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

Theorem

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 $n = 6 k + 5$

\(\text {(1)}: \quad\)

\(\ds n\)

\(=\)

\(\ds 6 k\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(=\)

\(\ds \paren {6 k} \paren {6 k + 1} \paren {2 \paren {6 k} + 1}\)

\(\ds \)

\(=\)

\(\ds 6 \paren {k \paren {6 k + 1} \paren {12 k + 1} }\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(\equiv\)

\(\ds 0 \pmod 6\)

$\Box$

\(\text {(2)}: \quad\)

\(\ds n\)

\(=\)

\(\ds 6 k + 1\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(=\)

\(\ds \paren {6 k + 1} \paren {6 k + 2} \paren {2 \paren {6 k + 1} + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {6 k + 1} \paren {6 k + 2} \paren {12 k + 3}\)

\(\ds \)

\(=\)

\(\ds \paren {6 k + 1} \paren {2 \paren {3 k + 1} } \paren {3 \paren {4 k + 1} }\)

\(\ds \)

\(=\)

\(\ds 6 \paren {6 k + 1} \paren {3 k + 1} \paren {4 k + 1}\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(\equiv\)

\(\ds 0 \pmod 6\)

$\Box$

\(\text {(3)}: \quad\)

\(\ds n\)

\(=\)

\(\ds 6 k + 2\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(=\)

\(\ds \paren {6 k + 2} \paren {6 k + 3} \paren {2 \paren {6 k + 2} + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {2 \paren {3 k + 1} } \paren {3 \paren {2 k + 1} } \paren {12 k + 5}\)

\(\ds \)

\(=\)

\(\ds 6 \paren {3 k + 1} \paren {2 k + 1} \paren {12 k + 5}\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(\equiv\)

\(\ds 0 \pmod 6\)

$\Box$

\(\text {(4)}: \quad\)

\(\ds n\)

\(=\)

\(\ds 6 k + 3\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(=\)

\(\ds \paren {6 k + 3} \paren {6 k + 4} \paren {2 \paren {6 k + 3} + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {3 \paren {2 k + 1} } \paren {2 \paren {3 k + 2} } \paren {12 k + 7}\)

\(\ds \)

\(=\)

\(\ds 6 \paren {2 k + 1} \paren {3 k + 2} \paren {12 k + 7}\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(\equiv\)

\(\ds 0 \pmod 6\)

$\Box$

\(\text {(5)}: \quad\)

\(\ds n\)

\(=\)

\(\ds 6 k + 4\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(=\)

\(\ds \paren {6 k + 4} \paren {6 k + 5} \paren {2 \paren {6 k + 4} + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {6 k + 4} \paren {6 k + 5} \paren {12 k + 9}\)

\(\ds \)

\(=\)

\(\ds \paren {2 \paren {3 k + 2} } \paren {6 k + 5} \paren {3 \paren {4 k + 3} }\)

\(\ds \)

\(=\)

\(\ds 6 \paren {3 k + 2} \paren {6 k + 5} \paren {4 k + 3}\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(\equiv\)

\(\ds 0 \pmod 6\)

$\Box$

\(\text {(6)}: \quad\)

\(\ds n\)

\(=\)

\(\ds 6 k + 5\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(=\)

\(\ds \paren {6 k + 5} \paren {6 k + 6} \paren {2 \paren {6 k + 5} + 1}\)

\(\ds \)

\(=\)

\(\ds 6 \paren {6 k + 5} \paren {k + 1} \paren {12 k + 11}\)

\(\ds \leadsto \ \ \)

\(\ds n \paren {n + 1} \paren {2 n + 1}\)

\(\equiv\)

\(\ds 0 \pmod 6\)

$\blacksquare$

Sources

• 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $4$