Goes back to the same digit

We have that $a^b \equiv b \mod 10$ and $b^a \equiv a \mod 10$ . First, notice that if $a = 1$ then $1 \equiv b \mod 10 \implies b = 1$ which contradicts the fact that they must be distinct. So $a,b \neq 1$ .

Because $10 = 2 \times 5$ this means $a^b \equiv b \mod 2$ but all the elements in $\mathbb{Z} / 2 \mathbb{Z}$ (0 and 1) have the property that $x^y = x$ so what this implies is $a \equiv b \mod 2$ . (They must have the same parity).

Then notice that in $\mathbb{Z} / 10 \mathbb{Z}$ , $x^5 \equiv x \mod 10$ so if $a,b = 5$ that would imply $a = b$ which contradicts the fact that they must be distinct.

Now suppose that $b = 4$ . Then $a^4 \equiv 4 \mod 5 \implies 1 \equiv 4 \mod 5$ which is a contradiction. The same happens for $b=8$ so we can say $a,b \neq 4,8$ . Similarly, if $b = 9$ we would have $a^9 \equiv 9 \mod 5 \implies a \equiv 9 \mod 5$ . This means $a$ would have to be $4$ or $9$ but we proved they must have the same parity so $a = 9$ which contradicts the fact that they must be distinct.

So we have proven $a,b \neq 1,4,5,8,9$ . Let's suppose that $a,b$ were even. Then necessarily one would have to be $2$ and the other $6$ . We can check my computation that this is not the case. Now let's suppose that $a,b$ are odd. Then necessarily one would have to be $3$ and the other $7$ and this pair works. So the solution is $3 \times 7 = 21$ .