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The 2021 Problem

Posted 10 months ago
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Notice that $2021$ is a concatenation of consecutive integers: $20\sim 21$
Also $2021$ is a product of consecutive primes: $43\times 47$.

What is the next number with both of these properties?
$24502451$ is close, $4943\times 4957$ but $4951$ is in between.
$2484224843$ is close, $49831\times 49853$ but $49843$ is in between.
$715353612\sim 715353607$ isn't quite there but is $845785793\times 845785799$

With Mathematica, we can find more solutions.
794018604377235322848433897872605582 ~
794018604377235322848433897872605583 =
891077215721081784886888257701070827 ×
891077215721081784886888257701070829

2518711810848159770018909254809359591672377471484881441744436703324716 ~
2518711810848159770018909254809359591672377471484881441744436703324717 =
5018676928083894672666012088036109843105301546773725102790665815794437 ×
5018676928083894672666012088036109843105301546773725102790665815794441

353879205744237011544616255111782082608671961515039134082358165448687146 ~
353879205744237011544616255111782082608671961515039134082358165448687147 =
594877471202462845078583461328011525336167267541426222873827376039101347 ×
594877471202462845078583461328011525336167267541426222873827376039101401

Some code to use is the following:

modPass[k_] := Abs[Mod[k, 6, -1]] == 1;
solve2021[exp_,add_]:=
Module[{cases},
cases=Solve[{10^exp  a +(a+1) == b (b+add),10^(exp-1)< b<10^exp}, Integers];
If[Length[cases]>0,
Select[Select[b/.cases,modPass[#] && modPass[#+add]&], PrimeQ[#] && PrimeQ[#+add]&]]]

solve2021[36, 2]
{891077215721081784886888257701070827}
solve2021[70, 4]
{5018676928083894672666012088036109843105301546773725102790665815794437}
solve2021[72, 54]
{594877471202462845078583461328011525336167267541426222873827376039101347}

Still a bit more to check, there may be intervening primes. For example, solve2021[19, 76] has intervening primes.

What are other solutions?

3 Replies
Posted 10 months ago

Here's a list of hits and near misses. The first hit is (43+0)(43+4) = 2021. The first near miss is (53+0)(53+8) = 3233.

{{43,{0,4}}, {4999493,{0,14}}, {891077215721081784886888257701070827,{0,2}}, {5018676928083894672666012088036109843105301546773725102790665815794437,{0,4}}, {594877471202462845078583461328011525336167267541426222873827376039101347,{0,54}}, {53,{0,6,8}}, {4943,{0,8,14}}, {15541,{0,10,18}}, {49831,{0,12,22}}, {4437314920661962295348881728019584825055609007902269768049678912707397,{0,30,76}}, {39236533463,{0,86,114,128}}, {97090540284081007,{0,82,214,252}}, {3879052403643149143,{0,10,66,76}}, {61843,{0,18,28,36,66}}, {89925533,{0,20,26,48,60}}, {69783320419929621602037383,{0,6,84,120,230}}, {1706791026491912957048478532183,{0,46,54,66,106}}, {7955646631992522480129538315515466631309,{0,98,294,402,452}}}

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Posted 9 months ago

In fact, there are many more such examples, though very few of them involve twin primes. With the terminology in the reply above, up to $a<10^{323}$, there appears to be only one such $(a,b,add=2)$: 794018604377235322848433897872605582794018604377235322848433897872605583 = 891077215721081784886888257701070829 × 891077215721081784886888257701070827.

Using 40 Miller-Rabin iterations, I have obtained the following equalities, all with a LHS involving up to 644 digits:

25187118108481597700189092548093595916723774714848814417444367033247162518711810848159770018909254809359591672377471484881441744436703324717 = 5018676928083894672666012088036109843105301546773725102790665815794441 × 5018676928083894672666012088036109843105301546773725102790665815794437 (shown above)

101936573393229113746896078599158434015231091149521033461905493584141767240967286279147302756939077798218255227325457025854937718962600442639675514396101936573393229113746896078599158434015231091149521033461905493584141767240967286279147302756939077798218255227325457025854937718962600442639675514397 = 319275074807334613749250748026616288651767499918792546380748503672549086703203277071612100663412425258886294180722873606632627934423491347284352444051 × 319275074807334613749250748026616288651767499918792546380748503672549086703203277071612100663412425258886294180722873606632627934423491347284352444047

93919382143599443874360974179992284882076026690881005214561418008150790494258654795862966202627730222978744961557082451286192284357751247078089025462416769391938214359944387436097417999228488207602669088100521456141800815079049425865479586296620262773022297874496155708245128619228435775124707808902546241677 = 9691201274537612443672374967144909095364760038874719917817704809652025255645361670391758426040026736596675168244667860182899459759546967572708946339679961 × 9691201274537612443672374967144909095364760038874719917817704809652025255645361670391758426040026736596675168244667860182899459759546967572708946339679957

981718780415872785585564201031992816866441710921031298607709720433328082774757576994927683441599794030113676908258383896762642955368235301572285359675451074772206407951599583811274848830644325337802597091580076981718780415872785585564201031992816866441710921031298607709720433328082774757576994927683441599794030113676908258383896762642955368235301572285359675451074772206407951599583811274848830644325337802597091580077 = 990817228562297853620579821430841029476907621168021829447531348075063345099849542958089591111031777996054793781667515002540761697546809287372732881673066506036575450357354487060783218299272638998714542277310011 × 990817228562297853620579821430841029476907621168021829447531348075063345099849542958089591111031777996054793781667515002540761697546809287372732881673066506036575450357354487060783218299272638998714542277310007

et cetera.


The StackExchange post A 2021 problem: 20∼21 and 43×47 suggests a heuristic that the number of expected solutions with $2n$ digits is $O(1/n)$. Judging by the sheer number of solutions, this heuristic is probably incorrect. An alternative model may have to be created.

In approaching this problem I formulated it through expressing it as $a~(a+1)=p(p+2k)$, where $p+2k$ is the prime after $p$. Simplification yields $100...01a+1=(p+k)^2-k^2$, so $100...01a+1+k^2$ becomes a square. The problem thus amounts to showing that $1+k^2$ is a quadratic residue modulo $10^n+1$, and when complete, finding all modular square roots of $1+k^2$ and checking that the primality requirements are satisfied. (Namely, that $p$ and $p+2k$ are prime and all numbers in between are composite.)

One clear pattern found is that there are many such pairs with the same number of digits ( $n$). For instance, with $k=2$ (cousin primes) there are 27 solutions with $n=210$ and none with $210<n<270$. This is something that the heuristic mentioned cannot explain, but it is readily explained with this implementation. For $n=210$, $10^{210}+1$ has 22 distinct prime factors, all of them having 5 as a Q.R.. This means that 5 has $2^{22}\approx 4\times 10^6$ square roots mod $10^{210}+1$. The density of primes at around $10^{210}$ is approximately 1 out of $210\ln10\approx 480$, so the approximate expected number of solutions satisfying the two primality conditions is $\frac{2^{22}}{480^2}\approx 18$. Of course, this is also a heuristic as the true fluctuations of primes cannot be accounted for.

Another noteworthy pattern is that there are far more such pairs with $k=2$ (prime gap 4) than for any other choice of $k$. Among the 4 solutions Ed Pegg has found, 2 of them have prime gap 4, and one has prime gap 2. As it turns out, for $n<323$ there is only 1 solution with prime gap 2 but 40 solutions with prime gap 4. This may be related to our use of base 10, as prime factors of $10^n+1$ tend to readily be congruent to 1 or 9 mod 10, so 5 is a QR mod $10^n+1$ for many $n$. The same might not hold true for other $k$, as it is unlikely that all 22 prime factors of $10^{210}+1$ should be such that an arbitrary number is a quadratic residue modulo these primes. (when this happens, which is exceedingly rare, there may be several solutions.

Overall, this is an extremely fascinating and deep problem. May 2021 be a uniquely good year!


Extra info: I have obtained that for $k=38$, $1+k^2=1445=5\times 17^2$, so that 1445 is a Q.R. mod $10^n+1$ if 5 is one. Therefore, one can observe that for $k=38$ or $682$, it is expected that there would be multiple solutions with $n=210$ or $270$, where there were many solutions for $k=2$. (For $k=682$, there is a chance that the numbers between the two primes may not be prime, but considering that heuristically reduces the number by only a factor of 4.)

POSTED BY: F. Lai
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