Problem 6

An ordered pair

(x,y)

of integers is called a primitive point if

\gcd(x,y)=1

. Given a finite set

S

of primitive points, prove that there exist a positive integer

n

and integers

a_0,a_1,\ldots,a_n

such that for every

(x,y)\in S

, we have

a_0x^n+a_1x^{n-1}y+a_2x^{n-2}y^2+\cdots+a_{n-1}xy^{n-1}+a_ny^n=1

Step 4 of 8: Prove the crucial coprimality

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In plain words

Reduction modulo a prime turns a zero determinant into proportional vectors.

\gcd(P(x_k,y_k),D)=1

Detailed analysis

Fix a prime $p$ dividing both $P(x_k,y_k)$ and one $d_i=y_ix_k-x_iy_k$ . Since both points are primitive, their reductions modulo $p$ are nonzero vectors. The zero determinant makes them proportional, so for some nonzero residue $\lambda$ we have $(x_i,y_i)\equiv\lambda q\pmod p$ . Homogeneity then gives $P(x_i,y_i)=1$ $\equiv\lambda^nP(x_k,y_k)\equiv0\pmod p$ , a contradiction. Therefore no prime divides both $P(x_k,y_k)$ and any $d_i=y_ix_k-x_iy_k$ , so $\gcd(P(x_k,y_k),D)=1$ .