Is there a collection of distinct positive integers $(k_1, k_2, k_3, p_1, p_2, p_3)$ such that:

  • $p_1, p_2, p_3$ are odd primes, and $k_1, k_2, k_3$ are odd

  • $(k_1 + 2) p_1 = k_2 p_2$ and $(k_2 + 2) p_2 = k_3 p_3$

  • $k_i \geq p_i$

Additionally, is it possible that there are infinitely many such collections? Or, that there are collections with arbitrarily many $p_i$ and $k_i$? What if we restrict to cases where $p_i < p_{i+1}$?

  • $\begingroup$ $(97+2)7=693=63\cdot11$; $(63+2)11=715=55\cdot13$ $\endgroup$ – almagest May 19 '16 at 20:35
  • $\begingroup$ What is your last condition? It cannot be $p_i<p_i+1$ surely! Is it $p_i<p_{i+1}$? $\endgroup$ – almagest May 19 '16 at 20:36
  • $\begingroup$ My impression is that it would be easy to generate infinitely many such collections, but I am off to do something else now! $\endgroup$ – almagest May 19 '16 at 20:38
  • $\begingroup$ @almagest You are correct, that was a typo on my part. $\endgroup$ – P... May 19 '16 at 20:44

Note that $p_1\mid k_2$ and $p_2 \mid k_3$. Let $k_2 = a_1p_1, k_3 = a_2p_2$.

  1. Choose $p_3, p_2, a_2$.
  2. $k_2 = a_2p_3 - 2$.
  3. $k_3 = a_2p_2$.
  4. Choose $p_1$ from the prime factors of $k_2$.
  5. $a_1 = k_2 / p_1$.
  6. $k_1 = a_1p_2 - 2$.

Now the requirement that $k_3 \ge p_3$ requires that $a_2 \ge p_3/p_2$. That $k_2 \ge p_2$ requires $a_1 \ge p_2/p_1$, but substituting the formulas for $a_1$ and $k_2$ gives $a_2p_3 - 2 \ge p_2$ or $a_2 \ge \frac{p_2 - 2}{p_3}$. And the requirement that $k_1 > p_1$ requires that $p_1$ be chosen so that $$p_1 \le 1 + \sqrt{1 + k_2p_2}$$ Since $1 + k_2p_2 > k_2$, we can be sure that such $p_1$ always exist.

So meeting your original conditions just requires choosing two arbitrary primes $p_2, p_3$, then choosing $$a_2 \ge \max\left\{\frac{p_3}{p_2}, \frac{p_2 -2}{p_3}\right\}$$ Then calculating $k_2$ and choosing a prime fractor $p_1$ that also satisfies $$p_1 \le 1 + \sqrt{1 + k_2p_2}$$

Since $p_2, p_3$ are arbitrary, enforcing $p_3 > p_2$ is not restrictive, and it seems obvious that choosing $p_1 < p_2$ is not difficult either.

So there are infinitely many solutions.

  • $\begingroup$ I'll go ahead and accept, but you're missing two subtleties. You ignore the restriction that the different $p_i$ are distinct (which causes problems for choices of $a_2$ with $k_2 = p_2^i p_3^j$ and your method outright fails when $k_2$ is prime (since then $p_2 - 2 = k_1 > p_1 = k_2 > p_2$). I'm not sure whether or not it's possible to account for the former case, and while the prime number theorem shows that for large enough $a_2$ you can ensure that $k_2$ is composite with probability tending to 1, I don't think you can say for any given $p_2$ and $p_3$ how large $a_2$ must be. $\endgroup$ – P... May 20 '16 at 18:37
  • $\begingroup$ The question I was answering is "do infinitely many solutions exist", not "how can you unfailingly generate solutions".I only pushed this until it was obvious infinitely many solutions were possible. I did not ignore the distinct requirement. It is clearly stated in the penultimate paragraph. Since $p_3$ does not divide 2, $k_2$ cannot have $p_3$ as a factor. $k_2=p_2^i$ is indeed possible, but this is a fringe case only rarely encountered. The case of $k_2$ being prime is also seen to not hold in an infinite number of cases, even without invoking the prime number theorem. $\endgroup$ – Paul Sinclair May 20 '16 at 19:38
  • $\begingroup$ To late to edit that comment, but it would have better said "the distinct requirement is discussed in the penultimate paragraph". As I said, once it was obvious that you could generate infinitely many solutions, I left it there. You can easily bypass the pitfalls mentioned in a number of ways, such as starting with a $k_2$ which is not the power of a prime, then choosing $p_3$ from the factors of $k_2 + 2$, etc. $\endgroup$ – Paul Sinclair May 20 '16 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.