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Let $p_1$ through $p_6$ be consecutive primes in ascending order, and consider the simultaneous equation $$p_1x+p_2y=p_3\\p_4x+p_5y=p_6$$

Motivating Question: For what $p_1$ does the system provide integer solutions $(x,y)$?

Solving the system, we have $$x=\frac{p_3p_5-p_2p_6}{p_1p_5-p_2p_4}\quad\text{and}\quad y=\frac{p_3-p_1x}{p_2}=\frac{p_1p_6-p_3p_4}{p_1p_5-p_2p_4}$$ so it is equivalent to asking when $p_1p_5-p_2p_4$ divides $p_3p_5-p_2p_6$.

I asked this on MSE a few months ago and user @RGS has created a script for the first $100,000$ primes.

I give the results here.

for p1 = 2 we have (x,y) = (16.0,-9.0) for p1 = 5 we have (x,y) = (-9.0,8.0) for p1 = 19 we have (x,y) = (-13.0,12.0) for p1 = 59 we have (x,y) = (-3.0,4.0) for p1 = 151 we have (x,y) = (-1.0,2.0) for p1 = 487 we have (x,y) = (-2.0,3.0) for p1 = 587 we have (x,y) = (-1.0,2.0) for p1 = 739 we have (x,y) = (-2.0,3.0) for p1 = 881 we have (x,y) = (-2.0,3.0) for p1 = 1097 we have (x,y) = (-1.0,2.0) for p1 = 1453 we have (x,y) = (-2.0,3.0) for p1 = 1697 we have (x,y) = (-5.0,6.0) for p1 = 2213 we have (x,y) = (-2.0,3.0) for p1 = 2239 we have (x,y) = (-2.0,3.0) for p1 = 2243 we have (x,y) = (-2.0,3.0) for p1 = 3037 we have (x,y) = (-2.0,3.0) for p1 = 3217 we have (x,y) = (-2.0,3.0) for p1 = 4019 we have (x,y) = (-3.0,4.0) for p1 = 4073 we have (x,y) = (-2.0,3.0) for p1 = 4093 we have (x,y) = (-2.0,3.0) for p1 = 4241 we have (x,y) = (-5.0,6.0) for p1 = 5477 we have (x,y) = (-2.0,3.0) for p1 = 6199 we have (x,y) = (-2.0,3.0) for p1 = 6793 we have (x,y) = (-2.0,3.0) for p1 = 7669 we have (x,y) = (-2.0,3.0) for p1 = 7877 we have (x,y) = (-2.0,3.0) for p1 = 8291 we have (x,y) = (-2.0,3.0) for p1 = 8819 we have (x,y) = (-5.0,6.0) for p1 = 9041 we have (x,y) = (-3.0,4.0) for p1 = 9059 we have (x,y) = (-3.0,4.0) for p1 = 9181 we have (x,y) = (-2.0,3.0) for p1 = 9461 we have (x,y) = (-2.0,3.0) for p1 = 9811 we have (x,y) = (-2.0,3.0) for p1 = 11159 we have (x,y) = (-5.0,6.0) for p1 = 11467 we have (x,y) = (-3.0,4.0) for p1 = 11621 we have (x,y) = (-2.0,3.0) for p1 = 12821 we have (x,y) = (-3.0,4.0) for p1 = 13877 we have (x,y) = (-2.0,3.0) for p1 = 15149 we have (x,y) = (-1.0,2.0) for p1 = 16657 we have (x,y) = (-3.0,4.0) for p1 = 19421 we have (x,y) = (-2.0,3.0) for p1 = 19457 we have (x,y) = (-1.0,2.0) for p1 = 19571 we have (x,y) = (-1.0,2.0) for p1 = 20021 we have (x,y) = (-3.0,4.0) for p1 = 20101 we have (x,y) = (-1.0,2.0) for p1 = 21011 we have (x,y) = (-2.0,3.0) for p1 = 22039 we have (x,y) = (-1.0,2.0) for p1 = 22271 we have (x,y) = (-2.0,3.0) for p1 = 23059 we have (x,y) = (-2.0,3.0) for p1 = 23761 we have (x,y) = (-1.0,2.0) for p1 = 24071 we have (x,y) = (-1.0,2.0) for p1 = 25583 we have (x,y) = (-2.0,3.0) for p1 = 25997 we have (x,y) = (-2.0,3.0) for p1 = 29567 we have (x,y) = (-2.0,3.0) for p1 = 30643 we have (x,y) = (-2.0,3.0) for p1 = 31153 we have (x,y) = (-3.0,4.0) for p1 = 31177 we have (x,y) = (15590.0,-15587.0) for p1 = 31477 we have (x,y) = (-2.0,3.0) for p1 = 32377 we have (x,y) = (-5.0,6.0) for p1 = 32503 we have (x,y) = (-6.0,7.0) for p1 = 33149 we have (x,y) = (-5.0,6.0) for p1 = 33767 we have (x,y) = (-2.0,3.0) for p1 = 40459 we have (x,y) = (-1.0,2.0) for p1 = 40823 we have (x,y) = (-2.0,3.0) for p1 = 40927 we have (x,y) = (-1.0,2.0) for p1 = 41143 we have (x,y) = (-2.0,3.0) for p1 = 41467 we have (x,y) = (-1.0,2.0) for p1 = 42179 we have (x,y) = (-3.0,4.0) for p1 = 42181 we have (x,y) = (-1.0,2.0) for p1 = 42299 we have (x,y) = (-2.0,3.0) for p1 = 43781 we have (x,y) = (-2.0,3.0) for p1 = 44501 we have (x,y) = (-2.0,3.0) for p1 = 44809 we have (x,y) = (-2.0,3.0) for p1 = 46229 we have (x,y) = (-3.0,4.0) for p1 = 47501 we have (x,y) = (-1.0,2.0) for p1 = 47699 we have (x,y) = (-5.0,6.0) for p1 = 48761 we have (x,y) = (-2.0,3.0) for p1 = 50261 we have (x,y) = (-5.0,6.0) for p1 = 51131 we have (x,y) = (-2.0,3.0) for p1 = 51157 we have (x,y) = (-2.0,3.0) for p1 = 52289 we have (x,y) = (-5.0,6.0) for p1 = 54091 we have (x,y) = (-2.0,3.0) for p1 = 55331 we have (x,y) = (-2.0,3.0) for p1 = 58367 we have (x,y) = (-5.0,6.0) for p1 = 58537 we have (x,y) = (-1.0,2.0) for p1 = 58733 we have (x,y) = (-2.0,3.0) for p1 = 60089 we have (x,y) = (-5.0,6.0) for p1 = 60257 we have (x,y) = (-6.0,7.0) for p1 = 60733 we have (x,y) = (-5.0,6.0) for p1 = 62927 we have (x,y) = (-5.0,6.0) for p1 = 63487 we have (x,y) = (-1.0,2.0) for p1 = 63521 we have (x,y) = (-1.0,2.0) for p1 = 66643 we have (x,y) = (-3.0,4.0) for p1 = 67477 we have (x,y) = (-2.0,3.0) for p1 = 69653 we have (x,y) = (-2.0,3.0) for p1 = 70583 we have (x,y) = (-3.0,4.0) for p1 = 70783 we have (x,y) = (-3.0,4.0) for p1 = 71861 we have (x,y) = (-2.0,3.0) for p1 = 71881 we have (x,y) = (-2.0,3.0) for p1 = 71987 we have (x,y) = (-1.0,2.0) for p1 = 72089 we have (x,y) = (-5.0,6.0) for p1 = 72817 we have (x,y) = (-6.0,7.0) for p1 = 73471 we have (x,y) = (-1.0,2.0) for p1 = 75209 we have (x,y) = (-3.0,4.0) for p1 = 76079 we have (x,y) = (-5.0,6.0) for p1 = 77587 we have (x,y) = (-5.0,6.0) for p1 = 79811 we have (x,y) = (-2.0,3.0) for p1 = 80681 we have (x,y) = (-2.0,3.0) for p1 = 81019 we have (x,y) = (-2.0,3.0) for p1 = 81671 we have (x,y) = (-2.0,3.0) for p1 = 83071 we have (x,y) = (-2.0,3.0) for p1 = 83089 we have (x,y) = (-2.0,3.0) for p1 = 85601 we have (x,y) = (-2.0,3.0) for p1 = 87557 we have (x,y) = (-12.0,13.0) for p1 = 91159 we have (x,y) = (-5.0,6.0) for p1 = 91423 we have (x,y) = (-2.0,3.0) for p1 = 92107 we have (x,y) = (-2.0,3.0) for p1 = 93047 we have (x,y) = (-1.0,2.0) for p1 = 93937 we have (x,y) = (-2.0,3.0) for p1 = 94477 we have (x,y) = (-5.0,6.0) for p1 = 95957 we have (x,y) = (-6.0,7.0) for p1 = 96353 we have (x,y) = (-1.0,2.0) for p1 = 96553 we have (x,y) = (-6.0,7.0) for p1 = 97511 we have (x,y) = (-2.0,3.0) for p1 = 97577 we have (x,y) = (-2.0,3.0) for p1 = 100391 we have (x,y) = (-5.0,6.0) for p1 = 101267 we have (x,y) = (-1.0,2.0) for p1 = 101873 we have (x,y) = (-2.0,3.0) for p1 = 102451 we have (x,y) = (-2.0,3.0) for p1 = 103651 we have (x,y) = (-2.0,3.0) for p1 = 103811 we have (x,y) = (-12.0,13.0) for p1 = 103991 we have (x,y) = (-2.0,3.0) for p1 = 104009 we have (x,y) = (-1.0,2.0) total count is 133

Some interesting things can be observed.

Firstly, $\text{sgn}(x)\neq\text{sgn}(y)$. We expect this since the negative value compensates for the large positive value to give a smaller positive value (namely $p_3$ and $p_6$).

Next, we see that apart from five exceptions $p_1=2,5,19,59,31177$, $y$ is always greater than $-x$ by $1$. Also, the same set of values of $(x,y)$ keep on reappearing.

Question 1: Is there an explanation for the regularity with which $$x+y=1\quad?$$

If we substitute this into the simultaneous equation, we get that $$\frac{p_1-p_2}{p_2-p_3}=\frac{p_4-p_5}{p_5-p_6}$$ This might be related to the twin prime conjecture.

Finally, the proportion of primes that satisfy the system is $133/10000=1.33\%$. If we consider the first $90,000, 80,000$ or $70,000$ primes we still see that the proportion hovers around $1.3\%$.

Question 2: What is the value of $$\lim_{x\to\infty}\frac{\pi'(x)}{\pi(x)}$$ where $\pi'(x)$ is the number of primes $p_1\le x$ that satisfy the simultaneous equation?

Note that in a second answer in MSE, it was shown that proving Dickson's Conjecture would prove the infinitude of these integer solutions.

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  • $\begingroup$ Note that $p_1p_5-p_2p_4$ divides $p_3p_5-p_2p_6$ if and only if it divides $p_1p_6-p_3p_4$, since $$p_1(p_3p_5-p_2p_6)-p_2(p_1p_6-p_3p_4)=p_3(p_1p_5-p_2p_4).$$ $\endgroup$
    – Greg Martin
    Jun 6, 2018 at 17:10
  • $\begingroup$ @GregMartin Well spotted. Edited. $\endgroup$
    – TheSimpliFire
    Jun 6, 2018 at 18:45
  • $\begingroup$ Is there something similar in the case of system $p_1x+p_2y+p_3z=p_4$, $p_5x+p_6y+p_7z=p_8$, $p_9x+p_{10}y+p_{11}z=p_{12}$? (of course, we will still have $x+y+z=1$ with only finitely many exceptions, but maybe something interesting beside that?..) $\endgroup$
    – Alexander Kalmynin
    Jun 6, 2018 at 22:20
  • $\begingroup$ @AsymptotiacK Unfortunately no, I haven't been able to find any for the first hundred or so. What might be interesting is that for both the $2\times2$ ans $3\times3$ systems of equations, even the non-integer solutions are quite 'simple' fractions like $5/4$ or $7/11$. $\endgroup$
    – TheSimpliFire
    Jun 7, 2018 at 7:53

1 Answer 1

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This is not a good forum for this question. Since it touches upon prime gaps, I will provide a perspective to help explain.

Since the problem deals with consecutive primes in order, with $p_1$ being the smallest prime, it aids to subtract off $p_1$. Doing this for the first equation and representing differences with $d$s, we can rewrite the first equation as $$p_1(x-1) + (p_1 + d_1)y = d_2.$$

Here $d_1$ and $d_2$ are expected to be small compared to $p_1$, and for sufficiently large primes, are guaranteed to be smaller than the $3/5$ power of $p_1$. The $d_j$ are often less than $\log p_1$ and that is the intuition to hold for what follows.

Since the differences are small, $x-1$ and $y$ have to be about the same in absolute value to get a solution, especially if both $x$ and $y$ are much smaller than $p_1$. This should motivate why the solutions satisfy $x+y =1$. Note that the relation $yd_1 = d_2$ then needs to be observed, and as $d_1$ is smaller than $d_2$, $y \gt 1$.

Now subtract the second equation from the first. $$d_4x + d_5y = d_6$$. Again the differences are small numbers, and if we are looking also for integral solutions satisfying $x+y=1$, then $d_6 - d_5$ must be a multiple of $d_4 - d_5$, which relates to the divisibility conditions discussed in comments. Again the $d$s are small and the condition essentially is one on (a part of) a sextuple of consecutive primes. There may be analytic methods to predict the frequency of these tuples. Although it is not guaranteed, observing their frequency among the first million examples is a good clue as to the expected distribution over all examples.

Again, this is not a good forum for this question.

Gerhard "Sextuple Makes Me Feel Young" Paseman, 2018.06.06.

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  • $\begingroup$ I think, it is possible to deduce that $x+y=1$ for all but finitely many examples from the (hypothetical) estimate $p_{n+1}-p_n=o(p_n^{1/2})$, because in this case we have $|y|\approx 1+\frac{d_5-d_2}{d_4-d_1}$, which cannot be very large, while $|y|$ has to be large if $x+y\neq 1$ $\endgroup$
    – Alexander Kalmynin
    Jun 6, 2018 at 22:14
  • $\begingroup$ I agree with your heuristic argument. Where do you think this question is best suited? $\endgroup$
    – TheSimpliFire
    Jul 7, 2018 at 16:20
  • $\begingroup$ I suggest math.stackexchange, but read their How To Ask pages first. Gerhard "Post May Need Some Tweaks" Paseman, 2018.07.07. $\endgroup$
    – Gerhard Paseman
    Jul 7, 2018 at 16:41
  • $\begingroup$ I have already asked that question there a few months ago: see here $\endgroup$
    – TheSimpliFire
    Jul 7, 2018 at 18:44

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