Ukrainian National Mathematical Olympiad

Let's rewrite the conditions in form of a system: there exist natural numbers $k, m, n$, for which equalities are true:
$$2a-1=kb, \quad 2b-1=nc \text{ and } 2c-1=ma.$$

It is obvious, that all the numbers $k, m, n$ and $a, b, c$ are odd.
Then we have, that
$$b = \frac{2a-1}{k} \Rightarrow 2b-1 = \frac{4a-2}{k}-1 = nc \Rightarrow c = \frac{4a-2-k}{kn} \Rightarrow 2c-1 = \frac{8a-4-2k}{kn}-1 = ma \Rightarrow \\ 8a-4-2k-kn = knma \Rightarrow a = \frac{4+2k+kn}{8-knm}.$$
Now we see, that an equality must be true: $knm<8$. Then from symmetry of conditions and oddness of numbers $k, m, n$ it follows, that the following variants are possible (with accuracy to cycle).

Variant 1: $k=m=n=1$. Then we have $c=2b-1 \Rightarrow a=2c-1=4b-3 \Rightarrow b=2a-1=8b-7 \Rightarrow a=b=c=1$, but numbers $a, b, c$ must be different.

Variant 2: $k=3, m=n=1$. Then we have $c=2b-1 \Rightarrow a=2c-1=4b-3 \Rightarrow 3b=2a-1=8b-7 \Rightarrow 5b=7$ – contradiction, because $b$ is not integer.

Variant 3: $k=5, m=n=1$. Then we have $c=2b-1 \Rightarrow a=2c-1=4b-3 \Rightarrow 5b=2a-1=8b-7 \Rightarrow 3b=7$ – contradiction, because $b$ is not integer.

Variant 4: $k=7, m=n=1$. Then we have $c=2b-1 \Rightarrow a=2c-1=4b-3 \Rightarrow 7b=2a-1=8b-7 \Rightarrow b=7 \Rightarrow c=13, a=25$ and this set satisfies the condition.