Fibonacci terms coincide with polynomial

Given the polynomial

$$p(x,y)=2y^4x+y^3x^2-2y^2x^3-y^5-yx^4+2y$$

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.

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

• 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.) – Rodrigo Dec 18 '19 at 17:06
• I noticed. I figured the questions were related, which was key to figuring out the workable expression for the polynomial. – Andrés E. Caicedo Dec 18 '19 at 17:24