1. **State the problem:** Determine whether the function $f(x) = x^2$ is one-to-one (injective) when the domain and codomain are both the set of integers. 2. **Recall the definition of one-to-one function:** A function $f$ is one-to-one if for every pair of distinct inputs $a$ and $b$, the outputs are distinct, i.e., if $f(a) = f(b)$ then $a = b$. 3. **Apply the definition to $f(x) = x^2$:** Suppose $f(a) = f(b)$ for integers $a$ and $b$. Then: $$a^2 = b^2$$ 4. **Solve the equation:** This implies: $$a^2 - b^2 = 0$$ $$ (a - b)(a + b) = 0$$ 5. **Analyze the factors:** For the product to be zero, either: $$a - b = 0 \quad \text{or} \quad a + b = 0$$ 6. **Interpret the results:** - If $a - b = 0$, then $a = b$. - If $a + b = 0$, then $a = -b$. 7. **Check if $a = -b$ can happen with $a \neq b$:** For example, $f(2) = 4$ and $f(-2) = 4$, but $2 \neq -2$. 8. **Conclusion:** Since $f(a) = f(b)$ does not always imply $a = b$, the function $f(x) = x^2$ is **not** one-to-one from integers to integers. **Final answer:** $f(x) = x^2$ is not one-to-one on the set of integers.