(a) Determine all real numbers $a>0$ for which $\sqrt{a^{2}+a}=\frac{2}{3}$.

(b) For each positive integer $m$, determine the difference between $\left(m+\frac{1}{2}\right)^{2}+\left(m+\frac{1}{2}\right)$ and the nearest perfect square.

(c) For every positive integer $n$, prove that the number of positive integers $c$ with $n<\sqrt{c+\sqrt{c}}<n+1$ is even.
(a) We begin by noting that $\sqrt{a^{2}+a}=\frac{2}{3}$ is equivalent to $a^{2}+a=\frac{4}{9}$.

Multiplying this equation by 9 and re-arranging, we obtain the equivalent equation $9 a^{2}+9 a-4=0$.

Factoring, we obtain $(3 a-1)(3 a+4)=0$ and so $a=\frac{1}{3}$ or $a=-\frac{4}{3}$.

Since $a>0$, then $a=\frac{1}{3}$.

Substituting into the original equation, we obtain

$$
\sqrt{a^{2}+a}=\sqrt{\left(\frac{1}{3}\right)^{2}+\frac{1}{3}}=\sqrt{\frac{1}{9}+\frac{1}{3}}=\sqrt{\frac{4}{9}}=\frac{2}{3}
$$

and so $a=\frac{1}{3}$ is the only solution to the equation when $a>0$.

(b) Expanding and simplifying,

$$
\left(m+\frac{1}{2}\right)^{2}+\left(m+\frac{1}{2}\right)=m^{2}+m+\frac{1}{4}+m+\frac{1}{2}=m^{2}+2 m+\frac{3}{4}
$$

Since $m$ is a positive integer, then $m^{2}+2 m+\frac{3}{4}>m^{2}$.

Also, $m^{2}+2 m+\frac{3}{4}\frac{1}{4}$, and so $\left(m+\frac{1}{2}\right)^{2}+\left(m+\frac{1}{2}\right)$ is closest to $(m+1)^{2}$ and their difference is $\frac{1}{4}$.
(c) A fact that will be important throughout this solution is that the expresssion $c+\sqrt{c}$ increases as $c$ increases.

In other words, when $04$ (because $\sqrt{3}>1$ ).

Thus, when $n=1$, there are 2 values of $c$ that work.

When $n=2$, the inequality is $22$ ).

Since $c+\sqrt{c}$ is increasing, the inequality $4n^{2}-2 n+1$ and so $\sqrt{n(n-1)+1}>\sqrt{n^{2}-2 n+1}=n-1$.

Also, when $n \geq 1$, we have $n(n-1)+1=n^{2}-n+1 \leq n^{2}$ and so $\sqrt{n(n-1)+1} \leq \sqrt{n^{2}}=n$. Thus, when $c=n(n-1)+1$, we see that

$c+\sqrt{c}=n(n-1)+1+\sqrt{n(n-1)+1}=n^{2}-n+1+\sqrt{n(n-1)+1}>n^{2}-n+1+(n-1)=n^{2}$

Also, when $c=n(n-1)+1$, we see that

$$
c+\sqrt{c}=n(n-1)+1+\sqrt{n(n-1)+1} \leq n^{2}-n+1+n=n^{2}+1n^{2}$, we can update $(*)$ to

$$
n^{2}<n(n+1)+\sqrt{n(n+1)}<(n+1)^{2}
$$

which is item (III).

We have now shown the desired inequalities for $c=n(n-1), c=n(n-1)+1, c=n(n+1)$, and $c=n(n+1)+1$, which means that the number of positive integers $c$ that satisfy the inequality for a given $n$ is $2 n$, which is even, as required.