Question about how to prove $x^5\equiv x \pmod {10}$

Question

I was trying to prove why $x^5\equiv x \pmod {10}$ for all natural numbers $x$. I saw a proof where they applied Euler's theorem to show this. They said that the totient function for $10$ is $4$. ($φ(10) = 4$). Therefore,

$x^5 ≡ 1^4 \cdot x^1 ≡ x\ \pmod { 10}$

However my question is: This isn't a correct proof is it? Because it requires $x$ to be relatively prime with $10$ for this to work. So it's a partial proof only (it would prove for $1$, $3$, $7$, and $9$ but not for the other $x$'s)?

If this is true, is there a way to generalize using Euler's Theorem to prove for all $x$? Or is there another way to go about doing this proof?

You are correct--that proof only works for $x$ coprime to $10$. I'm trying to think of a "nice" way of proving it, other than brute-forcing the non-coprime values of $x$. — apnorton, Dec 27 '14 at 00:46

score 3 · Accepted Answer · answered Dec 27 '14 at 01:20

It's fairly easy using the Chinese Remainder Theorem: The ring $\mathbf Z/10 \mathbf Z$ is isomorphice to the product of rings $ \mathbf Z/2 \mathbf Z\times \mathbf Z/5\mathbf Z$. In each of these rings (actually fields, since 2 and 5 are prime numbers), it is true that $x^5=x$ for any $x$ (use little Fermat), hence it is true in their product.

Question about how to prove $x^5\equiv x \pmod {10}$

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