Question Number 82034 by Dah Solu Tion last updated on 17/Feb/20
$$\boldsymbol{{P}}{rove}\:\:{by}\:\:{maths}\:\:{induction}\:\:{tbat} \\ $$$$\boldsymbol{{n}}^{\mathrm{5}} \:−\:\boldsymbol{{n}}^{\mathrm{3}} \:\:\boldsymbol{{is}}\:\boldsymbol{{divisible}}\:\boldsymbol{{by}}\:\mathrm{24}. \\ $$
Commented by MJS last updated on 17/Feb/20
$$\mathrm{without}\:\mathrm{induction}\:\mathrm{24}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3} \\ $$$${n}^{\mathrm{5}} −{n}^{\mathrm{3}} =\left({n}−\mathrm{1}\right){n}^{\mathrm{3}} \left({n}+\mathrm{1}\right) \\ $$$$\forall{n}\in\mathbb{Z}:\:\mathrm{3}\mid\left({n}−\mathrm{1}\right)\:\mathrm{xor}\:\mathrm{3}\mid{n}\:\mathrm{xor}\:\mathrm{3}\mid\left({n}+\mathrm{1}\right) \\ $$$$\Rightarrow\:\mathrm{3}\mid\left({n}^{\mathrm{5}} −{n}^{\mathrm{3}} \right) \\ $$$$\forall{n}\in\mathbb{Z}:\:\mathrm{2}\mid\left({n}−\mathrm{1}\right)\wedge\mathrm{2}\mid\left({n}+\mathrm{1}\right)\:\mathrm{xor}\:\mathrm{2}\mid{n} \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{2}\mid{n}\:\Rightarrow\:\mathrm{8}\mid{n}^{\mathrm{3}} }\\{\mathrm{2}\nmid{n}\:\Rightarrow\:{n}=\mathrm{2}{k}+\mathrm{1}\:\Rightarrow\:\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)=\mathrm{4}{k}\left({k}+\mathrm{1}\right)\:\Rightarrow\:\mathrm{8}\mid\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}\end{cases} \\ $$$$\Rightarrow\:\mathrm{8}\mid\left({n}^{\mathrm{5}} −{n}^{\mathrm{3}} \right) \\ $$$$ \\ $$$$\Rightarrow\:\mathrm{24}\mid\left({n}^{\mathrm{5}} −{n}^{\mathrm{3}} \right) \\ $$