Inverse of half the modular

Question

I have been given a question about why not all numbers have an inverse in mod 26. With only odd numbers having inverse as they are coprime with 26 except 13. Is there any other reason that 13 doesn't have an inverse in mod 26 other than that 26 and 13 have factors of 13.

Answer 1

An integer $a$ has an inverse, mod $m$, if and only if there exists an integer $x$ such that $$ax \equiv 1\;(\text{mod}\;m)$$

Thus, if $13$ was invertible, mod $26$, the congruence $$13x \equiv 1\;(\text{mod}\;26)$$ would have a solution. \begin{align*} \text{But}\;\;&13x \equiv 1\;(\text{mod}\;26)\\[4pt] \implies\;&13x \equiv 1\;(\text{mod}\;13)\\[4pt] \implies\;&0 \equiv 1\;(\text{mod}\;13)\\[4pt] \implies\;&13{\,\mid\,}1\\[4pt] \end{align*} contradiction.

Thus, the reason why $13$ is not invertible, mod $26$, is precisely because $13$ and $26$ have a common factor greater than $1$.