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

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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 1

Let $a$ be as defined.

Then:

\(\ds 24\) \(\divides\) \(\ds \paren {a^2 - 1}\) Square Modulo 24 of Odd Integer Not Divisible by 3
\(\ds \leadsto \ \ \) \(\ds 24\) \(\divides\) \(\ds \paren {a^2 - 1 + 24}\)
\(\ds \leadsto \ \ \) \(\ds 24\) \(\divides\) \(\ds \paren {a^2 + 23}\)

$\blacksquare$


Proof 2

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:

\(\ds a^2 + 23\) \(=\) \(\ds \paren {6 k + 5}^2 + 23\)
\(\ds \) \(=\) \(\ds 36 k^2 + 60 k + 48\)
\(\ds \) \(=\) \(\ds 24 \paren {\dfrac {k \paren {3 k + 5} } 2 + 2}\)

$\blacksquare$


Sources

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