I have the following problem:
Determine a function $f(x,y)$ defined on the positive integers and satisfying each of the following:
(1) $f(x, y)$ is a positive integer
(2) $f(2,3)=31$ and $f(4,1)=17.$
(3) $f(x,y)$ is a polynomial of minimal degree, with integer coefficients.
(4) For each of the following, there exists a unique positive integer $x$ or $y$ such that $f(x,4)=89, f(5, y)=26, f(x,2)=129, f(14, y)=223, f(x,5)=126, f(17,y)=316.$
I think this is about all I know about this function. As to whether it exists, or if it does whether the solution is unique, I have no clue. However, I have shown that there are no solutions of first degree. (See below.) Naturally, one would want to start examining polynomials of second degree in $x$ and $y,$ but clearly the method I used in the first degree case isn't very helpful here, so I wondered if there were a neater way to get at a solution, or even to show that one does exist uniquely.
Thank you.
Proof that there is no solution of degree one
Let $f(x,y)=ax+by+c$ be a solution. Then from condition #2 it follows that $a=2-\frac c5, \,b=9-\frac c5.$ Since $a, b$ are integral, then $c=5k,$ with $k\in\mathrm Z,$ so that $a=2-k, \,b=9-k.$
Now from the first equation in condition #4, we obtain $x=\frac{k-53}{k-2}=1-\frac{51}{k-2}.$ Since $x>0,$ then $k-2<0$ or $k-2>51.$ Since $x$ is an integer, then $k-2$ divides $51,$ so that $k=1,-1,-15,-49.$ From the second equation, we find that $y=-\frac{16}{k-9},$ and the requirement that $y$ be a positive integer means $k-9<0$ and divides $16,$ so that we can reduce the solution set of $k$ to be $k=1.$ From the third equation, and using $k=1,$ we find that $x=108.$ However, the fourth equation and $k=1$ give $y=25.5,$ so that $k$ does not exist satisfying all the given conditions. The claim then follows.