Proving that the sequence terms ${T_{n+1}}=T_n^2-T_n+1$ are relatively prime

Question

Given that $T_1=2$ and ${T_{n+1}}=T_n^2-T_n+1$, $n>0$.

prove that $T_n$ and $T_m$ are relatively prime whenever $n \neq m$

I tried by calculating the first few terms in the sequence, but I couldn't recognize a pattren. I am also thinking of doing an induction proof on $k$, where $n+m=k$. But don't have idea where to start.

Any hits or solution will be appreciated

Answer 1

Thanks to Naruki Masuda hint, I was able to answer the question.

Prove by induction

Base Case : Since $T_{n+1}=T_n^2-T_n+1$

Then $T_{n+1}\equiv 1 \pmod{T_n}$

Induction hypothesis : assume it's true for some $k\in\mathbb{N}$

$T_{n+k}\equiv 1 \pmod{T_n}$
Then $T_{n+k+1}={T_{n+k}}^2-T_{n+k}+1\equiv 1^2-1+1=1\pmod{T_n}$

Thus by mathematical induction, it's true for all elements in this sequence that $T_n$ and $T_m$ are relatively prime whenever $n \neq m$