## Task A-4.3.

Determine all triples of natural numbers $(a, b, c)$ for which

$$
2^{a}+2021=3^{b} \cdot 25^{c}
$$
## Solution.

Let's look at the remainders that the expressions on the left and right sides of the given equation give when divided by 3. The right side is divisible by 3, the number 2021 gives a remainder of 2 when divided by 3, so the number $2^{a}$ must give a remainder of 1. Looking at the remainders that $2^{a}$ gives when divided by 3 for different values of $a$, we see that the remainders 2 and 1 repeat with a period of 2. We conclude that $a$ must be an even number, i.e., $a=2a_{1}$ for some natural number $a_{1}$.

In particular, $a \geqslant 2$, so $2^{a}$ is divisible by 4. The number 2021 gives a remainder of 1 when divided by 4, so the right side must also give this remainder. The number $25^{c}$ gives a remainder of 1, while the expression $3^{b}$ gives remainders of 3 and 1 periodically when divided by 4. We conclude that $b$ is also an even number, i.e., $b=2b_{1}$ for some natural number $b_{1}$.

From the given equation, we have

$$
2021=3^{2b_{1}} \cdot 5^{2c}-2^{2a_{1}}=\left(3^{b_{1}} \cdot 5^{c}-2^{a_{1}}\right)\left(3^{b_{1}} \cdot 5^{c}+2^{a_{1}}\right)
$$

Since the factors on the right side are odd numbers whose difference is $2^{a_{1}+1}$, we conclude that they are relatively prime. Considering that the prime factorization of 2021 is $2021=43 \cdot 47$, and that the first factor on the right side of the equation is strictly less than the second, we have two cases: the first factor is 1 and the second is 2021, or the first factor is 43 and the second is 47.

In the first case, $3^{b_{1}} \cdot 5^{c}-2^{a_{1}}=1$ and $3^{b_{1}} \cdot 5^{c}+2^{a_{1}}=2021$. Subtracting the equations gives $2^{a_{1}+1}=2022$, which has no solution.

In the second case, $3^{b_{1}} \cdot 5^{c}-2^{a_{1}}=43$ and $3^{b_{1}} \cdot 5^{c}+2^{a_{1}}=47$. Subtracting the equations gives $2^{a_{1}+1}=4$, so $a_{1}=1$ and $a=2$. Substituting into either of the remaining equations gives $3^{b_{1}} \cdot 5^{c}=45$, so $b_{1}=2$ (and thus $b=4$) and $c=1$.

Finally, the only solution to the given equation is $(a, b, c)=(2,4,1)$.