Euler-Fermat extended

An often useful theorem in number theory is Euler's theorem (also known as the Fermat–Euler theorem or Euler's totient theorem), which says that:

If $a$ and $n$ are coprime positive integers, then

$a^{\phi(n)}\equiv 1 (\text{mod }n)$

where $\phi(n)$ is Euler's totient function, counting how many of the integers $1,2,\ldots,n$ are coprime with $n$.

If, however, $a$ and $n$ are not coprime, we know for sure that the above does not hold. After all, if $a$ and $n$ share a common factor (greater than 1), any power of $a$ will also share this factor and will not equal 1 modulo $n$. However, if we formulate the equivalence as:

$a^{\phi(n)+1}\equiv a (\text{mod }n)$

(which, as it turns out, is equivalent to the original for $a,n$ coprime), there are certain cases where this relation still holds even if $a$ and $n$ do share a common factor. A sufficient (though not necessary) condition is for $n$ to be squarefree. Thus we obtain the following theorem:

Theorem. If $a,n$ are positive integers where $n$ is squarefree, then:

$a^{\phi(n)+1}\equiv a (\text{mod }n)$

where $\phi(n)$ is Euler's totient function.

Proof.

Let $n=p_1p_2\dots p_k$, where all the $p_i$ are distinct primes. We can write $a=bp_{c_1}p_{c_2}\dots p_{c_m}$, where the $c_i\in\{1,2,\ldots,k\}$ are all distinct and $b$ and $n$ are coprime. Modulo $p_{c_i}$ (for any $i=1,2,\ldots, m$) we have $a^{\phi(n)+1}\equiv 0 \equiv a (\text{mod }p_{c_1})$ since, of course, $p_{c_1}$ divides $a$. Analogously we know the same for all the other $p_{c_i}$. Furthermore, we have:

$a^{\phi(n)}\equiv 1 \left(\text{mod }\frac{n}{p_{c_1}\dots p_{c_m}}\right)$

since $a$ and $\frac{n}{p_{c_1}\dots p_{c_m}}$ are coprime, and $\phi\left(\frac{n}{p_{c_1}\dots p_{c_m}}\right)$ divides $\phi(n)$. Hence we obtain the following system:

$\begin{aligned} a^{\phi(n)+1} &\equiv a (\text{mod }p_{c_1})\\ &\vdots\\ a^{\phi(n)+1} &\equiv a (\text{mod }p_{c_m})\\ a^{\phi(n)+1} &\equiv a \left(\text{mod }\frac{n}{p_{c_1}\dots p_{c_m}}\right). \end{aligned}$

Since all the modulo classes on the right are coprime, we can use the Chinese remainder theorem to see that there exists a unique solution for this system, namely

$a^{\phi(n)+1} \equiv a \left(\text{mod }p_{c_1}\dots p_{c_m}\frac{n}{p_{c_1}\dots p_{c_m}}\right)$

or, in other words:

$a^{\phi(n)+1} \equiv a (\text{mod } n).\qquad\square$

This shows us, for example, that $a^5\equiv a (\text{mod }10)$ for any $a$, since $\phi(10)=4$ and $10$ is squarefree. Hence the last digit of any number is equal to the last digit of its fifth power, a very useful fact in for example the problem which inspired this note.