[Algebra]
Problem: Suppose that $f(x)=x^{2}+(2 n-1) x+\left(n^{2}-22\right)$ for some integer $n$. What is the smallest positive integer $n$ for which $f(x)$ has no real roots?

Solution: The quadratic function $f(x)=x^{2}+(2 n-1) x+\left(n^{2}-22\right)$ has no real roots exactly when its discriminant, $\Delta$, is negative.

The discriminant of this function is

$$
\begin{aligned}
\Delta & =(2 n-1)^{2}-4(1)\left(n^{2}-22\right) \\
& =\left(4 n^{2}-4 n+1\right)-\left(4 n^{2}-88\right) \\
& =-4 n+89
\end{aligned}
$$

We have $\Delta<0$ exactly when $-4 n+89<0$ or $4 n>89$.

This final inequality is equivalent to $n>\frac{89}{4}=22 \frac{1}{4}$.

Therefore, the smallest positive integer that satisfies this inequality, and hence for which $f(x)$ has no real roots, is $n=23$.

Final answer: 23