## Task A-2.4.

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

$$
2^{a} \cdot 5^{b}-1=11 \cdot 3^{c}
$$
## Solution.

Let's look at the remainders that the expressions on the left and right sides of the given equation give when divided by 5. The first addend on the left side is divisible by 5, so the left side gives a remainder of 4 when divided by 5. On the other hand, the number 11 gives a remainder of 1 when divided by 5. Finally, the expression $3^{c}$, depending on the value of the number $c$, gives remainders $3, 4, 2, 1$, which then cyclically repeat with a period of 4. We conclude that, for the right side to also give a remainder of 4 when divided by 5, the number $c$ must be of the form $4c_{1} + 2$, for some non-negative integer $c_{1}$. In particular, $c$ is an even number.

Now let's look at the remainders that the same expressions give when divided by 8. Since $c$ is an even number, the number $11 \cdot 3^{c} = 11 \cdot 9^{2c_{1} + 1}$ gives a remainder of 3 when divided by 8. Therefore, the number $2^{a} \cdot 5^{b} = 1 + 11 \cdot 3^{c}$ must give a remainder of 4 when divided by 8. This is only possible if $a = 2$ (if $a = 1$, the left side gives a remainder of 2; if $a \geq 3$, the left side is divisible by 8).

Finally, let's look at the remainders that these expressions give when divided by 3. The right side is divisible by 3, so the number $2^{a} \cdot 5^{b} = 4 \cdot 5^{b}$ must give a remainder of 1. Since $5^{b}$ gives periodic remainders 2 and 1, we conclude that $b$ is even, i.e., $b = 2b_{1}$, for some natural number $b_{1}$.

The left side can now be factored using the difference of squares:

$$
\left(2 \cdot 5^{b_{1}} - 1\right)\left(2 \cdot 5^{b_{1}} + 1\right) = 11 \cdot 3^{c}
$$

On the left side of the equation, we have the product of two consecutive odd numbers, and therefore they are relatively prime. Neither factor can be equal to 1, since they are at least $2 \cdot 5^{1} - 1 = 9$. Therefore, one factor must be equal to 11, and the other to $3^{c}$.

In the first case, we have $2 \cdot 5^{b_{1}} - 1 = 11$, which has no solution.

In the second case, we have $2 \cdot 5^{b_{1}} + 1 = 11$, from which $b_{1} = 1$, i.e., $b = 2$. Then $2 \cdot 5^{b_{1}} - 1 = 9 = 3^{c}$, i.e., $c = 2$.

Thus, the only solution to the equation is $(a, b, c) = (2, 2, 2)$.