58th 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: $$2a-1=kb,2b-1=nc\text{ and }2c-1=ma.$$

It is obvious, that all the numbers $k,m,n$ and $a,b,c$ are odd. Then we have, that $$\begin{gathered} b=\frac{2a-1}{k}\Rightarrow 2b-1=\frac{4a-2}{k}-1=nc\Rightarrow c=\frac{4a-2-k}{kn}\Rightarrow 2c-1=\frac{8a-4-2k}{kn}-1=ma\Rightarrow \\ 8a-4-2k-kn=knma\Rightarrow a=\frac{4+2k+kn}{8-knm}. \end{gathered}$$ 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.