### Oblong-Triangular Relationship

After students noted the relationship between square and oblong numbers depicted in Figure 8, the next task was to model any oblong number \(n(n+1)\) as the sum of two equal triangular numbers; see Figure 9a from Thomas Heath's *A History of Greek Mathematics: From Thales to Euclid* (1921). All students immediately noted the *one-half* relation between an oblong number and the corresponding triangular number (both analytically and visually) and suggested an oblong number modeling based on two congruent triangular numbers (see Figures 9b and 9c).

**Figure 9:** Oblong-triangular relationship with Cuisenaire rods

Theorem of Theon of Smyrna

Upon establishing the oblong-triangular relationship, students focused on the theorem of Theon, which states that any square number is expressible as the sum of two consecutive triangular numbers. As Thomas Heath explained, referring to Figure 10a:

*In this case, as is seen from the figure, the sides of the triangles differ by unity, and of course* \[{\frac{1}{2}}n(n-1)+\frac{1}{2}n(n+1)=n^2\quad{\sf{\small{(1921, pp. 83{\mbox{-}}84).}}}\]

Figures 10b and 10c depict Cuisenaire rod representations of Theon’s theorem.

**Figure 10:** Theon’s theorem, the sum of two consecutive triangular numbers is a square, or \(\frac{1}{2}n(n-1)+\frac{1}{2}n(n+1)=n^2,\) with Cuisenaire rods

Triangular Numbers and Squares: Another Connection

As "quoted by Plutarch and used by Diophantus" (1921, p. 84), the theorem stating

*that 8 times any triangular number plus 1 makes a square …* *is equivalent to the formula*

\[{8\cdot\frac{1}{2}}n(n+1)+1=4n(n+1)+1={(2n+1)}^2\quad{\sf\small{{(1921, p. 84).}}}\]

Students first verified the formula given in the text; they then focused on generating Cuisenaire rod patterns modeling the given template (see Figure 11a):

*It may easily have been proved by means of a figure made up of dots in the usual way. Two equal triangles make up an oblong figure of the form *\(n(n+1),\)* as above. Therefore we have to prove that four equal figures of this form with one more dot make up *\({(2n+1)}^2.\)* The annexed figure representing *\(7^2\)* shows how it can be divided into four 'oblong' figures *\(3\cdot4\)* leaving *\(1\)* over.* (1921, p. 84)

Figures 11b and 11c depict Cuisenaire rod representations of this ancient theorem. Whereas one group of students focused on the “4 oblong plus 1” approach (see Figure 11b), other students were more explicit in their representations by embracing the “8 triangular plus 1” approach (see Figure 11c).

**Figure 11:** "Eight times any triangular number plus one makes a square" with Cuisenaire rods