Square Factors 58Febf

1. **State the problem:** We need to determine if Thomas's claim that all square numbers (except 0) have an odd number of factors is correct. 2. **Recall the rule about factors:** Normally, factors come in pairs. For example, for a number $n$, if $a$ is a factor, then $\frac{n}{a}$ is also a factor. 3. **Special case for square numbers:** For square numbers, one of these pairs is actually the same number repeated because $\sqrt{n} \times \sqrt{n} = n$. This means the factor $\sqrt{n}$ is counted only once. 4. **Conclusion:** Because of this repeated factor, square numbers have an odd number of factors. 5. **Example:** Consider $16$, which is $4^2$. Its factors are $1, 2, 4, 8, 16$. There are $5$ factors, which is odd. **Final answer:** Thomas is correct; all square numbers (except 0) have an odd number of factors because their square root factor is counted only once, making the total count odd.