Theorem: N (n + 1) (2n + 1) over 6 is Integer

Let $n \in \Z$ be an integer.
Then $\dfrac {n \paren {n + 1} \paren {2 n + 1} } 6$ is also an integer.

Proof:
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}}