4. Let $a, b, c$ be integers, each between 1 and 2021 (inclusive), that satisfy the equation

$$
\sqrt{a}+\sqrt{b}=\sqrt{a+c \sqrt{b}} .
$$

How many distinct values of $c$ are possible?
(A) 130
(B) 132
(C) 133
(D) 1936
(E) 2025
4. The answer is (A). Squaring the equation from the text, we obtain the equivalent condition $a+2 \sqrt{a b}+b=a+c \sqrt{b}$. Subtracting $a$ and dividing by $\sqrt{b}$, we arrive at

$$
2 \sqrt{a}+\sqrt{b}=c.
$$

For $c$ to be an integer, both $a$ and $b$ must be perfect squares. This can be observed by squaring the equality $2 \sqrt{a}=c-\sqrt{b}$, which yields $4 a=c^{2}+b-2 c \sqrt{b}$; thus, $\sqrt{b}$ is a rational number, so $b$ is a perfect square, and in particular, $\sqrt{b}$ is an integer. Returning to the original equation $2 \sqrt{a}+\sqrt{b}=c$, it immediately follows that $a$ must also be a perfect square.

Given the condition $a, b \leq 2021$, we must count how many values the expression $2 x+y$ can take with $1 \leq x, y \leq 44$.

It is easy to see that all and only the values between 3 and $3 \cdot 44$ inclusive are taken: with $x=1$, the interval [3,46] is filled, with $x=23$ the interval $[47,88]$ is taken, and finally with $x=44$ the numbers in the interval $[89,132]$ are also reached.