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Let p1 and p2 be prime numbers and p1 < p2. Proove that there is a natural number n for which:

  • p1 + n(p2 - p1) is prime AND
  • p2 + n(p2 - p1) is not prime

    I've checked for n = 1, n = p1 and all of the answers that sounded reasonable. I tried also subtracting p1 + n(p2 - p1) from p2 + n(p2 - p1) but I'm not getting anywhere. Any leads?

    • Billy Joel
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    • If you take $n=kp_2$ you can find a value $k$ for a prime in the first case by Dirichlet's theorem on primes in arithmetic progression. – Mark Bennet Nov 11 '17 at 15:44
    • I can only find proof that there are infinitely many primes in Dirichlet's arithmetic progression. But that doesn't really help me, or does it? – Billy Joel Nov 11 '17 at 16:12
    • You find a prime in the first sequence. The value of the second expression is automatically divisible by $p_2$. – Mark Bennet Nov 11 '17 at 16:25

    1 Answers1

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    Trying $n=1$ is a good instinct - it would often work. The only trouble is that sometimes $2p_2-p_1$ is prime. But then you can try $n=2$ and only fail if $3p_2-2p_1$ is prime - and continue like so. Basically, if you examine the reason why it fails, you'll see that you can just increment $n$ until you have the desired property.

    Observe that $$p_2+n(p_2-p_1)=p_1+(n+1)(p_2-p_1).$$ In particular, if we define $s_n=p_1+n(p_2-p_1)$, then we are trying to show that there is some $n$ such that $s_n$ is prime and $s_{n+1}$ is not. Both $s_0$ and $s_1$ are prime. Clearly, some $s_n$ is not prime, since the sequence contains zeros mod any number coprime to $p_2-p_1$. Thus, consider the first $m$ such that $s_m$ that is not prime. Then, let $n=m-1$ and you're done.

    Milo Brandt
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