Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Our platform connects you with professionals ready to provide precise answers to all your questions in various areas of expertise. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

Let $a$ be an integer such that $0 \le a \le 7$ and $a^2 \equiv a \pmod{8}$. If $a \neq 0,$ then find the value of $a$.

Sagot :

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 ‍

I hope this will help you