I believe that for every prime $p\geq 5$ there exists at least one prime $q$ that is both congruent to 1 mod $p$ and congruent to 5 mod 6. It's well known that there are an infinite number of primes congruent to 1 mod $p$ and there are an infinite number of primes congruent to 5 mod 6, but the conjunction of the two conditions is baffling me. Any ideas? (The conjecture is true for $p < 437,077$ because programming).
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$q\equiv 1!\mp 2p\pmod{6p}$ for $, p\equiv \pm 1 \pmod{6},,$ see $ $ Sophie Germain primes. – Bill Dubuque Feb 12 '15 at 23:32
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Thanks, I did look at Sophie Germain primes briefly, but no ideas came to mind. – Ian Parberry Feb 13 '15 at 00:02
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Using the Chinese Remainder Theorem, you will find that $$\left\{ \begin{array}{ccc} x&\equiv &5\pmod{6}\\ x&\equiv &1\pmod{p}\end{array}\right.$$ Has a unique solution $\pmod{6p}$. Also since $5$ and $1$ are relatively prime to $6$ and $p$ respectively, the solution will be relatively prime to $6p$.
Then by Dirichlet's Theorem on primes in arithmetic progression, the congruence class of the solution will contain (infinitely many) primes.
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Awesome. I got as far as CRT myself, but Dirichlet's Theorem is just the missing piece I need. Thanks! – Ian Parberry Feb 13 '15 at 00:03
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