Why does $x^{103} ≡ x^3$ (mod $11$) equal $x^3 ≡ 4$ (mod $11)$?

Question

Why $x^{103} ≡ x^3\pmod{11}$ equal x^3 ≡ 4(mod 11) I see it at p.69 in I understand "x^10 ≡ 1 (mod 11) => x^100 ≡ 1 (mod 11) => x^103 ≡ x^3 (mod 11)" but I do not know "why A answer of x^103 ≡ x^3 (mod 11) is equal a answer of x^3 ≡ 4(mod 11)"

Below is the text, from p. 67 of Silverman's book A friendly introduction to number theory.

Answer 1

If $x^3\equiv 4\pmod{11}$ and $x^{103}\equiv x^3\pmod{11}$, then $x^{103}\equiv4\pmod{11}$

In general, if $a\equiv b\pmod{n}$ and $b\equiv c\pmod{n}$, then $a\equiv c\pmod{n}$

Answer 2

$\!\bmod 11\!:\, $ he is solving $\,x^{\large 103}\equiv 4\ $ by using little Fermat to reduce the exponent from $\,103\,$ to $\,3.\,$ Note $\,0^{\large 103}\!\not\equiv 4,\,$ so $\,x\not\equiv 0,\,$ so little Fermat $\,\Rightarrow\, x^{\large 10}\!\equiv 1.\,$ Raising this to power $10$ using Congruence Rules

$$\begin{align} x^{\large 10}&\equiv 1\ \ \ \ \ {\rm by\ little\ Fermat\ and}\ x\not\equiv 0\\[.3em] \smash{\overset{(\ \ )^{\Large 10}}\Longrightarrow}\ \ (x^{\large 10})^{\large 10}&\equiv 1^{\large 10}\ \ {\rm by\ the\ Congruence\ Power\ Rule}\\[.3em] \Longrightarrow\ \ \ \ \ \ x^{\large 100}&\equiv 1\ \ \ \ \ {\rm by\ exponent\ laws}\\[.4em] \smash{\overset{\large \times\ x^{\Large 3}}\Longrightarrow}\ \ \ \ \ \ \color{#c00}{x^{\large 103}}&\equiv \color{#c00}{x^{\large 3}}\ \ \ {\rm by\ the\ Congruence\ Product\ Rule} \end{align}$$

Written more concisely: $\,\ x^{\large 103}\equiv x^{\large 3}(x^{\large 10})^{\large 10}\equiv x^{\large 3}(1^{\large 10})\equiv x^{\large 3}\ $

Thus from $\ \color{#c00}{x^{\large 103}\equiv x^{\large 3}}\ $ we infer $\ \color{#c00}{x^{\large 103}}\equiv 4\iff \color{#c00}{x^{\large 3}}\equiv 4\,\pmod{\!11}$

Remark $ $ The same idea works generally, i.e. if we know that $\, x^{\large n}\equiv 1\,$ then all exponents on $\,x\,$ can be reduced $\!\bmod n,\,$ i.e. $\,x^{\large k}\equiv x^{\large k\bmod n},\,$ see modular order reduction.