Question

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

Answer 1

$n=1$ -----> $n(n²-1)=1(1-1)=1\left(0\right)=0$
$n=3$ -----> $n(n²-1)=3(9-1)=3\left(8\right)=24$
We can see that this is true for $n=1$ and $3$
Assume statement is true for $n=2k-1$. Then we must show it is true for$n=2k+1$
Statement true for $n=2k-1$

$(2k-1)\left(\right(2k-1)²-1)=24m$, for some integer m
$(2k-1)\left(\right(2k-1)²-1)=24m$
$(2k-1)\left(\right(2k-1-1\left)\right(2k-1+1\left)\right)=24m$
$(2k-1)(2k-2)\left(2k\right)=24m$
$8k³-12k²+4k=24m$
When$n=2k+1:$
$n(n^2+1)=(2k+1)\left(\right(2k+1{)}^2-1)$
$=(2k+1)\left(\right(2k+1-1\left)\right(2k+1+1\left)\right)$
$=(2k+1)\left(2k\right)(2k+2)$
$=8k^3+12k^3+4k$
$=(8k^3-12k^2+4k)+24k^2$
$=24m+24k^2$
$=24(m+k²)----->divisibleby24$
Statement true for $n=2k-1$ statement true for $n=2k+1$
Therefore, by principles of mathematical induction, statement is true for all odd positive integers n.