Let x_n and y_n be sequences such that x_0=y_0=1, x_1=y_1=13 and x_n+2=38x_{n+1}-x_n, y_{n+2}=20y_{n+1}-y_n for n >= 0.

No common terms in two sequences defined by linear quadratic recurrence relations
Let ${x}_{n}$ and ${y}_{n}$ be sequences such that ${x}_{0}={y}_{0}=1,{x}_{1}={y}_{1}=13$ and
${x}_{n+2}=38{x}_{n+1}-{x}_{n},$
${y}_{n+2}=20{y}_{n+1}-{y}_{n}$
for $n\ge 0$. I want to show that there is no common terms when $n\ge 2$. In other words, there are no $n,m\ge 2$ such that ${x}_{n}={y}_{m}$.
I encountered this problem when tried to make an olympiad-like problem. So I am not sure it is true, but I have checked it with computer for $n\le {10}^{200000}$. (Since I am not a good coder, I can't make sure that it does not have an error such as overflow. Sorry...)
Is there any strategy to solve these kind of problems? If there is, and if you gave me just a hint or a related concept
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Step 1
Yes, put ${x}_{n}={a}^{n}$ to get ${a}^{2}-38a+1=0$ let ${a}_{1}$ and ${a}_{2}$ be the roots the ${x}_{n}={C}_{1}{a}_{1}^{n}+{C}_{2}{a}_{2}^{n}$. you can calculate ${C}_{1}$ and ${C}_{2}$ from the conditions ${x}_{0}=0$ abd ${x}_{1}=13$.
Step 2
This method works if you get the a-equation which is free of n.