, 2019

Assume, for contradiction, that
$$a^{2} + \left\lceil \frac{4 a^{2}}{b} \right\rceil = (a + k)^{2}, \quad \text{or} \quad \left\lceil \frac{(2a)^{2}}{b} \right\rceil = (2a + k)k.$$
Clearly, $k \geqslant 1$. In other words, the equation
$$\left\lceil \frac{c^{2}}{b} \right\rceil = (c + k)k \tag{1}$$
has a positive integer solution $(c, k)$, with an even value of $c$.

Choose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From
$$\frac{c^{2}}{b} > \left\lceil \frac{c^{2}}{b} \right\rceil - 1 = c k + k^{2} - 1 \geqslant c k$$
and
$$\frac{(c - k)(c + k)}{b} < \frac{c^{2}}{b} \leqslant \left\lceil \frac{c^{2}}{b} \right\rceil = (c + k)k$$
it can be seen that $c > b k > c - k$, so
$$c = k b + r \quad \text{with some} \ 0 < r < k.$$
By substituting this in (1) we get
$$\left\lceil \frac{c^{2}}{b} \right\rceil = \left\lceil \frac{(b k + r)^{2}}{b} \right\rceil = k^{2} b + 2 k r + \left\lceil \frac{r^{2}}{b} \right\rceil$$
and
$$(c + k)k = (k b + r + k)k = k^{2} b + 2 k r + k(k - r),$$
so
$$\left\lceil \frac{r^{2}}{b} \right\rceil = k(k - r) \tag{2}$$
Notice that relation (2) provides another positive integer solution of (1), namely $c' = r$ and $k' = k - r$, with $c' > 0$ and $0 < k' < k$. That contradicts the minimality of $k$, and hence finishes the solution.

Suppose that
$$a^{2} + \left\lceil \frac{4 a^{2}}{b} \right\rceil = c^{2}$$
with some positive integer $c > a$, so
$$\begin{align*} & c^{2} - 1 < a^{2} + \frac{4 a^{2}}{b} \leqslant c^{2} \\ & 0 \leqslant c^{2} b - a^{2}(b + 4) < b \tag{3} \end{align*}$$
Let $d = c^{2} b - a^{2}(b + 4)$, $x = c + a$ and $y = c - a$; then we have $c = \frac{x + y}{2}$ and $a = \frac{x - y}{2}$, and (3) can be re-written as follows:
$$\begin{align*} \left( \frac{x + y}{2} \right)^{2} b - \left( \frac{x - y}{2} \right)^{2} (b + 4) & = d, \\ x^{2} - (b + 2) x y + y^{2} + d = 0 ; \quad 0 & \leqslant d < b. \tag{4} \end{align*}$$
So, by the indirect assumption, the equation (4) has some positive integer solution $(x, y)$.

Fix $b$ and $d$, and take a pair $(x, y)$ of positive integers, satisfying (4), such that $x + y$ is minimal. By the symmetry in (4) we may assume that $x \geqslant y \geqslant 1$.

Now we perform a usual "Vieta jump". Consider (4) as a quadratic equation in variable $x$, and let $z$ be its second root. By the Vieta formulas,
$$x + z = (b + 2) y, \quad \text{and} \quad z x = y^{2} + d$$
so
$$z = (b + 2) y - x = \frac{y^{2} + d}{x}$$
The first formula shows that $z$ is an integer, and by the second formula $z$ is positive. Hence $(z, y)$ is another positive integer solution of (4). From
$$\begin{aligned} (x - 1)(z - 1) & = x z - (x + z) + 1 = (y^{2} + d) - (b + 2) y + 1 \\ & < (y^{2} + b) - (b + 2) y + 1 = (y - 1)^{2} - b(y - 1) \leqslant (y - 1)^{2} \leqslant (x - 1)^{2} \end{aligned}$$
we can see that $z < x$ and therefore $z + y < x + y$. But this contradicts the minimality of $x + y$ among the positive integer solutions of (4).