Prime power decomposition

Question

$x^{147} \equiv (((x^{7})^{7})^{3})\equiv x^{3}(mod7)$

How does $x^{147}$ simplify into $x^{3}(mod7)$

What Corollary is responsible for this?

Edit:

Fermat's Little Theorem is needed:

147 = 3 * 7 * 7

$x^{6} \equiv 1(mod7)$

$((x^{}*x^{6})^{7})^{3} \equiv ((1*x)^{7})^{3} (mod7)$

$((x^{}*1)^{7})^{3}\equiv ((x)^{7})^{3} (mod7)$

$(x^{7})^{3} \equiv ((x)^{7})^{3} (mod7)$

$(x^{}*x^{6})^{3} \equiv (x^{2})^{3}(mod7)$

$x^{147} \equiv(x^{}*1)^{3}\equiv x^{5}(mod7)$

My attempt is off by a degree of 2. What did I do wrong?

Answer 1

We will use the following fact:

Fact: For any finite group $G$ of order $n$ and any element $g\in G$, we have that $g^n=e$.

Now, consider the specific case of $G=\mathbb{Z}/7\mathbb{Z}$ under multiplication. Our fact tells us that $x^6\equiv 1\pmod{7}$ since $G$ has six elements. From this, we can multiply on both sides by $x$ to obtain Fermat's Little Theorem, namely that $x^7\equiv x\pmod{7}$. Therefore, we may simplify $x^{147}$ in the following way:

$$x^{147}=(x^{7})^{7\cdot3}\equiv (x)^{7\cdot3} = (x^7)^3\equiv x^3\pmod{7}.$$

Edit: Looking at your calculations, in your fifth line you state that $(x\cdot x^6)^3\equiv (x^2)^3\pmod{7}$, but we know that $x^6\equiv 1\pmod{7}$, so $(x\cdot x^6)^3\equiv x^3\pmod{7}$. Seems like just a computation slip-up!