This is equivalent to proving that $n \paren {n + 1} \paren {2 n + 1}$ is a multiple of $6$. There are $6$ cases to consider: :$(1): \quad n \equiv 0 \pmod 6$: we have $n = 6 k$ :$(2): \quad n \equiv 1 \pmod 6$: we have $n = 6 k + 1$ :$(3): \quad n \equiv 2 \pmod 6$: we have $n = 6 k + 2$ :$(4): \quad n \equiv 3 \pmod 6$: we have $n = 6 k + 3$ :$(5): \quad n \equiv 4 \pmod 6$: we have $n = 6 k + 4$ :$(6): \quad n \equiv 5 \pmod 6$: we have $n = 6 k + 5$ {{begin-eqn}} {{eqn | n = 1       | l = n       | r = 6 k       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | r = \paren {6 k} \paren {6 k + 1} \paren {2 \paren {6 k} + 1}       | c =  }} {{eqn | r = 6 \paren {k \paren {6 k + 1} \paren {12 k + 1} }       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | o = \equiv       | r = 0 \pmod 6       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 2       | l = n       | r = 6 k + 1       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | r = \paren {6 k + 1} \paren {6 k + 2} \paren {2 \paren {6 k + 1} + 1}       | c =  }} {{eqn | r = \paren {6 k + 1} \paren {6 k + 2} \paren {12 k + 3}       | c =  }} {{eqn | r = \paren {6 k + 1} \paren {2 \paren {3 k + 1} } \paren {3 \paren {4 k + 1} }       | c =  }} {{eqn | r = 6 \paren {6 k + 1} \paren {3 k + 1} \paren {4 k + 1}       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | o = \equiv       | r = 0 \pmod 6       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 3       | l = n       | r = 6 k + 2       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | r = \paren {6 k + 2} \paren {6 k + 3} \paren {2 \paren {6 k + 2} + 1}       | c =  }} {{eqn | r = \paren {2 \paren {3 k + 1} } \paren {3 \paren {2 k + 1} } \paren {12 k + 5}       | c =  }} {{eqn | r = 6 \paren {3 k + 1} \paren {2 k + 1} \paren {12 k + 5}       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | o = \equiv       | r = 0 \pmod 6       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 4       | l = n       | r = 6 k + 3       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | r = \paren {6 k + 3} \paren {6 k + 4} \paren {2 \paren {6 k + 3} + 1}       | c =  }} {{eqn | r = \paren {3 \paren {2 k + 1} } \paren {2 \paren {3 k + 2} } \paren {12 k + 7}       | c =  }} {{eqn | r = 6 \paren {2 k + 1} \paren {3 k + 2} \paren {12 k + 7}       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | o = \equiv       | r = 0 \pmod 6       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 5       | l = n       | r = 6 k + 4       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | r = \paren {6 k + 4} \paren {6 k + 5} \paren {2 \paren {6 k + 4} + 1}       | c =  }} {{eqn | r = \paren {6 k + 4} \paren {6 k + 5} \paren {12 k + 9}       | c =  }} {{eqn | r = \paren {2 \paren {3 k + 2} } \paren {6 k + 5} \paren {3 \paren {4 k + 3} }       | c =  }} {{eqn | r = 6 \paren {3 k + 2} \paren {6 k + 5} \paren {4 k + 3}       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | o = \equiv       | r = 0 \pmod 6       | c =  }} {{end-eqn}} {{qed|lemma}} {{begin-eqn}} {{eqn | n = 6       | l = n       | r = 6 k + 5       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | r = \paren {6 k + 5} \paren {6 k + 6} \paren {2 \paren {6 k + 5} + 1}       | c =  }} {{eqn | r = 6 \paren {6 k + 5} \paren {k + 1} \paren {12 k + 11}       | c =  }} {{eqn | ll= \leadsto       | l = n \paren {n + 1} \paren {2 n + 1}       | o = \equiv       | r = 0 \pmod 6       | c =  }} {{end-eqn}} {{qed}}