# is 1001 the only sum of two positive cubes that is the product of three consecutive odd primes?

That is $$\ 10^3+1^3=7.11.13$$.

I could find no other examples. So I am looking to see if there are any more solutions to $$x^3+y^3=p.q.r$$, where $$x, y$$ are positive integers and $$p are consecutive odd primes. Now $$x^3 + y^3 =LQ$$ where $$L=x+y$$, and $$Q = L^2 -3xy$$.

After playing with the problem for a while I thought maybe I should try and express the triple of factors in terms of the gaps between them. So $$g_1 = q-p$$, and $$g_2=r-q$$, and I was thinking what was the biggest possible factor that divided the resulting sum of cubes after choosing the gaps.

Using elementary arguments I found that, $$L=q$$, or $$L=r$$ , so dealing with the later case put $$L=r$$, and $$Q=pq= L^2 -3xy$$, eliminate $$p$$, and $$q$$, and put $$x = (L+d)/2$$, $$y=(L-d)/2$$ for $$d$$ an odd positive integer. Giving $$L^2 - (g_1+2g_2)L+g_1(g_1+g_2)=L^2-3(L^2-d^2)/4$$. Completing the square and simplifying gives $$z^2-9d^2 = 4f$$, where $$z= 3L-2(g_1+2g_2)$$, and $$f = g_1^2+g_1g_2+g_2^2$$, the gaps are even so suitable factors can always be found, that is $$a$$, $$b$$, such that $$4f= ab$$, and we get using the difference of squares $$z= (a+b)/2$$, and so $$L=\frac{z+2(g_1+2g_2)}{3}$$. Now the largest $$z$$ is $$f+1$$, and so the largest,

$$L=\frac{g_1^2+g_1g_2+g_2^2+1+2g_1+4g_2}{3}$$ and chosing $$g$$ to be the bigger of the two gaps gives a largest $$L=\frac{3g^2+6g+1}{3}<(g+1)^2$$

so for both cases the largest possible "prime" factor in any triple of solutions was bounded above by roughly the square of the largest gap. Looking at Wikipedia prime gaps then Dr T Nicely's site on first occurrence of prime gaps https://faculty.lynchburg.edu/~nicely/gaps/gaplist.html I noticed that apart from a number of small cases that the primes where gaps first occurred where larger than $$(g+1)^2$$ suggesting that 1001 is the only case for the range of known first occurrences of prime gaps. I also saw somewhere but cannot remember at the moment something called Shank's conjecture, which is something like the first occurrence of a gap is after a prime that is about $$e^\sqrt g$$, which is obviously for large enough gap much bigger than $$(g+1)^2$$. See section 7 Marek Wolf 'Some heuristics on the gaps between consecutive primes' https://arxiv.org/pdf/1102.0481v2.pdf.

So this is roughly where I have got with the problem. I suspect I have missed a simpler solution?

35 is the only sum of two cubes equal to the product of two consecutive odd primes, using the notation above, let $$x^3 +y^3 = pq$$, and $$8 be consecutive odd primes then $$L< Q = L^2 -3xy$$, Put $$p=L$$, $$q=Q$$ and $$x=(L+d)/4$$, $$y = (L-d)/4$$, then $$(p^2+3d^2)/4=q$$, so $$(p^2+3)/4 \le q$$ but for $$p> 8$$, $$q>2p$$, which contradicts Bertrand's Postulate whereby if we have consecutive primes then $$p< q<2p$$ https://en.wikipedia.org/wiki/Bertrand's_postulate

You mention prime gaps and prime races; this addresses both. For prime $$p \geq 11,$$ with next prime $$p + g,$$ as far as we have been able to compute we find $$g < \log^2 p$$ with logarithm base $$e \approx 2.71828$$

https://en.wikipedia.org/wiki/Cram%C3%A9r%27s_conjecture

I took consecutive primes $$p \equiv q \equiv 1 \pmod 6,$$ allowing for possible primes in between that are $$6n-1.$$ Then I produced the representations of $$p$$ and of $$q$$ as $$u^2 - uv+ v^2,$$ combining those by Gauss composition into $$pq = x^2 - xy + y^2$$ Then $$(x+y) pq = x^3 + y^3$$ is what you were requesting. I printed out when $$p - 4 \log^2 q < x+y < q + 4 \log^2 q$$ If we call $$o$$ the prime just before $$p,$$ and $$r$$ the prime just after $$q,$$ I printed the word Interesting when $$o \leq x+y \leq r$$ I paid no attention to factoring $$x+y.$$ Sometimes it is prime. Anyway, the printout dies out when $$q > 46000.$$

It has just reached Mon Oct 26 15:17:47 PDT 2020 progress 5580013. or 5 million and change.

Let me put just the good bits. The last INTERESTING line is

109*127 x 121 y 7 next 131 x+y 128 middle 1 INTERESTING Note $$128 < 131$$

Mon Oct 26 15:02:31 PDT 2020

Mon Oct 26 15:02:31 PDT 2020
progress 13

7*13  x 6 y -5 x+y  1  previous 5  middle  1
7*13  x 9 y -1 x+y  8 between  INTERESTING
7*13  x 10 y 1 x+y  11 between  INTERESTING
7*13  x 10 y 9 next 17 x+y  19  middle  1
7*13  x 11 y 5 next 17 x+y  16  middle  1 INTERESTING
7*13  x 11 y 6 next 17 x+y  17  middle  1 INTERESTING
13*19  x 11 y -7 x+y  4  previous 11  middle  1
13*19  x 14 y -3 x+y  11  previous 11  middle  1 INTERESTING
13*19  x 17 y 3 next 23 x+y  20  middle  1 INTERESTING
13*19  x 17 y 14 next 23 x+y  31  middle  1
13*19  x 18 y 7 next 23 x+y  25  middle  1
13*19  x 18 y 11 next 23 x+y  29  middle  1
31*37  x 22 y -17 x+y  5  previous 29  middle  0
31*37  x 27 y -11 x+y  16  previous 29  middle  0
31*37  x 38 y 11 next 41 x+y  49  middle  0
31*37  x 38 y 27 next 41 x+y  65  middle  0
31*37  x 39 y 17 next 41 x+y  56  middle  0
31*37  x 39 y 22 next 41 x+y  61  middle  0
37*43  x 25 y -21 x+y  4  previous 31  middle  1
37*43  x 31 y -14 x+y  17  previous 31  middle  1
37*43  x 45 y 14 next 47 x+y  59  middle  1
37*43  x 45 y 31 next 47 x+y  76  middle  1
37*43  x 46 y 21 next 47 x+y  67  middle  1
37*43  x 46 y 25 next 47 x+y  71  middle  1
61*67  x 46 y -27 x+y  19  previous 59  middle  0
61*67  x 53 y -18 x+y  35  previous 59  middle  0
61*67  x 71 y 18 next 71 x+y  89  middle  0
61*67  x 73 y 27 next 71 x+y  100  middle  0
61*67  x 73 y 46 next 71 x+y  119  middle  0
67*73  x 54 y -25 x+y  29  previous 61  middle  1
67*73  x 65 y -9 x+y  56  previous 61  middle  1
67*73  x 74 y 9 next 79 x+y  83  middle  1
67*73  x 79 y 25 next 79 x+y  104  middle  1
73*79  x 53 y -34 x+y  19  previous 71  middle  0
73*79  x 66 y -17 x+y  49  previous 71  middle  0
73*79  x 83 y 17 next 83 x+y  100  middle  0
73*79  x 87 y 34 next 83 x+y  121  middle  0
79*97  x 86 y -3 x+y  83 between  INTERESTING
79*97  x 89 y 3 x+y  92 between  INTERESTING
97*103  x 94 y -11 x+y  83  previous 89  middle  1
97*103  x 105 y 11 next 107 x+y  116  middle  1
97*103  x 115 y 49 next 107 x+y  164  middle  1
103*109  x 87 y -31 x+y  56  previous 101  middle  1
103*109  x 118 y 31 next 113 x+y  149  middle  1
103*109  x 122 y 53 next 113 x+y  175  middle  1
109*127  x 107 y -19 x+y  88  previous 107  middle  1
109*127  x 114 y -7 x+y  107  previous 107  middle  1 INTERESTING
109*127  x 121 y 7 next 131 x+y  128  middle  1 INTERESTING
109*127  x 126 y 19 next 131 x+y  145  middle  1
139*151  x 132 y -23 x+y  109  previous 137  middle  1
139*151  x 155 y 23 next 157 x+y  178  middle  1
151*157  x 173 y 51 next 163 x+y  224  middle  0
157*163  x 129 y -50 x+y  79  previous 151  middle  0
157*163  x 146 y -25 x+y  121  previous 151  middle  0
157*163  x 171 y 25 next 167 x+y  196  middle  0
157*163  x 179 y 50 next 167 x+y  229  middle  0
181*193  x 163 y -41 x+y  122  previous 179  middle  1
181*193  x 204 y 41 next 197 x+y  245  middle  1
199*211  x 180 y -43 x+y  137  previous 197  middle  0
199*211  x 197 y -15 x+y  182  previous 197  middle  0
199*211  x 212 y 15 next 223 x+y  227  middle  0
199*211  x 223 y 43 next 223 x+y  266  middle  0
223*229  x 217 y -17 x+y  200  previous 211  middle  1
223*229  x 234 y 17 next 233 x+y  251  middle  1
271*277  x 241 y -57 x+y  184  previous 269  middle  0
271*277  x 253 y -38 x+y  215  previous 269  middle  0
271*277  x 291 y 38 next 281 x+y  329  middle  0
271*277  x 298 y 57 next 281 x+y  355  middle  0
277*283  x 270 y -19 x+y  251  previous 271  middle  1
277*283  x 289 y 19 next 293 x+y  308  middle  1
307*313  x 291 y -35 x+y  256  previous 293  middle  1
307*313  x 326 y 35 next 317 x+y  361  middle  1
307*313  x 339 y 70 next 317 x+y  409  middle  1
331*337  x 298 y -63 x+y  235  previous 317  middle  0
331*337  x 311 y -42 x+y  269  previous 317  middle  0
331*337  x 353 y 42 next 347 x+y  395  middle  0
331*337  x 361 y 63 next 347 x+y  424  middle  0
373*379  x 343 y -59 x+y  284  previous 367  middle  0
373*379  x 402 y 59 next 383 x+y  461  middle  0
397*409  x 364 y -69 x+y  295  previous 389  middle  1
397*409  x 433 y 69 next 419 x+y  502  middle  1
571*577  x 534 y -73 x+y  461  previous 569  middle  0
571*577  x 607 y 73 next 587 x+y  680  middle  0
601*607  x 578 y -49 x+y  529  previous 599  middle  0
601*607  x 627 y 49 next 613 x+y  676  middle  0
631*643  x 676 y 87 next 647 x+y  763  middle  1
661*673  x 652 y -29 x+y  623  previous 659  middle  0
661*673  x 681 y 29 next 677 x+y  710  middle  0
727*733  x 714 y -31 x+y  683  previous 719  middle  0
727*733  x 745 y 31 next 739 x+y  776  middle  0
739*751  x 700 y -83 x+y  617  previous 733  middle  1
739*751  x 783 y 83 next 757 x+y  866  middle  1
823*829  x 809 y -33 x+y  776  previous 821  middle  1
823*829  x 842 y 33 next 839 x+y  875  middle  1
1033*1039  x 1017 y -37 x+y  980  previous 1031  middle  0
1033*1039  x 1054 y 37 next 1049 x+y  1091  middle  0
1051*1063  x 1004 y -99 x+y  905  previous 1049  middle  1
1051*1063  x 1103 y 99 next 1069 x+y  1202  middle  1
1123*1129  x 1091 y -67 x+y  1024  previous 1117  middle  0
1123*1129  x 1158 y 67 next 1151 x+y  1225  middle  0
1153*1171  x 1121 y -78 x+y  1043  previous 1151  middle  1
1153*1171  x 1199 y 78 next 1181 x+y  1277  middle  1
1483*1489  x 1446 y -77 x+y  1369  previous 1481  middle  1
1483*1489  x 1523 y 77 next 1493 x+y  1600  middle  1
1567*1579  x 1532 y -79 x+y  1453  previous 1559  middle  1
1567*1579  x 1611 y 79 next 1583 x+y  1690  middle  1
1579*1597  x 1547 y -79 x+y  1468  previous 1571  middle  1
1579*1597  x 1626 y 79 next 1601 x+y  1705  middle  1
1657*1663  x 1611 y -94 x+y  1517  previous 1637  middle  0
1657*1663  x 1705 y 94 next 1667 x+y  1799  middle  0
1663*1669  x 1642 y -47 x+y  1595  previous 1657  middle  1
1663*1669  x 1689 y 47 next 1693 x+y  1736  middle  1
2551*2557  x 2502 y -101 x+y  2401  previous 2549  middle  0
2551*2557  x 2603 y 101 next 2579 x+y  2704  middle  0
2659*2671  x 2612 y -103 x+y  2509  previous 2657  middle  1
2659*2671  x 2715 y 103 next 2677 x+y  2818  middle  1
2791*2797  x 2731 y -122 x+y  2609  previous 2789  middle  0
2791*2797  x 2853 y 122 next 2801 x+y  2975  middle  0
2797*2803  x 2769 y -61 x+y  2708  previous 2791  middle  1
2797*2803  x 2830 y 61 next 2819 x+y  2891  middle  1
3229*3253  x 3183 y -113 x+y  3070  previous 3221  middle  1
3229*3253  x 3296 y 113 next 3257 x+y  3409  middle  1
3307*3313  x 3251 y -115 x+y  3136  previous 3301  middle  0
3307*3313  x 3366 y 115 next 3319 x+y  3481  middle  0
3541*3547  x 3483 y -119 x+y  3364  previous 3539  middle  0
3541*3547  x 3602 y 119 next 3557 x+y  3721  middle  0
3547*3559  x 3492 y -119 x+y  3373  previous 3541  middle  1
3547*3559  x 3611 y 119 next 3571 x+y  3730  middle  1
3943*3967  x 3891 y -125 x+y  3766  previous 3931  middle  1
3943*3967  x 4016 y 125 next 3989 x+y  4141  middle  1
5113*5119  x 5043 y -143 x+y  4900  previous 5107  middle  0
5113*5119  x 5186 y 143 next 5147 x+y  5329  middle  0
5197*5209  x 5161 y -83 x+y  5078  previous 5189  middle  0
5197*5209  x 5244 y 83 next 5227 x+y  5327  middle  0
5683*5689  x 5642 y -87 x+y  5555  previous 5669  middle  0
5683*5689  x 5729 y 87 next 5693 x+y  5816  middle  0
7723*7741  x 7681 y -101 x+y  7580  previous 7717  middle  1
7723*7741  x 7782 y 101 next 7753 x+y  7883  middle  1
10987*10993  x 10929 y -121 x+y  10808  previous 10979  middle  0
10987*10993  x 11050 y 121 next 11003 x+y  11171  middle  0
13297*13309  x 13236 y -133 x+y  13103  previous 13291  middle  0
13297*13309  x 13369 y 133 next 13313 x+y  13502  middle  0
18049*18061  x 17977 y -155 x+y  17822  previous 18047  middle  1
18049*18061  x 18132 y 155 next 18077 x+y  18287  middle  1
20947*20959  x 20869 y -167 x+y  20702  previous 20939  middle  0
20947*20959  x 21036 y 167 next 20963 x+y  21203  middle  0
21937*21943  x 21854 y -171 x+y  21683  previous 21929  middle  0
21937*21943  x 22025 y 171 next 21961 x+y  22196  middle  0
26821*26833  x 26732 y -189 x+y  26543  previous 26813  middle  0
26821*26833  x 26921 y 189 next 26839 x+y  27110  middle  0
26863*26881  x 26777 y -189 x+y  26588  previous 26861  middle  1
26863*26881  x 26966 y 189 next 26891 x+y  27155  middle  1
30307*30313  x 30209 y -201 x+y  30008  previous 30293  middle  0
30307*30313  x 30410 y 201 next 30319 x+y  30611  middle  0
30937*30949  x 30841 y -203 x+y  30638  previous 30931  middle  1
30937*30949  x 31044 y 203 next 30971 x+y  31247  middle  1
34033*34039  x 33929 y -213 x+y  33716  previous 34031  middle  0
34033*34039  x 34142 y 213 next 34057 x+y  34355  middle  0
35977*35983  x 35870 y -219 x+y  35651  previous 35969  middle  0
35977*35983  x 36089 y 219 next 35993 x+y  36308  middle  0
36637*36643  x 36529 y -221 x+y  36308  previous 36629  middle  0
36637*36643  x 36750 y 221 next 36653 x+y  36971  middle  0
45439*45481  x 45337 y -245 x+y  45092  previous 45433  middle  0
45439*45481  x 45582 y 245 next 45491 x+y  45827  middle  0
Mon Oct 26 15:02:35 PDT 2020
progress 60013

Mon Oct 26 15:02:38 PDT 2020
progress 120013

Mon Oct 26 15:02:43 PDT 2020
progress 180013

Mon Oct 26 15:02:47 PDT 2020
progress 240013

Mon Oct 26 15:02:54 PDT 2020
progress 300013

Mon Oct 26 15:02:58 PDT 2020
progress 360013


this is the C++ program in its current state. Uses GMP and my own collection of useful classes

#include <iostream>
#include <stdlib.h>
#include <fstream>
#include <strstream>
#include <list>
#include <set>
#include <math.h>
#include <iomanip>
#include <string>
#include <algorithm>
#include <iterator>
#include <gmp.h>
#include <gmpxx.h>
#include "form.h"

using namespace std;

//   g++  -o two_cubes two_cubes.cc  -lgmp -lgmpxx

//   g++  -o two_cubes two_cubes.cc  -lgmp -lgmpxx

int main()
{
cout << endl;
system("date");
cout << endl;
mpz_class oldp = 7;
mpz_class p = 7;

set<mp_pair>  oldpairs;
set<mp_pair>  pairs;
set<mp_pair>  compositepairs;

for(mpz_class x = 1;  3 * x * x <= 4*  p; ++x)
{
if( mp_SquareQ( 4*p - 3 * x * x )  )
{
mpz_class w = mp_Sqrt( 4*p - 3 * x * x  );

mpz_class y = ( x + w) / 2 ;
mp_pair xy;
xy.setFields(x,y); oldpairs.insert(xy);
xy.SetNegative(); oldpairs.insert(xy);
xy.setFields(y,x); oldpairs.insert(xy);
xy.SetNegative(); oldpairs.insert(xy);
y = ( x - w) / 2 ;
xy.setFields(x,y); oldpairs.insert(xy);
xy.SetNegative(); oldpairs.insert(xy);
xy.setFields(y,x); oldpairs.insert(xy);
xy.SetNegative(); oldpairs.insert(xy);

}// if square
} // for x

mpz_class bound = 100000;

bound *= bound;

for(  p = 13; p <= bound; p += 6)
{
if( p % 9000 == 13 ) cerr << " progress " << p << endl;
if( p % 60000 == 13 ) { system("date") ; cout << " progress " << p  << endl << endl; }
if( mp_PrimeQ(p)  )
{
// cout << endl;
//  cout << p * oldp << "   " ;
mpz_class middle = 0;
for(mpz_class u = oldp + 1; u < p; ++u)
{
if( mp_PrimeQ(u) )   ++middle;

}
//  cout << "  fax  "  << Factored(p * oldp) << endl;
double ll = mp_Log(p);
ll *= ll;
int l2 = (int) ceil(ll) ;
//  cout << p << " ceil " << l2 <<  endl;
int boo = 1;
boo = boo & middle < 2;

pairs.clear();

for(mpz_class x = 1;  3 * x * x <= 4*  p; ++x)
{
if( mp_SquareQ( 4*p - 3 * x * x )  )
{
mpz_class w = mp_Sqrt( 4*p - 3 * x * x  );

mpz_class y = ( x + w) / 2 ;
mp_pair xy;
xy.setFields(x,y); pairs.insert(xy);
xy.SetNegative(); pairs.insert(xy);
xy.setFields(y,x); pairs.insert(xy);
xy.SetNegative(); pairs.insert(xy);
y = ( x - w) / 2 ;
xy.setFields(x,y); pairs.insert(xy);
xy.SetNegative(); pairs.insert(xy);
xy.setFields(y,x); pairs.insert(xy);
xy.SetNegative(); pairs.insert(xy);

}// if square
} // for x

compositepairs.clear();

set<mp_pair>::iterator iter1,iter2, iter;
for(iter1 = oldpairs.begin(); iter1 != oldpairs.end(); ++iter1) {
for(iter2 = pairs.begin(); iter2 != pairs.end(); ++iter2) {
mp_pair oldpair = *iter1;
mp_pair currentpair = *iter2;

mpz_class x = oldpair.GetX();
mpz_class y = oldpair.GetY();
mpz_class z = currentpair.GetX();
mpz_class w = currentpair.GetY();
mp_pair newpair( x*z - y*w,  x*w + y*z - y*w   );

compositepairs.insert( newpair);
}} // iter1 iter2

for(iter = compositepairs.begin(); iter != compositepairs.end(); ++iter) {

mp_pair newpair = *iter;
if( newpair.GetX() > 0 && newpair.GetX() + newpair.GetY()  > 0 && newpair.GetX() > newpair.GetY() ){

//  cerr <<  oldp << "  "  << p << "  "  << newpair.GetX()  << "  "  << newpair.GetY() << endl;
mpz_class x = newpair.GetX();
mpz_class y = newpair.GetY();

if( (x+y >= oldp - 3 * l2)  &&(x+y <= p + 3 * l2))  //  mp_PrimeQ(x+y) &&
{
mpz_class t;
mpz_class previous;
mpz_class next;
if( x+y <= oldp && boo  )
{
t = oldp - 2 ;
while( !mp_PrimeQ(t) ) --t;
previous = t;
cout  << oldp << "*" << p <<   "  x " << x << " y " << y <<  " x+y  "  << x+y << "  previous " << previous  << "  middle  " << middle ;
if(x+y >= previous ) cout << " INTERESTING " ;
cout << endl;
} // if less
else  if( x+y >= p  && boo )
{
t = p + 2 ;
while( !mp_PrimeQ(t) ) ++t;
next = t;
cout << oldp << "*" << p  <<   "  x " << x << " y " << y  <<  " next " << next <<  " x+y  "  << x+y  << "  middle  " << middle ;
if(x+y <= next ) cout << " INTERESTING " ;
cout << endl;
} // if more
else if( oldp < x + y && x+y < p)  cout << oldp << "*" << p  <<   "  x " << x << " y " << y <<  " x+y  "  << x+y <<  " between  INTERESTING  "  << endl;

} // between logs

}  // if newpair
} // for composite
oldp = p;
oldpairs.clear();

for(iter = pairs.begin(); iter != pairs.end(); ++iter) {
mp_pair oldpair = *iter;
oldpairs.insert( oldpair);
}  // for iter

}  // if p prime

} // for p
cout << endl << endl;
system("date");
return 0;
}

//   g++  -o two_cubes two_cubes.cc  -lgmp -lgmpxx

• Thanks, so I take it 1001 is the only example up to a prime factor of 4600? Oct 27, 2020 at 12:21
• I think I can show that 35 is the only sum of two positive cubes equal to the product of two consecutive odd primes, what about products of 4 or more consecutive primes? Oct 27, 2020 at 12:23
• @pauldjackson recommend you copy my program, if you have C++ on your computer run it, get some idea of what you are dealing with. As far as the figure 46000, I checked up to about 15,000,000 yesterday, this program. The printout stopped at about 46000, then it is just the "progress" reports to keep a record of how far it had checked. This program would be easier to render in Python, for example. Gauss composition is a key part of this, you want to represent the product of $n-1$ primes as $x^2 - xy + y^2$ in all possible ways, then $x+y$ the final prime. Eisenstein integers... Oct 27, 2020 at 13:47
• @pauldjackson read the answer carefully before you ask for more work on my part. If, as appears, you lack the background to absorb it, ask separate questions until you have absorbed it. Oct 27, 2020 at 13:50
• @DavidDiaz the sum of two cubes factors as $(x+y)(x^2 - xy + y^2).$ We are asking about that product being squarefree. That requires $\gcd(x,y) = 1.$ That condition says the $x^2 - xy + y^2$ be a product of primes $1 \pmod 3$ or three times such a product. Once we pass the smallest consecutive primes, we use the fact that $x+y$ is considerably smaller than $x^2 - xy + y^2$ to conclude that we have $x^2 - xy + y^2 = pq$ and $x+y=r$ are the three primes. I concentrated on quickly finding all expressions for $x^2 - xy + y^2 = pq$ with $x > |y|$ Oct 28, 2020 at 13:05

Since the product of three consecutive odd primes is odd, the sum of the two cubes is odd and $$a$$, $$b$$ have opposite parity.

OP comments that $$35=3^3+2^3=5\cdot7$$ is the only sum of two cubes equal to the product of two consecutive odd primes. I note that it is also the least that is the product of two distinct odd primes. The next one is $$4^3+1^3=5\cdot13$$. Similarly here, besides being the product of three consecutive odd primes, $$10^3+1^3=7\cdot11\cdot13$$ is the least $$a^3+b^3$$ that is the product of three distinct primes. The next one is $$12^3+1^3=7\cdot13\cdot19$$.

I. In seeking a second solution for$$a^3+b^3=pqr$$supposing $$a>b$$ and $$p$$, $$q$$, $$r$$ consecutive odd primes, note that since$$a^3+b^3=(a+b)(a^2-ab+b^2)$$then for any given $$a$$, $$3$$ will be a factor of $$a^3+b^3$$ for $$(a+b)\equiv 0 \mod 3$$, that is for every third odd $$b$$ if $$a$$ is even, or even $$b$$ if $$a$$ is odd. Further, $$5$$ will be a factor when $$(a+b)\equiv 0\mod5$$, i.e. for every fifth odd/even $$b$$ when $$a$$ is even/odd. And similarly for $$7$$, $$11$$, $$13$$, and all odd primes.

But it appears that $$7$$ also divides $$(a^2-ab+b^2)$$ twice for every seven consecutive $$b$$, and thus divides $$(a^3+b^3)$$ for three out of every seven consecutive $$b$$. I gather this from inspection, but assume it can be proven.

For example, with $$a=73$$ and $$b=2, 4, 6, 8,...,72$$, then, in six rows of six, $$a^3+b^3=$$

$$(3^2\cdot5^2\cdot7\cdot13\cdot19), (7\cdot11\cdot31\cdot163), (13\cdot79\cdot379), (3^5\cdot7\cdot229), (37\cdot83\cdot127), (5\cdot17\cdot4597)$$$$(3^2\cdot19\cdot29\cdot79), (7\cdot89\cdot631),(7\cdot13\cdot4339), (3^2\cdot31\cdot1423), (5\cdot7\cdot19\cdot601), (97\cdot4153)$$$$(3^3\cdot11\cdot37^2), (13\cdot101\cdot313), (7\cdot103\cdot577), (3^2\cdot5\cdot7\cdot13\cdot103), (107\cdot4003), (7\cdot109\cdot571)$$$$(3^2\cdot31\cdot37\cdot43), (19\cdot113\cdot211), (5\cdot23\cdot4027), (3^3\cdot7\cdot13\cdot193), (7\cdot17\cdot61\cdot67), (11^2\cdot4129)$$$$(3^2\cdot7\cdot41\cdot199), (5^3\cdot19\cdot223), (13\cdot127\cdot331), (3^2\cdot43\cdot1459), (7^3\cdot13\cdot131), (7\cdot19\cdot4549)$$$$(3^4\cdot5\cdot1549), (7^2\cdot97\cdot137), (31\cdot139\cdot157), (3^2\cdot47\cdot1663), (11\cdot13\cdot5119), (5\cdot7\cdot29\cdot751)$$

II. Since neither $$3\cdot5\cdot7=105$$ nor $$5\cdot7\cdot11=385$$ is the sum of two cubes, and hence neither $$3$$ nor $$5$$ can be among the three consecutive prime factors of any solution, then for any given $$a$$ we can disregard one-third of all $$b$$, plus two-thirds of one-fifth of all $$b$$ (since one in three $$a+b$$ divisible by $$5$$ is also divisible by $$3$$ and so $$b$$ for that case has already been removed):$$\frac{1}{3}+\frac{2}{3}\cdot\frac{1}{5}=\frac{5}{15}+\frac{2}{15}=\frac{7}{15}>46\%$$of all $$b$$ are ruled out.

And since $$7$$ was the least of the three consecutive odd primes in the one known solution, it cannot appear in any greater solution, and the fraction of $$b$$ excluded will now be$$\frac{7}{15}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{3}{7}=\frac{49}{105}+\frac{24}{105}=\frac{73}{105}>69\%$$

Again, since $$11\cdot13\cdot17=12^3+9^3+3^3-1$$ is not a solution, $$11$$ cannot be a factor in a second solution.

And since $$13\cdot17\cdot19=14^3+11^3+5^3-1$$ is not a solution, we can likewise exclude $$13$$ from any other solution. And here again, as happens with $$7$$, and is also clear in the example above, $$13$$ divides $$a^3+b^3$$ not only for $$(a+b)\equiv 0\mod13$$, but also for another two of every thirteen consecutive $$b$$ (i.e. when $$13$$ divides $$a^2-ab+b^2$$).

With $$11$$ and $$13$$ ruled out, the portion of $$b$$ excluded increases to$$\frac{73}{105}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{1}{11}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{10}{11}\cdot\frac{3}{13}=\frac{805}{1001}>80\%$$

Finally, if we grant that $$17\cdot19\cdot23=19^3+8^3+58$$ and $$19\cdot23\cdot29=23^3+6^3+290$$ are not sums of two cubes, then we can rule out $$17$$ and $$19$$ from any solution, and the fraction of all $$b$$ excluded is$$\frac{805}{1001}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{10}{11}\cdot\frac{10}{13}\cdot\frac{1}{17}+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot\frac{10}{11}\cdot\frac{10}{13}\cdot\frac{16}{17}\cdot\frac{3}{19}=\frac{4114625}{4849845}\approx85\%$$

III. I know that taking this approach in response to OP's request for "a simpler solution," may seem like embarking on a long ground war with little prospect of a decisive victory. But I was surprised to find how great an advance is made as we exclude primes $$p=7, 13, 19,...\equiv 1\mod6$$, since the above example clearly suggests (and I assume it can be proven) that these are factors of $$a^3+b^3$$ for three of every $$p$$ consecutive $$b$$, instead of just one as happens when $$p=5, 11, 17,...\equiv 5\mod6$$. Small odd primes, especially those $$\equiv 1\mod6$$, are so plentiful as factors of $$a^3+b^3$$ that excluding just the first seven odd primes eliminates a large portion of possible solutions of$$a^3+b^3=pqr$$where $$p$$, $$q$$, $$r$$ are consecutive odd primes.