[math]T_{n+1}={T_n}^2-T_n+1[/math]

and T

_{0}=2,

prove that

[math]\forall p\neq q \textup{GCD}(T_p,T_q)=1[/math]

**Moderators:** gmalivuk, Moderators General, Prelates

Given

[math]T_{n+1}={T_n}^2-T_n+1[/math]

and T_{0}=2,

prove that

[math]\forall p\neq q \textup{GCD}(T_p,T_q)=1[/math]

[math]T_{n+1}={T_n}^2-T_n+1[/math]

and T

prove that

[math]\forall p\neq q \textup{GCD}(T_p,T_q)=1[/math]

Rhombic wrote:Given

[math]T_{n+1}={T_n}^2-T_n+1[/math]

and T_{0}=2,

prove that

[math]\forall p\neq q \textup{GCD}(T_p,T_q)=1[/math]

Let me write those formulæ in bbcode:

T

T

And you want to show by induction that the T

Is this homework?

Last edited by Qaanol on Sun Apr 05, 2015 7:40 pm UTC, edited 2 times in total.

wee free kings

Can you prove that T_{n} and T_{n+1} are relatively prime for any n? Can you prove that T_{n} and T_{n+2} are relatively prime for any n?

Also, have you tried listing out the first, say 10 or so terms to see if the conclusion holds for small n?

Because you should probably do that.

Edit: The proof is actually quite straightforward once you recognize the pattern.

Because you should probably do that.

Edit: The proof is actually quite straightforward once you recognize the pattern.

wee free kings

Like with many induction proofs, this problem can become simpler if you attempt to show a stronger statement. Specifically,

**Spoiler:**

(∫|p|^{2})(∫|q|^{2}) ≥ (∫|pq|)^{2}

Thanks, skeptical scientist, for knowing symbols and giving them to me.

Thanks, skeptical scientist, for knowing symbols and giving them to me.

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