Balkan Mathematical Olympiad competition problems

We write $p = |a - b|$ and assume for contradiction that $q = (a + b)^2 + 4$ is a prime number.
Since $(a, b) \mid [a, b]$, we have that $(a, b) \mid 2021^c$. As $(a, b)$ also divides $p = |a - b|$, it follows that $(a, b) \in \{1, 43, 47\}$. We will consider all 3 cases separately:

(1) If $(a, b) = 1$, then $1 + ab = 2021^c$, and therefore
$$q = (a + b)^2 + 4 = (a - b)^2 + 4(1 + ab) = p^2 + 4 \cdot 2021^c. \quad (1)$$

a. Suppose $c$ is even. Since $q \equiv 1 \pmod 4$, it can be represented uniquely (up to order) as a sum of two (non-negative) squares. But (1) gives potentially two such representations so in order to have uniqueness we must have $p = 2$. But then $4|q$ a contradiction.

b. If $c$ is odd then $ab = 2021^c - 1 \equiv 1 \pmod 3$. Thus $a \equiv b \pmod 3$ implying that $p = |a - b| \equiv 0 \pmod 3$. Therefore $p = 3$. Without loss of generality $b = a + 3$. Then $2021^c = ab + 1 = a^2 + 3a + 1$ and so
$$(2a + 3)^2 = 4a^2 + 12a + 9 = 4 \cdot 2021^c + 5.$$
So 5 is a quadratic residue modulo 47, a contradiction as
$$\left(\frac{5}{47}\right) = \left(\frac{47}{5}\right) = \left(\frac{2}{5}\right) = -1.$$

(2) If $(a, b) = 43$, then $p = |a - b| = 43$ and we may assume that $a = 43k$ and $b = 43(k + 1)$, for some $k \in \mathbb{N}$. Then $2021^c = 43 + 43k(k + 1)$ giving that
$$(2k + 1)^2 = 4k^2 + 4k + 4 - 3 = 4 \cdot 43^{c-1} \cdot 47 - 3.$$
So $-3$ is a quadratic residue modulo 47, a contradiction as
$$\left(\frac{-3}{47}\right) = \left(\frac{-1}{47}\right) \left(\frac{3}{47}\right) = \left(\frac{47}{3}\right) = \left(\frac{2}{3}\right) = -1.$$

(3) If $(a, b) = 47$ then analogously there is a $k \in \mathbb{N}$ such that
$$(2k + 1)^2 = 4 \cdot 43^c \cdot 47^{c-1} - 3.$$
If $c > 1$ then we get a contradiction in exactly the same way as in (2). If $c = 1$ then $(2k + 1)^2 = 169$ giving $k = 6$. This implies that $a + b = 47 \cdot 6 + 47 \cdot 7 = 47 \cdot 13 \equiv 1 \pmod 5$. Thus $q = (a + b)^2 + 4 \equiv 0 \pmod 5$, a contradiction.