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, 2016 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, 2016 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, 2016 at 20:38
  • $\begingroup$ @almagest You are correct, that was a typo on my part. $\endgroup$
    – P...
    May 19, 2016 at 20:44

1 Answer 1


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, 2016 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. $\endgroup$ May 20, 2016 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. $\endgroup$ May 20, 2016 at 19:54

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