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For every nonnegative integer $n$,For example, $1^3+2^3+3^3+4^3=1+8+27+64=100=T_4^2$. A short proof uses the triangular number recurrence and the triangular number square-complement identity:Summing this difference of squares identity gives a telescoping sum; all intermediate squared triangular numbers cancel. Nicomachus describes a related grouping of the odd integers: take one, then the next two, then the next three, producing $1,3+5,7+9+11,\dots$, which are $1^3,2^3,3^3,\dots$. This historical grouping is recorded in Introduction to Arithmetic by Nicomachus (Hahn teacher edition) (Book II, Chapter 20, §5, printed p. 75).

A squared triangular number is the square of a number $T_n$, where $T_n$ is a triangular number. Beginning at $n=0$, the values are $0,1,9,36,100,225,441,\dots$. Nicomachus's theorem gives them a second description as cumulative sums of cubes of numbers.

A square triangular number is simultaneously a triangular number and a perfect square. For example, $36=T_8=6^2$. This differs from a squared triangular number: $T_4^2=100$ is not itself triangular, by the recognition criterion for triangular numbers. Writing $T_n=m^2$ gives $(2n+1{)}^2-8m^2=1$, a Pell equation that explains why this intersection of two counting patterns has additional arithmetic structure.