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

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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:

\(\ds 1 \paren {1^2 - 1} \paren {3 \times 1 + 2}\) \(=\) \(\ds 0\)
\(\ds \) \(\equiv\) \(\ds 0\) \(\ds \pmod {24}\)


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:

\(\ds \paren {k + 1} \paren {\paren {k + 1}^2 - 1} \paren {3 \paren {k + 1} - 2}\) \(=\) \(\ds k \paren {k + 1} \paren {k + 2} \paren {3 k + 5}\)
\(\ds \) \(=\) \(\ds k \paren {k + 1} \paren {k - 1 + 3} \paren {3 k + 2 + 3}\)
\(\ds \) \(=\) \(\ds k \paren {k + 1} \paren {k - 1} \paren {3 k + 2 + 3} + 3 k \paren {k + 1} \paren {3 k + 2 + 3}\)
\(\ds \) \(=\) \(\ds k \paren {k + 1} \paren {k - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {k - 1} + 3 k \paren {k + 1} \paren {3 k + 5}\)
\(\ds \) \(=\) \(\ds k \paren {k^2 - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {k - 1 + 3 k + 5}\)
\(\ds \) \(=\) \(\ds k \paren {k^2 - 1} \paren {3 k + 2} + 3 k \paren {k + 1} \paren {4 k + 4}\)
\(\ds \) \(=\) \(\ds k \paren {k^2 - 1} \paren {3 k + 2} + 12 k \paren {k + 1}^2\)


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:

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

$\blacksquare$


Sources

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