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Prove, without a calculator, that

3^{8} - 2^{8}

is divisible by $5$ .

Hint

Note that we have

3^{8} - 2^{8} = {(3^{4})}^{2} - {(2^{4})}^{2}

which is a difference of two squares.

Hint

\begin{aligned} 3^{8} - 2^{8} & = {(3^{4})}^{2} - {(2^{4})}^{2} \\ = (3^{4} - 2^{4}) (3^{4} + 2^{4}) \end{aligned}

That $3^{4} - 2^{4}$ looks interesting...

Solution

We make several applications of the difference of two squares:

\begin{aligned} 3^{8} - 2^{8} & = {(3^{4})}^{2} - {(2^{4})}^{2} \\ = (3^{4} - 2^{4}) (3^{4} + 2^{4}) \\ = ({(3^{2})}^{2} - {(2^{2})}^{2}) (3^{4} + 2^{4}) \\ = (3^{2} - 2^{2}) (3^{2} + 2^{2}) (3^{4} + 2^{4}) \\ = 5 (3^{2} + 2^{2}) (3^{4} + 2^{4}) \end{aligned}

as required!