# 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:

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

$\blacksquare$