To solve for all values of $x$ in the range $0 \leq x \leq 29$ such that:

$x^2 \equiv x \pmod{30}$

we start by rearranging the equation:

$x^2 - x \equiv 0 \pmod{30}$

This can be factored as:

$x(x - 1) \equiv 0 \pmod{30}$

Thus, $x(x - 1)$ must be divisible by 30. Since $30 = 2 \cdot 3 \cdot 5$, $x(x - 1)$ must be divisible by 2, 3, and 5.

Step 1: Analyzing $x(x - 1)$ for Divisibility by 2, 3, and 5

Since $x$ and $x - 1$ are consecutive integers, one of them is always even, so $x(x - 1)$ is divisible by 2.

Similarly, one of $x$ or $x - 1$ must be divisible by 3.

Finally, for $x(x - 1)$ to be divisible by 5, either $x$ or $x - 1$ must be divisible by 5.

Step 2: Possible Solutions

The divisibility requirements imply that $x$ must satisfy the following conditions:

1. $x$ or $x - 1$ must be divisible by 2, 3, and 5.
2. This occurs when $x \equiv 0 \pmod{30}$ or $x \equiv 1 \pmod{30}$.

Step 3: Valid Solutions in the Given Range

Since we only need solutions for $0 \leq x \leq 29$, we check:

1. $x = 0$
2. $x = 1$

Verification

1. For $x = 0$: $x^2 = 0 \equiv 0 \pmod{30}$

2. For $x = 1$: $x^2 = 1 \equiv 1 \pmod{30}$

Both values satisfy the condition $x^2 \equiv x \pmod{30}$.

Conclusion

The solutions are: $x = 0 \text{ and } x = 1$

Would you like a deeper explanation of any part of this solution?

Here are five related questions to further explore modular arithmetic and divisibility:

1. How would you solve for $x$ in $x^3 = x \pmod{30}$?
2. Can you determine all values of $x$ such that $x^2 = x \pmod{n}$ for different values of $n$?
3. What are other interesting properties of solutions to equations of the form $x^k = x \pmod{n}$?
4. How can we use the Chinese Remainder Theorem to solve similar modular equations?
5. What happens to the solutions if we expand the range to $x < 60$?

Tip: Modular equations often simplify by factoring and analyzing each factor's divisibility conditions.