Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 4 / 1C (Numbers and Sets)

Let $T_n=n(n+1)/2$, the $n$th triangular number. Pairing the terms of the arithmetic progression proves ${\sum }_{r=1}^{n}r=T_n$. We prove the cube identity by mathematical induction. For $n=1$ both sides equal one. If ${\sum }_{r=1}^n{r}^3=T_n^2$, thenAdding $(n+1{)}^3$ establishes the next case. ThusEquivalently, the same displayed difference telescopes from $T_0=0$, providing a direct proof without induction.

An integer square is divisible by four exactly when its base is even: an even base has square $4j^2$, while an odd base has square $(2j+1{)}^2\equiv 1(\mathrm{mod}4)$. Apply this observation to the triangular number $T_n$ obtained above. Since $n(n+1)$ is even,Checking the four residue classes of $n$ modulo four, the products $n(n+1)$ have residues $0,2,2,0$, respectively. ThereforeThe original PDF has two alternatives modulo four; the TeX transcription incorrectly turns the second alternative into a modulus-three expression.