3. Determine all triples $(a, b, c)$ of positive integers for which

$$
2^{a+2 b+1}+4^{a}+16^{b}=4^{c} .
$$
3. We can rewrite the equation as

$$
2^{a+2 b+1}+2^{2 a}+2^{4 b}=2^{2 c} .
$$

Let $m=\min (a+2 b+1,2 a, 4 b)$. The left side of equation (2) can thus be written in the form

$$
2^{a+2 b+1}+2^{2 a}+2^{4 b}=2^{m}\left(2^{a+2 b+1-m}+2^{2 a-m}+2^{4 b-m}\right) .
$$

At least one of the exponents $(a+2 b+1-m, 2 a-m, 4 b-m)$ is zero, and the corresponding power of two is thus equal to 1. If the remaining two exponents were positive, the number in the parentheses would be odd, which would mean that $2^{2 c}$ is divisible by an odd number greater than 1. Therefore, at least two of the exponents are zero, so in the triplet $(a+2 b+1,2 a, 4 b)$, there are two identical numbers, which are at most equal to the third one.

If $a+2 b+1=2 a$, we would get $a=2 b+1$, which contradicts the assumed inequality $2 a \leqq 4 b$. Similarly, from the equality $a+2 b+1=4 b$ we get $a=2 b-1$, which contradicts the inequality $4 b \leqq 2 a$. Therefore, it must be that $2 a=4 b$, i.e., $a=2 b$.

The left side of the given equation thus has the form

$$
2^{a+2 b+1}+2^{2 a}+2^{4 b}=2^{4 b+1}+2^{4 b}+2^{4 b}=4 \cdot 2^{4 b}=2^{4 b+2} .
$$

Therefore, the equation is satisfied if and only if $2 c=4 b+2$ or $c=2 b+1$.

For any natural number $b$, the triplet $(a, b, c)=(2 b, b, 2 b+1)$ is a solution to the given equation, and no other solutions exist.

Another solution. We can modify the left side of the given equation as follows:

$$
2^{a+2 b+1}+4^{a}+16^{b}=2 \cdot 2^{a} \cdot 4^{b}+\left(2^{a}\right)^{2}+\left(4^{b}\right)^{2}=\left(2^{a}+4^{b}\right)^{2} .
$$

After taking the square root of both sides, we get the equivalent equation

$$
2^{a}+4^{b}=2^{c} \quad \text { or } \quad 2^{a}+2^{2 b}=2^{c} .
$$

From the uniqueness of the representation of a number in binary or in a similar way to the solution of problem B-I-1, we can determine that $a=2 b$, from which it follows that $c=2 b+1$. As in the previous solution, we have thus concluded that all solutions to the equation are of the form $(a, b, c)=(2 b, b, 2 b+1)$, where $b$ is any natural number.

For a complete solution, award 6 points. For the first approach, award 1 point for expressing the terms of the equation as powers of 2, 2 points for proving that two terms in the triplet $(a+2 b+1,2 a, 4 b)$ must be equal, 1 point for proving that $2 a=4 b$, 1 point for calculating $c$, and 2 points for providing the correct answer. For the second approach, award 3 points for expressing the left side of the equation as a square, 2 points for the reasoning leading to the equality $a=2 b$, and 1 point for calculating $c$ and the correct answer.