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.
 A: 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.
