To solve the congruence \(x^2 \equiv 0 \pmod{6}\), we need to find the values of \(x\) (from the set of residues \(\{0, 1, 2, 3, 4, 5\}\)) such that when \(x^2\) is taken modulo 6, we get 0.
We'll check each residue \(x\) to see if \(x^2 \equiv 0 \pmod{6}\):
1. If \(x = 0\):
\[
0^2 = 0 \equiv 0 \pmod{6} \quad \text{(satisfies the condition)}
\]
2. If \(x = 1\):
\[
1^2 = 1 \not\equiv 0 \pmod{6}
\]
3. If \(x = 2\):
\[
2^2 = 4 \not\equiv 0 \pmod{6}
\]
4. If \(x = 3\):
\[
3^2 = 9 \equiv 3 \pmod{6} \quad \text{(since } 9 - 6 = 3\text{)}
\]
5. If \(x = 4\):
\[
4^2 = 16 \equiv 4 \pmod{6} \quad \text{(since } 16 - 12 = 4\text{)}
\]
6. If \(x = 5\):
\[
5^2 = 25 \equiv 1 \pmod{6} \quad \text{(since } 25 - 24 = 1\text{)}
\]
We see that the only \(x\) which satisfies \(x^2 \equiv 0 \pmod{6}\) is \(x = 0\).
Thus, the solution to the given congruence is:
\[
\boxed{0}
\]
Which of the residues $0,$ $1,$ $2,$ $3,$ $4,$ $5$ satisfy the congruence $x^2 \equiv 0 \pmod{6}?$
Give your answer as a list, separated by commas, in order from least to greatest.
1 answer