Can somebody help me how to solve
$x* 23 \equiv_{60} 1 $ with $ x \in \mathbb{N} $ and $x > 100$
What would be a good approach?
I know that x = 107 would be a solution. However how can I find solutions and how to prove that there are no solutions, if there aren't any?
Euler's theorem tells you that for $n$ and $a$ coprime, you have $$a^{\phi(n)} \equiv 1 \bmod n$$ where $\phi(n)$ is Euler's totient which counts the numbers less than $n$ and coprime to it.
So in this case we would know that $$23^{\phi(60)} = 23^{16} \equiv 1 \bmod 60$$ so we can say that $$x\equiv 23^{15} \bmod 60 $$ gives us solutions.
There is also Carmichael's reduced totient function $\lambda$ which often gives a smaller exponent for the same purpose. In this case $\lambda(60) = 4$ and so we can use $x\equiv 23^{\lambda(60)-1}\equiv 23^3 \equiv 47 \bmod 60$.
Now this gives your value, since $107\equiv 47\bmod 60$, and is one of the possible choice of the integers $47+60k$ which obey the equivalence.
For a rapid method not using exponents, the extended Euclidean algorithm will both check that $a$ and $n$ are coprime and deliver the modular inverse of $a\bmod m$ (which is what your equation here is asking for, $ax\equiv 1 \bmod n$). Note that solving this also helps to solve other $\bmod n$ equations involving $a$.
$x\cdot a\equiv_{b} 1\implies x=\frac{-(b^{-1})\cdot b+1}{a}\bmod b$
Use the Euclidean algorithm.
Since $\gcd(23,60) = 1$ there exists integers $m,n$ such that $23m + 60n = 1.$
$m$ will be our inverse.
The Euclidean algorithm:
$\begin {array}{rrc} n&m&60n+23m\\ \hline 1&&60\\ &1&23\\ 1&-2&14\\ -1&3&9\\ 2&-5&5\\ 5&-13&1\\ \end{array}$
What are we doing? The first column are candidates for $n.$ The second column are candidates for $m.$ The third column is $60n + 23m.$
We then subtract the largest multiple in the lines below from lines above to generate smaller numbers for $60n + 23m.$
We get down to $-13$ as a multiplicative inverse. And if $-13$ is an inverse, so is $47.$
