Prove the assertions of the following problems
Prove that the expression $n^{3} - n$ is divisible by $24$ for any odd $n$ .

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Solution

$n = 1$ -----> $n (n ² - 1) = 1 (1 - 1) = 1 (0) = 0$
$n = 3$ -----> $n (n ² - 1) = 3 (9 - 1) = 3 (8) = 24$
We can see that this is true for $n = 1$ and $3$
Assume statement is true for $n = 2 k - 1$ . Then we must show it is true for $n = 2 k + 1$
Statement true for $n = 2 k - 1$
$(2 k - 1) ((2 k - 1) ² - 1) = 24 m$ , for some integer m
$(2 k - 1) ((2 k - 1) ² - 1) = 24 m$
$(2 k - 1) ((2 k - 1 - 1) (2 k - 1 + 1)) = 24 m$
$(2 k - 1) (2 k - 2) (2 k) = 24 m$
$8 k ³ - 12 k ² + 4 k = 24 m$
When $n = 2 k + 1 :$
$n (n^{2} + 1) = (2 k + 1) ((2 k + 1)^{2} - 1)$
$= (2 k + 1) ((2 k + 1 - 1) (2 k + 1 + 1))$
$= (2 k + 1) (2 k) (2 k + 2)$
$= 8 k^{3} + 12 k^{3} + 4 k$
$= (8 k^{3} - 12 k^{2} + 4 k) + 24 k^{2}$
$= 24 m + 24 k^{2}$
$= 24 (m + k ²) - - - - - > d i v i s i b l e b y 24$
Statement true for $n = 2 k - 1$ statement true for $n = 2 k + 1$
Therefore, by principles of mathematical induction, statement is true for all odd positive integers n.

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