I have to solve this congruence: $x^3 \equiv 1\,(\mod\;101)$ , but I I can't understand some steps, $$ x^3 \equiv 1\,(\mod\;101) \\ \gcd(3,100)=1\\ 3a+100b=1\\ x^1\equiv x^{3a+100b} \equiv1^a\cdot1^b\equiv 1\;(\mod\;101) $$ Solution:$x \equiv 1 \;(\mod\;101)$
In particular, I don't understand what the $3$ of the greatest common divisor represents(is it the exponent?) and why $x^{3a+100b}$ is equivalent to $1^a\cdot1^b$.
On
$101$ is prime, so the nonzero residues modulo $101$ form a cyclic group of order $100$ (under multiplication modulo $101$). Since $\gcd(3,100)=1$ it follows that the map that takes $x$ to $x^3$ is one-one and onto, so $x^3\equiv1\bmod{101}$ has exactly one solution. Since $x=1$ is obviously a solution, it is the only solution.
Yes, the $3$ in $\gcd(3,100)=1$ is the exponent in the equation (which is a cubic equation b.t.w.).
As to $x^{3a+100b}$, it is equal to $x^{3a}x^{100b}=(x^3)^a(x^{100})^b=1^a(x^{100})^b$, and $x^{100}\equiv 1\bmod 101$ by lil' Fermat, since $101$ is prime.