How to solve $a^7 \equiv a \pmod {42}$ involving congruences? For all integers $a$ prove that
$$a^7 \equiv a \pmod {42}.$$ 
There is no use telling you all what and how much I tried because I cannot even understand the problem itself left alone attempting it. It would be great help for me if I would be provided some hint on either solving the problem or understanding it.
But, I see one thing that $42=7\cdot6$ and somehow we can use Fermat theorem but how? I do not know it.
All suggestions/advice leading me to the problem would be greatly welcomed.

You need to show that 42 always divides $a^7-a.$ Since $42 = 2\cdot 3 \cdot 7,$ it suffices to show that $2,3,7$ divide $a^7-a.$ 


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*2 divides $a^7-a$ because $a^7$ has the same parity as $a.$

*3 divides $a^7-a = a(a^6-1) = a(a^3-1)(a^3+1)$ because $a^3 \equiv a \mod 3$ (we just check that by cases).

*7 divides $a^7-a$ by Fermat's little theorem.

