this question I have tried numbers till $30$. Is this conjecture valid?

StubbornAtom
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jnyan
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    Till $30$ is not very far at all... Write a program. – lhf Aug 31 '16 at 18:10
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    As said in the previous comment, you should write a program to know if this conjecture can be true. By the way, if $P=37$, $P^2+P+1=1047=3\times 7 \times 67$ so the conjecture can't be true – H. Potter Aug 31 '16 at 18:14
  • Even if you tried up to $2352342352342135132423523541324236235$ it would prove nothing. I don't have any examples but I would bet there are reasonable conjectures whose first counterexamples are larger than that. – Matt Samuel Aug 31 '16 at 18:14
  • I would expect (based on nothing concrete) that no polynomial yields only primes or semi primes when evaluated at primes, except for the trivial one $f(x)=x$. – lhf Aug 31 '16 at 18:18
  • @lhf $p^2$ is semi-prime, so $f(x) = px$ works too (and allowing $p^2$ is necessary for things as Chen's theorem – reuns Aug 31 '16 at 18:24
  • @user1952009, right, thanks. – lhf Aug 31 '16 at 18:27
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    Without electronic help, one can easily jump to false conclusions. Even Fermat falsely conjectured that $2^{2^n}+1$ is prime for all $n\ge 0$. Euler disproved this by showing $641|2^{32}+1$ – Peter Aug 31 '16 at 18:36
  • @Peter there is a famous false conjecture that is valid for up to $10^{23}$ – reuns Aug 31 '16 at 18:40
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    @MattSamuel The first counterexample to $n^{17}+9$ and $(n+1)^{17}+9$ being not coprime is $n=8424432925592889329288197322308900672459420460792433$ – wythagoras Aug 31 '16 at 19:01
  • @wythagoras, where does that come from? Oh, from https://oeis.org/A010034! And http://math.stackexchange.com/questions/111440/examples-of-apparent-patterns-that-eventually-fail/111939#111939. – lhf Sep 01 '16 at 02:09
  • Skewes' number has to do with another false conjecture about the prime counting function and the integral logarithm. – Peter Sep 02 '16 at 21:19

2 Answers2

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This is false for $p=37$ because $37^2+37+1=1407= 3 \cdot 7 \cdot 67$.

lhf
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If $p\equiv1$ mod $3$, then $p^2+p+1$ is divisible by $3$, and if $p\equiv2$ or $4$ mod $7$, then $p^2+p+1$ is divisible by $7$, so by Dirichlet's theorem on primes in arithmetic progressions (and the Chinese Remainder Theorem), there are infinitely many primes $p$ for which $p^2+p+1$ is divisible by both $3$ and $7$, hence neither prime nor semi-prime.

Barry Cipra
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