Problems & Puzzles: Puzzles

Puzzle 1066 4 consecutive triangular and sphenic numbers

G. L. Honaker, Jr. sent the following nice puzzle:

406 is the smaller number in the first set of three consecutive triangular numbers that are sphenic {406, 435, 465}.

406=28*29/2 = 2*7*29
435=29*30/2 = 3*5*29
465=30*31/2 = 3*5*31

Q. Does there exist a set of four consecutive triangular numbers that are sphenic?

During the week 4-10 Dec. 2021, contributions came from Adam Stinchcombe, Shyam Sunder Gupta, Fred Schneider, Giorgos Kalogeropoulos, Emmanuel Vantieghem, Oscar Volpatti


Adam wrote:

You can prove there are no four consecutive triangular, sphenic numbers with a modular argument.  Call n the index of the triangular number n*(n+1)/2.  Consider the smallest index of the four consecutive indices, and write it as 6n+k.  By cases:

k=0.  6n*(6n+1)/2 = 3*n*(6n+1) and to be sphenic 6n+1 must be prime.  From (6n+1)*(6n+2)/2 = (6n+1)*(3n+1) and 3n+1 is the product of two primes since 6n+1 was prime.  Then (6n+2)*(6n+3)/2 = (3n+1)*3*(2n+1) has too many prime factors.  Other arguments are similar.

k=1. (6n+2)(6n+3)/2=(3n+1)*3*(2n+1) and (6n+3)(6n+4)/2=3*(2n+1)*(3n+2) and 3n+1,3n+2 are consecutive numbers which are both prime.

k=2.  Same problem as k=1.

k=3. From (6n+5)(6n+6)/2 = (6n+5)*3*(n+1) so 6n+5 is prime. From (6n+4)(6n+5)/2=(3n+2)(6n+5), we see 3n+2 is the product of two primes.  Finally, (6n+3)(6n+4)/2=3(2n+1)(3n+2) has too many factors.

k=5.  From (6n+8)(6n+9)/2 = (3n+4)(3)(2n+3) and (6n+6)(6n+7)/2=3(n+1)(6n+7) both 3n+4 and 6n+7 are prime.  Then (6n+7)(6n+8)/2 = (6n+7)(3n+4) has too few prime factors.

The case k=4 is broken down into two cases: 4 mod 12 or 10 mod 12.

4 mod 12: (12n+4)(12n+5)/2 = 2(3n+1)(12n+5) and (12n+7)(12n+8)/2 =(12n+7)*2*(3n+2) and 3n+1,3n+2 are not both prime.

10 mod 12: (12n+11)(12n+12)/2 = (12n+11)*2*3*(n+1) is a product of at least 4 primes.

Any small "counterexamples" can be ruled out by an exhaustive search: for instance, 2 and 3 are consecutive numbers and prime, at n=0 (n+1) has no prime factors, etc.



Shyam wrote:

Puzzle 1066 4 consecutive triangular and sphenic numbers
Q. Does there exist a set of four consecutive triangular numbers that are sphenic?

In reference to puzzle 1066, My observations are as under:
Every fourth consecutive number is divisible by 4 so cannot be a sphenic number

but it is not necessary for triangular numbers. However I could not find any set of 4

consecutive sphenic triangular numbers up to 2*10^16.


Fred wrote:

Let's consider 3 consecutive triangular numbers first:

Among 4 consecutive numbers, two must be even and one must be a multiple of 4.  In this case, that multiple of 4, can not be a multiple of 8 because the triangular number(s) involved in this complication would have a repeated factor of 2 (2^3 / 2 = 2^2) and thus not sphenic.

a, b, c, d, e are all distinct prime numbers not equal to two.  Note:  The triangular numbers formed from each pair must all be sphenic (products of 3 distinct prime numbers).

v1 = 2 * a * b
v2 = c
v3 = 4 * d
v4 = e

This is the only way (barring rotation) of creating 3 consecutive triangular, sphenic numbers.

But, we have not considered multiples of 3, every set of 4 consecutive numbers must have at least one multiple of 3.  In this case, if v1 was a multiple of 3, so must v4 but v4 is prime.  So, the first number (or 4th) must not be a multiple of 3.

So, there are two possible scenarios, where the multiple of 3 is either 2nd or 3rd of the 4 numbers.  (a, b, c, and d are distinct primes.  3 is now fixed)

First case:
v1 = a
v2 = 2 * 3 * b
v3 = c
v4 = 4 * d

Second case:
v1 = 4 * a
v2 = b
v3 = 2 * 3 * c
v4 = d

Let's try to add a fifth number to the set of four consecutive numbers.  To avoid having a second multiple of 3, the lone multiple of 3 would have to be the middle number of the 5. 
In the first case, we need to prepend a number.
In the second case, append.

If an integer is a multiple of 4 but not a multiple of 8, it is of the form 8n + 4

When we prepend a number to the first case, we get a "v0" = 8n + 4 - 4 = 8n
When we append a number to the second case, we get a v5 = 8n + 4 + 4 = 8(n+1)

So, it is not possible to have a set of four consecutive triangular, sphenic numbers.



Giorgos wrote:

The answer is no. Such a set doesn't exist.
n(n+1)/2, (n+1)(n+2)/2, (n+2)(n+3)/2, (n+3)(n+4)/2 cannot be all sphenic.
1) n is even.
    If n is even then in order for n(n+1)/2 to be sphenic we have 2 cases:
    a) n=2*p1*p2  and  n+1=q1 
    b) n=2*p1       and  n+1=q1*q2 (where p1,p2,q1,q2 are all distinct primes)
    We will examine these 2 cases:
    1a)  (n+1)(n+2)/2 = (p1*p2+1)*q1
           In order to be sphenic, p1*p2+1=P1*P2 (where P1,P2 two distinct odd primes)
           This means that p1*p2+1 is odd and p1=2. So, n=4*p2 and n+2=6k (1)
           n+2= 2*(p1*p2+1) = 2*P1*P2 and from (1) P1=3
           Finally (n+1)(n+2)/2 = 3*q1*P2.
           Sphenic-> (n+2)(n+3)/2 = P1*P2*q2 = 3*P2*q2 (where {q1,q2} twin primes)
           Sphenic-> (n+3)(n+4)/2 = q2(2*p1*p2+4)/2 = q2(p1*p2+2) = 2* q2*(p2+1)
           This means that p2+1 should be an odd prime and p2=2.
           This is a contradiction because p2>p1=2.
   1b) (n+1)(n+2)/2 = (p1+1)*q1*q2
          In order to be sphenic p1+1 must be prime. This means p1=2 and n=4 which is not a solution.

2) n is odd.
    If n is odd then in order for n(n+1)/2 to be sphenic we have 2 cases:
    a) n=p1        and   n+1=2*q1*q2
    b) n=p1*p2  and    n+1= 2*q1
    2a) (n+1)(n+2)/2 = p1*q1*q2
          (n+2)(n+3)/2 = (q1*q2+1)(p1+2)
          In order to be sphenic q1*q2+1 = 2*P1 and p1+2=P2 (where P1,P2 odd primes and {p1,P2} twin primes)
          {n+3)(n+4)/2 = (2*q1*q2+2)(P2+1)/2 = (q1*q2+1)(P2+1) = 2*P1*(P2+1)
          In order to be sphenic P2+1 must be odd prime. This means P2=2 which is impossible because P2=p1+2.
    2b) (n+1)(n+2)/ 2 = q1(p1*p2+2). In order to be sphenic we have p1*p2+2 = P1*P2 (with P1,P2 odd primes).
          (n+2)(n+3)/2 = P1*P2*(2q1+2)/2 = P1*P2*(q1+1). In order to be sphenic q1+1 must be prime and q1=2.
          This means that n=3 which is not a solution.



Emmanuel wrote:

There are no such "quartets" !

Proof :
Let's denote by  T(m)  the  m-th triangular number  m(m+1)/2.
Regarding all numbers modulo 8 we find following table :
   m      0   1   2   3   4   5   6  7
  T(m)   0   1   3   2   2   3   1  0
We see immediately that if four consecutive triangular numbers are sphenic, they must be of the form
   T(m), T(m+1), T(m+2), T(m+3)  with  m  ===  1, 2 or 3  modulo 8.
This implies that in all three cases, T(8w+3) and T(8w+4) must be sphenic numbers.
The only possibility to obtain this is that  8w+4  must be  4p, with p prime
and  8w+3  and  8w+5  both prime.  That is only possible when  w = 1.
It is easily seen that this does not give four sphenic consecutive triangular numbers, QED.


Oscar wrote:

The answer is no; there can be sets of at most 3 consecutive triangular and sphenic numbers, either from T(24*q+4) to T(24*q+6) or from T(24*q+17) to T(24*q+19).
The given example T(28)=406 belongs to the first family, with q=1.

Triangular numbers T(8*k) and T(8*k+7) are divisible by 4, so they aren't squarefree.
T(8*k) = 4*k*(8*k+1)
T(8*k+7) = 4*(k+1)*(8*k+7)
Hence there can be blocks of length at most 6, from T(8*k+1) to T(8*k+6).

But consecutive triangular numbers T(8*k+3) and T(8*k+4) can be both sphenic only for k=1.
T(8*k+3) = 2*(2*k+1)*(8*k+3)
T(8*k+4) = 2*(2*k+1)*(8*k+5)
For k=0 we obtain two semiprimes, so let k>0; we need the odd factors 2*k+1, 8*k+3, 8*k+5 to be primes.
They belong to distinct residue classes modulo 3, as:
8*k+3 == 2*k+0 (mod 3)
8*k+5 == 2*k+2 (mod 3)
Hence at least one of them must be divisible by 3.

The only way to obtain three primes is when the smallest of them is 3 itself, for k=1.
Indeed T(11)=66 and T(12)=78 are sphenic numbers, but T(10)=55 and T(13)=91 are semiprimes, so in this case we obtain a block of length 2.

For k>1 there can be blocks of length at most 3:
from T(8*k+1) to T(8*k+3), if k=3*q+2, so that T(8*k+4) only is not sphenic;
from T(8*k+4) to T(8*k+6), if k=3*q, so that T(8*k+3) only is not sphenic.




Records   |  Conjectures  |  Problems  |  Puzzles