Given the polynomial


how would one prove that its positive values of $p$ (for $x=1,2,...,y=1,2...$) coincide with the Fibonacci numbers?

Tried something like $p(x-1, y-2)$ or $p(x-2, y-2) + p(x-1, y-1)$. But it didn't work, of course.

I've figured out that $p(f_{n-1}, f_n) = f_n$. So, Induction, perhaps?

I don't think it has anything to do with Fibonacci Polynomials.

  • $\begingroup$ then tag as a polynomial question. $\endgroup$ – user645636 Dec 18 '19 at 15:29
  • $\begingroup$ It doesn't seem $p(2,1)$ is a Fibonacci number $\endgroup$ – J. W. Tanner Dec 18 '19 at 15:29
  • $\begingroup$ @J.W.Tanner $p(2, 1)$ isn't positive. $\endgroup$ – John Gowers Dec 18 '19 at 15:30
  • $\begingroup$ $p(1,2) = 2$, on the other hand. $\endgroup$ – Rodrigo Dec 18 '19 at 15:30
  • $\begingroup$ polynomial remainder theorem may help. $\endgroup$ – user645636 Dec 18 '19 at 15:35

This is a perhaps surprising, really elegant, and not-quite-elementary result due to James Jones.

The key is to rewrite your polynomial appropriately, noting that $$ \begin{align}p(x,y)&=2y^4x+y^3x^2-2y^2x^3-y^5-yx^4+2y\\ &=y(2y^3x+y^2x^2-2yx^3-y^4-x^4+2)\\&=y(2-(y^4-2y^3x-y^2x^2+2yx^3+x^4))\\&=y(2-((y^4-2y^3x+y^2x^2)-2(y^2x^2-yx^3)+x^4))\\&=y(2-(y^2-yx-x^2)^2).\end{align} $$

From this it follows that, for $x,y$ positive, $p(x,y)>0$ if and only if $(y^2-yx-x^2)^2<2$, which means that if, in addition, $x,y$ are integers, then we must have $|y^2-yx-x^2|=0$ or $1$. The case where the expression is $0$ is handled in this question.

For the interesting case, I refer you to Jones's paper. In lemma 1, he shows that $f_{n+1}^2-f_{n+1}f_n-f_n^2=\pm1$. This is easily established by induction, and shows that all positive Fibonacci numbers are in the range of $p$ when its arguments are restricted to positive integers. Lemmas 2 and 3 prove the converse: If $x,y$ are positive integers and

  • $y^2-yx-x^2=1$, then $y=f_{2n+1}$, $x=f_{2n}$ for some $n$, while
  • if $y^2-yx-x^2=-1$, then $y=f_{2n}$, $x=f_{2n-1}$ for some $n$.

For the second, note that if $y^2-yx-x^2=-1$, then $(x+y)^2-(x+y)y-y^2=1$, so, arguing by induction, it suffices to consider the first case, which is then handled elegantly by Jones via induction and clever inequalities.

The reference (available online!) is

MR0382147 (52 #3035). Jones, James P. Diophantine representation of the Fibonacci numbers. Fibonacci Quart. 13 (1975), 84–88.

The result is related to Hilbert's tenth problem, the solution of which required establishing several similar results.

  • 1
    $\begingroup$ I had already worked things out alone. But this answer is amazing! Thanks for the help again! (The question you linked was mine, by the way.) $\endgroup$ – Rodrigo Dec 18 '19 at 17:06
  • $\begingroup$ I noticed. I figured the questions were related, which was key to figuring out the workable expression for the polynomial. $\endgroup$ – Andrés E. Caicedo Dec 18 '19 at 17:24

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