2

I recently found out about tritriduoprismic numbers (squares of triangular numbers), related to a 4D figure called tritriduoprism (which is a product of two triangles). The first few are 0, 1, 9, 36, 100, 225, 441, and so on. Notice that 0, 1 and 36 are triangular numbers themselves. In order to look for more, I wrote a Python script, as below :

import numpy as np

def square_tri(k):
    #Tests if k² is triangular


a = np.sqrt(2*k**2+0.25) - 0.5

return a==int(a)




c=0 #c is a counter, it will take all the integer values
s=0 #s is the c-th triangular number


for i in range(244452):
    if square_tri(s) :
        print(s)
    c+=1
    s+=c
print('The squares of these are triangular numbers')
print(f'Last number tested : {s}')

The output is as follows :

1
6
The squares of these are triangular numbers
Last number tested : 29878512378

0 isn't included because of rounding errors, and bigger numbers aren't well managed either. The point is, there aren't any such numbers (meaning triangular numbers that are squares of triangular numbers) below 29878512378, besides 0, 1 and 36. My conjecture is there aren't any at all, but I can't manage to prove it. Does anyone have ideas ?

Guilarai
  • 31
  • So you're asking for solutions to $ \frac{ n(n+1) } { 2 } = [ \frac{ m(m+1) } { 2 } ] ^2 $? $\quad$ As a first step, are you familiar with how to solve $ \frac{ n(n+1) } { 2} = k^2$? – Calvin Lin Jan 14 '25 at 19:38
  • This will result in a Diophantine equation involving two odd squares. Basically, $\frac{(n(n+1)}2=\frac{(2n+1)^2-1}8.$ The equation is of degree $4,$ so I don't if there is an elementary way to find all solutions. – Thomas Andrews Jan 14 '25 at 19:52
  • 2
    The equation seems to become $$(q^2-1)^2-8p^2=-8,$$ with $p,q$ odd natural numbers. – Thomas Andrews Jan 14 '25 at 20:01
  • 1
    If there is an elementary way to describe the solutions, my intuition says it would be to find all of a finite set of solutions and be able to prove there are no others. An infinite set of solutions seems much harder to give. That is only an intuition, though. – Thomas Andrews Jan 14 '25 at 20:06
  • The obvious solutions are $0=0^2$ and $1=1^2,$ which amounts to $(p,q)=(1,1)$ or $(3,3).$ – Thomas Andrews Jan 14 '25 at 20:12
  • The solutions to $u^2-8p^2=-8$ can (relatively) easily be described, but it is exponential, and finding when $u$ is one less than an odd square seems hard. – Thomas Andrews Jan 14 '25 at 20:15
  • You can avoid rounding errors by computing the square root with a integral version of Heron’s method. – Christophe Boilley Jan 14 '25 at 20:53
  • @CalvinLin if you're referring to Euler's formula, I came across it while looking for ideas about this problem. The solutions are expressed as a square, so I would just have to check if the base is triangular... but I don't see how to do that either. – Guilarai Jan 15 '25 at 19:00
  • @ThomasAndrews could you elaborate about how you come to this $(p,q)$ equation ? I tried to find it myself, and I came to $(2n+1)^2-8p^2=1$ with $p=\frac{m(m+1)}{2}$. – Guilarai Jan 15 '25 at 19:14
  • @ChristopheBoilley could you maybe point me to some reference about the use of this method ? – Guilarai Jan 15 '25 at 19:16
  • Well, turns out there are indeed no such numbers besides 0, 1 and 36. A proof is this is available in this reference : "Integral points on certain elliptic curves" by J.W.S. Cassels, Proc. Lond. Math. Soc., III. Ser. 14 A (1965), 55-57, but alas I couldn't access it. – Guilarai Jan 31 '25 at 16:52

0 Answers0