An ordered pair $(x,y)$ of integers is a primitive point if $\gcd (x,y)=1$. Given a finite set $S$ of primitive points, prove that there exists a positive integer $n$ and integers $a_0,a_1,\dots ,a_n$ such that, for each $(x,y)$ in $S$, we have:

In other words, there exists a homogeneous polynomial $f(x,y)\in \mathbb{Z}[x,y]$ of positive degree such that $f(x,y)=1$ for all $(x,y)\in S$.

Step 1 (Key idea): For each primitive point $(a,b)$ with $\gcd (a,b)=1$, by Bézout's identity there exist integers $c,d$ such that $ad-bc=1$. The linear form $L_{(a,b)}(x,y)=dx-cy$ satisfies $L_{(a,b)}(a,b)=da-cb=1$.

Step 2 (Single point): For a single primitive point $(a,b)$, the polynomial $f(x,y)=(dx-cy{)}^n$ for any $n\ge 1$ satisfies $f(a,b)=1$.

Step 3 (Multiple points): For $S=\{(a_1,b_1),\dots ,(a_k,b_k)\}$, we need a single homogeneous polynomial that equals 1 at all points. For each point $(a_i,b_i)$, let $L_i(x,y)=d_{i}x-c_{i}y$ where $a_i{d}_i-b_i{c}_i=1$.

Note that $L_i(a_i,b_i)=1$ but $L_i(a_j,b_j)$ may not be 1 for $j\ne i$.

Step 4 (Construction): Choose a large enough $n$. By CRT-type arguments and properties of homogeneous polynomials, one can construct $f$ as a suitable combination. Specifically, consider $f(x,y)={\sum }_i{P}_i(x,y)\cdot {\prod }_{j\ne i}(L_j(x,y){)}^{n_j}$ where the $P_i$ are chosen so that $f(a_i,b_i)=1$ for each $i$. The homogeneity constraint requires careful degree-balancing.

Step 5 (Conclusion): The construction works because $\gcd (a_i,b_i)=1$ ensures Bézout coefficients exist, and the polynomial can be made homogeneous of a single degree by multiplying terms by appropriate powers of $x^2+y^2$ or similar forms. $■$