24 divides Square of Odd Integer Not Divisible by 3 plus 23/Proof 2

Theorem

Let $a \in \Z$ be an integer such that:

$2 \nmid a$

$3 \nmid a$

where $\nmid$ denotes non-divisibility.

Then:

$24 \divides \paren {a^2 + 23}$

where $\divides$ denotes divisibility.

Proof

Let $a$ be as defined.

We have that $a$ is of the form:

$a = 6 k + 1$

or:

$a = 6 k + 5$

Hence:

\(\ds a^2 + 23\)

\(=\)

\(\ds \paren {6 k + 1}^2 + 23\)

\(\ds \)

\(=\)

\(\ds 36 k^2 + 12 k + 24\)

\(\ds \)

\(=\)

\(\ds 24 \paren {\dfrac {k \paren {3 k + 1} } 2 + 1}\)

or:

\(\ds a^2 + 23\)

\(=\)

\(\ds \paren {6 k + 5}^2 + 23\)

\(\ds \)

\(=\)

\(\ds 36 k^2 + 60 k + 48\)

\(\ds \)

\(=\)

\(\ds 24 \paren {\dfrac {k \paren {3 k + 5} } 2 + 2}\)

$\blacksquare$