Subfield: Number Theory


Question: Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for each $(x, y)$ in the set $S$.

Note: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form

$$
f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} .
$$

Solution:
First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\left(x_{1}, y_{1}\right)$ through $\left(x_{n}, y_{n}\right)$. If any two of these lattice points $\left(x_{i}, y_{i}\right)$ and $\left(x_{j}, y_{j}\right)$ lie on the same line through the origin, then $\left(x_{j}, y_{j}\right)=\left(-x_{i},-y_{i}\right)$ because both of the points are irreducible. We then have $f\left(x_{j}, y_{j}\right)= \pm f\left(x_{i}, y_{i}\right)$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points.

Consider the homogenous polynomials $\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define

$$
g_{i}(x, y)=\prod_{j \neq i} \ell_{j}(x, y)
$$

Then $\ell_{i}\left(x_{j}, y_{j}\right)=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}\left(x_{j}, y_{j}\right)=0$ for all $j \neq i$. Define $a_{i}=g_{i}\left(x_{i}, y_{i}\right)$, and note that $a_{i} \neq 0$.

Note that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties:

1. $g_{i}\left(x_{j}, y_{j}\right)=0$ if $j \neq i$.
2. $g_{i}\left(x_{i}, y_{i}\right)=a_{i}$.

For any $N \geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}\left(x_{i}, y_{i}\right)=1$, which exists since $\left(x_{i}, y_{i}\right)$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$.

We may now reduce the problem to the following claim:

Claim: For each positive integer a, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1 , such that $f_{a}(x, y) \equiv 1(\bmod a)$ for all relatively prime $(x, y)$.

To see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}(1 \leqslant i \leqslant n)$. Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial.

We prove the claim by factoring $a$. First, if $a$ is a power of a prime $\left(a=p^{k}\right)$, then we may choose either:

- $f_{a}(x, y)=\left(x^{p-1}+y^{p-1}\right)^{\phi(a)}$ if $p$ is odd;
- $f_{a}(x, y)=\left(x^{2}+x y+y^{2}\right)^{\phi(a)}$ if $p=2$.

Now suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that

$$
\frac{a}{q_{i}} F_{q_{i}}(x, y) \equiv \frac{a}{q_{i}}(\bmod a)
$$

for any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\frac{a}{q_{i}}$ that equals 1 . Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \equiv 1$ $(\bmod a)$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree.
As in the previous solution, label the irreducible lattice points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ and assume without loss of generality that no two of the points are collinear with the origin. We induct on $n$ to construct a homogenous polynomial $f(x, y)$ such that $f\left(x_{i}, y_{i}\right)=1$ for all $1 \leqslant i \leqslant n$.

If $n=1$ : Since $x_{1}$ and $y_{1}$ are relatively prime, there exist some integers $c, d$ such that $c x_{1}+d y_{1}=1$. Then $f(x, y)=c x+d y$ is suitable.

If $n \geqslant 2$ : By the induction hypothesis we already have a homogeneous polynomial $g(x, y)$ with $g\left(x_{1}, y_{1}\right)=\ldots=g\left(x_{n-1}, y_{n-1}\right)=1$. Let $j=\operatorname{deg} g$,

$$
g_{n}(x, y)=\prod_{k=1}^{n-1}\left(y_{k} x-x_{k} y\right)
$$

and $a_{n}=g_{n}\left(x_{n}, y_{n}\right)$. By assumption, $a_{n} \neq 0$. Take some integers $c, d$ such that $c x_{n}+d y_{n}=1$. We will construct $f(x, y)$ in the form

$$
f(x, y)=g(x, y)^{K}-C \cdot g_{n}(x, y) \cdot(c x+d y)^{L}
$$

where $K$ and $L$ are some positive integers and $C$ is some integer. We assume that $L=K j-n+1$ so that $f$ is homogenous.

Due to $g\left(x_{1}, y_{1}\right)=\ldots=g\left(x_{n-1}, y_{n-1}\right)=1$ and $g_{n}\left(x_{1}, y_{1}\right)=\ldots=g_{n}\left(x_{n-1}, y_{n-1}\right)=0$, the property $f\left(x_{1}, y_{1}\right)=\ldots=f\left(x_{n-1}, y_{n-1}\right)=1$ is automatically satisfied with any choice of $K, L$, and $C$.

Furthermore,

$$
f\left(x_{n}, y_{n}\right)=g\left(x_{n}, y_{n}\right)^{K}-C \cdot g_{n}\left(x_{n}, y_{n}\right) \cdot\left(c x_{n}+d y_{n}\right)^{L}=g\left(x_{n}, y_{n}\right)^{K}-C a_{n}
$$

If we have an exponent $K$ such that $g\left(x_{n}, y_{n}\right)^{K} \equiv 1\left(\bmod a_{n}\right)$, then we may choose $C$ such that $f\left(x_{n}, y_{n}\right)=1$. We now choose such a $K$.

Consider an arbitrary prime divisor $p$ of $a_{n}$. By

$$
p \mid a_{n}=g_{n}\left(x_{n}, y_{n}\right)=\prod_{k=1}^{n-1}\left(y_{k} x_{n}-x_{k} y_{n}\right)
$$

there is some $1 \leqslant k<n$ such that $x_{k} y_{n} \equiv x_{n} y_{k}(\bmod p)$. We first show that $x_{k} x_{n}$ or $y_{k} y_{n}$ is relatively prime with $p$. This is trivial in the case $x_{k} y_{n} \equiv x_{n} y_{k} \not \equiv 0(\bmod p)$. In the other case, we have $x_{k} y_{n} \equiv x_{n} y_{k} \equiv 0(\bmod p)$, If, say $p \mid x_{k}$, then $p \nmid y_{k}$ because $\left(x_{k}, y_{k}\right)$ is irreducible, so $p \mid x_{n}$; then $p \nmid y_{n}$ because $\left(x_{k}, y_{k}\right)$ is irreducible. In summary, $p \mid x_{k}$ implies $p \nmid y_{k} y_{n}$. Similarly, $p \mid y_{n}$ implies $p \nmid x_{k} x_{n}$.

By the homogeneity of $g$ we have the congruences

$$
x_{k}^{d} \cdot g\left(x_{n}, y_{n}\right)=g\left(x_{k} x_{n}, x_{k} y_{n}\right) \equiv g\left(x_{k} x_{n}, y_{k} x_{n}\right)=x_{n}^{d} \cdot g\left(x_{k}, y_{k}\right)=x_{n}^{d} \quad(\bmod p)
\tag{1.1}
$$

and

$$
y_{k}^{d} \cdot g\left(x_{n}, y_{n}\right)=g\left(y_{k} x_{n}, y_{k} y_{n}\right) \equiv g\left(x_{k} y_{n}, y_{k} y_{n}\right)=y_{n}^{d} \cdot g\left(x_{k}, y_{k}\right)=y_{n}^{d} \quad(\bmod p)
\tag{1.2}
$$

If $p \nmid x_{k} x_{n}$, then take the $(p-1)^{s t}$ power of (1.1); otherwise take the $(p-1)^{s t}$ power of (1.2); by Fermat's theorem, in both cases we get

$$
g\left(x_{n}, y_{n}\right)^{p-1} \equiv 1 \quad(\bmod p)
$$

If $p^{\alpha} \mid m$, then we have

$$
g\left(x_{n}, y_{n}\right)^{p^{\alpha-1}(p-1)} \equiv 1 \quad\left(\bmod p^{\alpha}\right)
$$

which implies that the exponent $K=n \cdot \varphi\left(a_{n}\right)$, which is a multiple of all $p^{\alpha-1}(p-1)$, is a suitable choice. (The factor $n$ is added only so that $K \geqslant n$ and so $L>0$.)