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Answer:
Since $a^2 \equiv a \pmod{8}$, we can write:
$a^2 - a = 8k$
for some integer $k$.
Factoring the left-hand side, we get:
$a(a - 1) = 8k$
Since $a$ and $a - 1$ are consecutive integers, one of them must be even, and therefore divisible by $2$. Hence, either $a$ or $a - 1$ is divisible by $4$ (since they differ by $1$).
This means that either $a$ or $a - 1$ is equal to $4k'$ for some integer $k'$.
Since $0 \le a \le 7$, the only possibility is $a = 4$.
Therefore, the value of $a$ is $\boxed{4}$.
Step-by-step explanation:
Here's a step-by-step solution:
1. Write the congruence equation: $a^2 \equiv a \pmod{8}$
2. Subtract $a$ from both sides: $a^2 - a \equiv 0 \pmod{8}$
3. Factor the left-hand side: $a(a - 1) \equiv 0 \pmod{8}$
4. Since $a$ and $a - 1$ are consecutive integers, one of them must be even (divisible by $2$).
5. Therefore, either $a$ or $a - 1$ is divisible by $4$ (since they differ by $1$).
6. Let's consider the cases:
a. If $a$ is divisible by $4$, then $a = 4k$ for some integer $k$.
Since $0 \le a \le 7$, the only possibility is $a = 4$.
b. If $a - 1$ is divisible by $4$, then $a - 1 = 4k'$ for some integer $k'$.
Since $0 \le a \le 7$, the only possibility is $a - 1 = 4$, which means $a = 5$. However, this contradicts the condition $a^2 \equiv a \pmod{8}$, since $5^2 = 25 \not\equiv 5 \pmod{8}$.
1. Therefore, the only possibility is $a = \boxed{4}$.
So, the value of $a$ is $4$.
finally done this answer
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