Baltic Way 2019

From the equation $5F_x=3F_y+1$ we have $$3F_y+1=5F_x>3F_x+1⟹y>x$$ On the other hand, if $y\ge x+2$ and $x>1$ then $$3F_y+1\ge 3F_{x+2}+1=3(F_{x+1}+3F_x)+1=6F_x+3F_{x-1}+1>5F_x$$ we have a contradiction. Therefore $y=x+1$ and we have to solve the equation becomes $3F_{x+1}+1=5F_x$ We will show by induction that $3F_{x+1}+1<5F_x$ for any $x\ge 7$. Indeed, for $x=7$ we have $F_7=13$, $F_8=21$. Therefore, $3F_8+1=3\cdot 21+1=64<5F_7=65$. For $x=8$ we have $3F_9+1=103<5F_8=105$. Assume $3F_{k+1}+1<5F_k$ and $3F_{k+2}+1<5F_{k+1}$ for some $k\ge 7$. We have then $3(F_{k+1}+F_{k+2})+2<5(F_k+F_{k+1})$. Thus $$3F_{k+3}+1<3F_{k+3}+2<5F_{k+2}$$ For $x<7$ we can check and see that $x\in \{3,5,6\}$.