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as before my question:problems: there's one step I don't understand $$a_{0}=0,a_{1}=a_{2}=1,a_{n+3}=a_{n+2}+a_{n+1}+a_{n}$$ then $$a_{n+168}=a_{n}\pmod {13}$$ it is said can use calculating polynomial discriminant .But I can't Thanks

math110
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  • easy enough to write a program. Indeed, the sequence (taken mod 13) strats over again at $168$ with $0,1,1,2,4,7,0,$ and so on. No evident projective repeats before that – Will Jagy Feb 14 '21 at 03:23
  • the relevant Pisano periods are listed at https://oeis.org/A046738/b046738.txt – Will Jagy Feb 14 '21 at 03:45

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Here's an idea from algebraic number theory. We solve the recursion in $\overline{\mathbb F_{13}}.$ Note that our characteristic polynomial in $\mathbb F_{13}[x]$ factors as $$x^3-x^2-x-1=(x+6)\left(x^2+6x+2\right).$$ Now fix $\alpha=-6$ with $\beta$ and $\gamma$ the roots of the quadratic. Because $\alpha,\beta,\gamma\in\mathbb F_{13^2}$ (they are, at worst, quadratic irrationals) we remark that it suffices to work in $\mathbb F_{13^2}$ from here on. In particular, we can solve the recursion as $$a_n=A\alpha^n+B\beta^n+C\gamma^n$$ for some constants $A,B,C\in\mathbb F_{13^2}.$ To finish, we note that $\alpha,\beta,\gamma\in\mathbb F_{13^2}^\times,$ so they have multiplicative order dividing into $\#\mathbb F_{13^2}^\times=168.$ Thus, $$a_{n+168}=A\alpha^{n+168}+B\beta^{n+168}+C\gamma^{n+168}=A\alpha^n+B\beta^n+C\gamma^n=a_n$$ in $\mathbb F_{13^2}.$ This finishes the proof.

Derive Foiler
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  • Why do you have overline here $\overline{\mathbb F_{13}}.$ What does it stands for? – nonuser May 27 '21 at 17:58
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    The overline is for the algebraic closure of $\mathbb{F}{13}.$ The point is that we want to start in a field where all polynomials have roots so that we can write $\alpha,\beta,\gamma$ without headaches; this is restricted to $\mathbb F{13^2}$ later. – Derive Foiler May 27 '21 at 18:00