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

$$
2^{a}+4^{b}=8^{c} .
$$
SOLUTION. The given equation can be rewritten as $2^{a}+2^{2 b}=2^{3 c}$. To be able to divide the expression on the left side by a power of two, we will use the notation $m=\min (a, 2 b), M=\max (a, 2 b), 0<m$ is obviously an odd number greater than 1, so it cannot be a divisor of the power $2^{3 c}$, which is on the right side of the given equation. Therefore, it must be that $M=m$, from which it follows that $a=2 b$ and by comparing both sides of the modified equation also $2 b+1=3 c$. Since $2 b+1$ is an odd number, the number $c$ must also be odd, so there exists a natural number $n$ for which $c=2 n-1$. From the equation $2 b+1=3 c$ we calculate $b=3 n-2$ and $a=2 b=6 n-4$.

For any natural number $n$, the triplet $(a, b, c)=(6 n-4,3 n-2,2 n-1)$ is a solution to the given equation, which can be verified by substitution, although this is not necessary in this process.

Note. From the notation $2^{a}+2^{2 b}=2^{3 c}$ and the uniqueness of the representation of a number in binary, it immediately follows that $a=2 b$, and thus $3 c=a+1$.

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