1. **Problem Statement:** Determine whether the function $f(x) = b x^2$ is one-to-one (injective) when the domain and codomain are both the set of integers. 2. **Definition of One-to-One Function:** A function $f$ is one-to-one if for every $x_1, x_2$ in the domain, whenever $f(x_1) = f(x_2)$, it implies that $x_1 = x_2$. 3. **Analyze the function:** The function is $f(x) = b x^2$, where $b$ is a constant integer. 4. **Check for injectivity:** Suppose $f(x_1) = f(x_2)$, then $$b x_1^2 = b x_2^2$$ Since $b$ is constant and nonzero (otherwise the function is trivial), divide both sides by $b$: $$x_1^2 = x_2^2$$ This implies $$x_1 = x_2 \quad \text{or} \quad x_1 = -x_2$$ 5. **Conclusion:** Because $f(x_1) = f(x_2)$ does not imply $x_1 = x_2$ uniquely (it could be $x_1 = -x_2$), the function is **not one-to-one** over the integers. 6. **Additional note:** The function is many-to-one since $f(x) = f(-x)$ for all integers $x$. Final answer: The function $f(x) = b x^2$ is not one-to-one from integers to integers.