Modulo Arithmetic/Examples/n(n^2-1)(3n-2) Modulo 24

Example of Modulo Arithmetic

 * $n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $k \paren {k^2 - 1} \paren {3 k + 2} \equiv 0 \pmod {24}$

from which it is to be shown that:
 * $\paren {k + 1} \paren {\paren {k + 1}^2 - 1} \paren {3 \paren {k + 1} + 2} \equiv 0 \pmod {24}$

Induction Step
This is the induction step:

By the induction hypothesis:
 * $k \paren {k^2 - 1} \paren {3 k + 2} = 24 r$

for some $r \in \Z$.

Take $12 k \paren {k + 1}^2$.

If $k$ is even, then $12 k$ and so $12 k \paren {k + 1}^2$ is divisible by $24$.

If $k$ is odd, then $k + 1$ is even and so $12 k \paren {k + 1}^2$ is again divisible by $24$.

Thus:
 * $12 k \paren {k + 1}^2 = 24 s$

for some $s \in \Z$.

Thus:
 * $k \paren {k^2 - 1} \paren {3 k + 2} + 12 k \paren {k + 1}^2 = 24 \paren {r + s}$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{>0}: n \paren {n^2 - 1} \paren {3 n + 2} \equiv 0 \pmod {24}$