1. **Problem Statement:** Consider the expansion of $(2 + x)^n$, where $n > 3$ and $n \in \mathbb{Z}$. The coefficient of $x^3$ is four times the coefficient of $x^5$. Find the value of $n$. 2. **Formula and Rules:** The general term in the expansion of $(a + b)^n$ is given by the binomial theorem: $$ T_{r+1} = \binom{n}{r} a^{n-r} b^r $$ where $r = 0, 1, 2, ..., n$. For $(2 + x)^n$, $a = 2$ and $b = x$. 3. **Identify the terms:** - The coefficient of $x^3$ corresponds to the term where $r=3$: $$ T_4 = \binom{n}{3} 2^{n-3} x^3 $$ - The coefficient of $x^5$ corresponds to the term where $r=5$: $$ T_6 = \binom{n}{5} 2^{n-5} x^5 $$ 4. **Set up the equation:** Given the coefficient of $x^3$ is four times the coefficient of $x^5$: $$ \binom{n}{3} 2^{n-3} = 4 \times \binom{n}{5} 2^{n-5} $$ 5. **Simplify the equation:** Divide both sides by $2^{n-5}$: $$ \binom{n}{3} 2^{2} = 4 \binom{n}{5} $$ Since $2^2 = 4$, this becomes: $$ 4 \binom{n}{3} = 4 \binom{n}{5} $$ Divide both sides by 4: $$ \binom{n}{3} = \binom{n}{5} $$ 6. **Use the property of binomial coefficients:** Recall that: $$ \binom{n}{k} = \binom{n}{n-k} $$ So, $$ \binom{n}{3} = \binom{n}{5} \implies 3 = n - 5 \implies n = 8 $$ 7. **Check the condition $n > 3$:** $n = 8$ satisfies this. **Final answer:** $$ \boxed{8} $$