How to see that the difference between two successive squared triangular numbers is a cube

		Shel Kaphan, 12 Jan 2023


Start with a triangular number T(n) of side length n.  
T(n) = n(n+1)/2.  For the diagrams, n=3.

                       o
 



                  o         o




             o         o         o


Square it by replacing each point with a copy of the triangle.
The resulting figure now has T(n)^2 points:

                       o
                      o o
                     o o o


                  o         o
                 o o       o o
                o o o     o o o


             o         o         o
            o o       o o       o o
           o o o     o o o     o o o


Add new points to form the next squared triangular number, T(n+1)^2, 
keeping track of the new points.  We must add n+1 triangles
that are n+1 on a side, and a new row of length n+1 to each of the
existing T(n) triangles of size T(n) each.

 
                       o
                      o o
                     o o o
                    x x x x

                  o         o
                 o o       o o
                o o o     o o o
               x x x x   x x x x

             o         o         o
            o o       o o       o o
           o o o     o o o     o o o
          x x x x   x x x x   x x x x
 
        x         x         x         x
       x x       x x       x x       x x
      x x x     x x x     x x x     x x x 
     x x x x   x x x x   x x x x   x x x x




Keeping only the newly added points, we get:
                       
                      
                     
                    x x x x

                  
                 
                
               x x x x   x x x x

             
            
           
          x x x x   x x x x   x x x x
 
        x         x         x         x
       x x       x x       x x       x x
      x x x     x x x     x x x     x x x 
     x x x x   x x x x   x x x x   x x x x



We can organize these new points into two "prisms", each of
length n+1. There are T(n) rows of length n+1 from the existing
T(n) triangles, and there are n+1 new triangles of edge length n+1
and size T(n+1).

                                       x
                             x        x x
                   x        x x      x x x
         x        x x      x x x
        x x      x x x
       x x x                               x                                         
                                 x        x x
                       x        x x      x x x
             x        x x      x x x    x x x x
            x x      x x x    x x x x
           x x x    x x x x
          x x x x


Now flip over the smaller prism:

                                     x x x
                           x x x      x x
                 x x x      x x        x
       x x x      x x        x
        x x        x                       x
         x                       x        x x
                       x        x x      x x x
             x        x x      x x x    x x x x
            x x      x x x    x x x x
           x x x    x x x x
          x x x x

 

Merge the prisms.  T(n) = n(n+1)/2 and T(n+1) = (n+1)(n+2)/2, so we have
a 3-d figure, a parallelepiped, the ends of which have
n(n+1)/2 + (n+1)(n+2)/2 = (n+1)(2n+2)/2 = n^2 + 2n + 1 = (n+1)^2
points and the length of which is n+1:

                                     x x x x 
                           x x x x    x x x x
                 x x x x    x x x x    x x x x
       x x x x    x x x x    x x x x    x x x x
        x x x x    x x x x    x x x x
         x x x x    x x x x
          x x x x


Straighten it out into a (n+1)^3 cube:

                                     x x x x 
                           x x x x   x x x x
                 x x x x   x x x x   x x x x
       x x x x   x x x x   x x x x   x x x x
       x x x x   x x x x   x x x x
       x x x x   x x x x
       x x x x



The difference between T(n)^2 and T(n+1)^2 is (n+1)^3.
so the squared triangular numbers are the sum of cubes.