1
$\begingroup$

Let $p^n$ be a prime power and, for integers $a,b,c$, let $Q(x,y)=ax^2+bxy+cy^2$ where $p\nmid b^2-4ac$. Define $$N(k,m):=|\{(i,j)\in\{0,\ldots,m-1\}^2: \gcd(i,j,m)=1, Q(i,j)\equiv k\pmod{m}\}|.$$

I wish to prove that, for all $k$, $N(k,p^n)=N(k,p)p^{n-1}$.

For example, the number of solutions of $2x^2-2xy+3y^2=k$ satisfy $N(k, 3)=4,2,2,4,2,2,4,2,2,\ldots$, while $N(k,27)=36,18,18,36,18,18,36,18,18,\ldots$ ($k=0,1,\ldots$).

This is a followup question to a previous question here.

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ In your notation $N(k,m)$, all you actually care about are solutions modulo prime powers $p^n$. Moreover, you are interested in solutions $(x,y) \bmod p^n$, where $\gcd(x,y,p^n) = 1$, which is equivalent to $\gcd(x,y,p) = 1$. Such solutions are called primitive because a vector $(a_1,\ldots,a_r) \bmod p^n$ where the $a_i$'s are not all divisible by $p$ is called a primitive vector. Every primitive solution mod $p^n$ reduces to one mod $p^{n-1}$. You want to lift primitive solutions mod $p^{n-1}$ to primitive solutions mod $p^{n}$. Try using Hensel's lemma. $\endgroup$
    – KConrad
    Commented Jun 20, 2022 at 0:10

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .