The 2 -adic valuation of the general degree-2 polynomial in 2 variables Shubham

: We define the p -adic valuation tree of a polynomial f ( x, y ) ∈ Z [ x, y ] by which we can find its p -adic valuation at any point. This work includes diverse 2 -adic valuation trees of certain degree-two polynomials in two variables. Among these, the 2 -adic valuation tree of x 2 + y 2 is most interesting. We use the observations from these trees to study the 2 -adic valuation tree of the general degree-two polynomial in 2 variables. We also study the 2 -adic valuation tree of the polynomial x 2 y + 5 .


Introduction
For n ∈ N, the exponent of the highest power of a prime p that divides n is called the p-adic valuation of the n.This is denoted by ν p (n). Legendre establishes the following result about p-adic valuation of n! in [11] , where a n = 2n n .
It follows from here that a n is always an even number and a n /2 is odd when n is a power of 2. This expression is called a closed form in [6].The definition of closed form depends on the context.This has been discussed in [4,8].
The work presented in [6] forms part of a general project initiated by Moll and his co-authors to analyse the set for a given sequence x = {x n } of integers.
The 2-adic valuation of {n 2 − a} is studied in [6].It is shown that n 2 − a, a ∈ Z has a simple closed form when a ̸ ≡ 4, 7 mod 8.For these two remaining cases the valuation is quite complicated.It is studied by the notion of the valuation tree.
Given a polynomial f (x) with integer coefficients, the sequence {ν 2 (f (n))} is described by a tree.This is called the valuation tree attached to the polynomial f .The vertices of this tree correspond to some selected subsets C m,j = {2 m i + j : i ∈ N}, 0 ≤ j < 2 m starting with the root node C 0,0 = N.The procedure to select classes is explained below in Example 1.1.Some notation for the vertices of this tree are introduced as follows: Otherwise, it is called non-terminal.The same terminology is given to vertices corresponding to the class C m,j .In the tree, terminal vertices are labelled by their constant valuation and non-terminal vertices are labelled by a * .
Example 1.1.Construction of valuation tree of x 2 + 5 is as follows: note that ν 2 (1 + 5) = 1 and ν 2 (2 + 5) is 0. So node v 0 is non terminating.Hence it splits into two vertices and forms the first level.These vertices correspond to the residue classes C 1,0 and C 1,1 .We can check that both these nodes are terminating with valuation 0 and 1.The valuation tree of x 2 + 5 is shown in Figure 1.
The main theorem of [6] is as follows.
Theorem 1.1.Let v be a non-terminating node at the k-th level for the valuation tree of n 2 + 7.
Then v splits into two vertices at the (k + 1)-level.Exactly one of them terminates, with valuation k.The second one has valuation at least k + 1.
The 2-adic valuation of the Stirling numbers is discussed in [2].The numbers S(n, k) are the number of ways to partition a set of n elements into exactly k non-empty subsets where n ∈ N and 0 ≤ k ≤ n.These are explicitly given by or, by the recurrence with initial condition S(0, 0) = 1 and S(n, 0) = 0 for n > 0.
The 2-adic valuation of S(n, k) can be easily determined for 1 ≤ k ≤ 4 and closed form expression is given as follows:

0, otherwise
The important conjecture described there is that the partitions of N in classes of the form m,j = {2 m i + j : i ∈ N and starts at the point where 2 m i + j ≥ k } leads to a clear pattern for v 2 (S(n, k)) for k ∈ N is fixed.We recall that the parameter m is called the level of the class.The main conjecture of [2] is now stated: This conjecture is only established for the case k = 5.A similar conjecture is given in [3] for the p-adic valuation of the Stirling numbers.
From the point of view of application, the 2-adic valuation tree is used to solve quadratic and cubic diophantine equations in [5].Apart from this, some other special cases are studied in [1,7,12].
In this work, we discuss the set where f (x, y) ∈ Z[x, y], by the generalized notion of the valuation tree.We believe that the p-adic valuation and the valuation tree of two variable polynomials have not been studied before.We define the p-adic valuation tree as follows: Definition 1.2.Let p be a prime number.Consider the integers f (x, y) for every (x, y) in Z 2 .The p-adic valuation tree of f is a rooted, labelled p 2 -ary tree defined recursively as follows: 1. Suppose that v 0 be a root vertex at level 0. There are p 2 edges from this root vertex to its p 2 children vertices at level 1.These vertices correspond to all possible residue classes (i 0 , j 0 ) mod p. Label the vertex corresponding to the class (i 0 , j 0 ) with 0 if f (i 0 , j 0 ) ̸ ≡ 0 (mod p) and with * if f (i 0 , j 0 ) ≡ 0 (mod p).
2. If the label of a vertex is 0, it does not have any children.
This process continues recursively so that at the l th level, there are p 2 children of any nonterminating vertex in the previous level (l − 1), each child of which corresponds to the residue classes Here i l−1 , j l−1 ∈ {0, 1, 2, . . ., p − 1} and (i 0 ) is the class of the parent vertex.Label the vertex corresponding to the class (i, j) with l − 1 if f (i, j) ̸ ≡ 0 (mod p l ) and * if f (i, j) ≡ 0 (mod p l ).

0, otherwise
We say the 2-adic valuation ν 2 (f (x, y)) of a polynomial f (x, y) has a closed-form if its 2-adic valuation tree has no infinite branch.Hence the 2-adic valuation of x 2 + y 2 + xy + x + y + 1 admits a closed form.
2 Some examples of 2-adic valuation tree Now we look at various diverse 2-adic valuation trees which can be described completely.
Example 2.1.The first three levels of the 2-adic valuation tree of x 2 + y 2 are shown in Figure 3.By investigating the nature of the above valuation tree, we find an interesting pattern.By using Theorem 2.1 we can find the label of next k levels once the label of k-th level is given.We need some notation to state the result: Consider the binary representation of b k and c k , Theorem 2.1.Let v be a vertex at the k-th level of the valuation tree of x 2 + y 2 .Let the pair ) and if w is a vertex descending from v at l-th level, then w is labelled with * whenever l ∈ {k + 1, k + 2, . . ., (2k − 1)} and with (2k − 1) for l = 2k.
Proof.We are given that On putting the expression for b k+l−1 and c k+l−1 in the above equation we get ), then the least power of 2 in equation ( 3) is 2 2k−1 .So vertices descending from v at k + 1, k + 2, . . ., (2k − 1)-th levels will be labelled with * and at the 2k-th level all vertices descending from v will be labelled with 2k − 1.
Example 2.2.The first few levels of the 2-adic valuation tree of x 2 + y 2 + xy + x + y has also a specific pattern as shown in where i 0 , i 1 , . . ., i k−1 , j 0 , j 1 , . . ., j k−1 ∈ {0, 1} and (i 0 , j 0 ) ̸ = (1, 1).We want to find ).On putting the expression of b k and c k in the above equation, we get We know that b , so the above equation becomes ⇒ i k (j 0 + 1) + j k (i 0 + 1) ≡ a (mod 2) Now if (i 0 , j 0 ) = (0, 0), then equation (2) becomes i k +j k ≡ a (mod 2).Hence there are two vertices labelled with k descending from v with i k + j k ̸ ≡ a (mod 2) and the other two vertices are not terminating labelled with * .
If (i 0 , j 0 ) = (1, 0), then equation ( 2) becomes i k ≡ a (mod 2).Hence there are two vertices labelled with k descending from v with i k ̸ ≡ a (mod 2) and the other two vertices are nonterminating labelled with * .Similarly for (i 0 , j 0 ) = (0, 1), equation (2) becomes j k ≡ a (mod 2).Hence there are two vertices labelled with k descending from v with j k ̸ ≡ a (mod 2) and the other two vertices are non-terminating labelled with * .By analysing the pattern in the above tree, we can state the following result: Theorem 2.3.Let v be a vertex labelled with * at the k-th level of the valuation tree of xy + x + y + 1, for k > 0. Then this vertex v splits into four vertices such that 1.If (i 0 , j 0 ) = (1, 1), then all four vertices will be labelled with * or k + 1.
The root vertex v 0 splits into four vertices, each labelled with * .
Proof.Let (b k−1 , c k−1 ) be associated with the vertex v at the k-th level of the valuation tree of xy where i 0 , i 1 , . . ., i k , j 0 , j 1 , . . ., j k ∈ {0, 1}.We want to find On putting the expression of b k and c k in the above equation, we get , so the above equation becomes Now if (i 0 , j 0 ) = (0, 0), then equation ( 5) becomes i k + j k ≡ a (mod 2).Hence there are two vertices labelled with k descending from v with i k + j k ̸ ≡ a (mod 2) and the other two vertices are non-terminating labelled with * .If (i 0 , j 0 ) = (1, 0), then equation ( 5) becomes i k ≡ a (mod 2).Hence there are two vertices labelled with k descending from v with i k ̸ ≡ a (mod 2) and the other two vertices are nonterminating labelled with * .
If (i 0 , j 0 ) = (0, 1) then equation ( 5) becomes j k ≡ a (mod 2).Hence there are two vertices labelled with k descending from v with j k ̸ ≡ a (mod 2) and the other two vertices are nonterminating labelled with * .
Remark.It may be pointed out there that the polynomials in the theorems 2.1, 2.2, 2.3 are symmetric.However, in section 4 and 5 we will see similar results for the asymmetric polynomials.
Then any (i, j) Proof.Let the sequence of indices generated to come an infinite branch of the tree at n-th level be Further, consider the sequence of the valuations We prove that: p .In the periodic case, the minimal period is a power of p.
Proof.Assume that f has no zeros in Z 2 p .If V p (f ) is not bounded there exists a sequence (n j , m j ) → ∞ such that V p (f ) → ∞.The compactness of Z 2 p gives a subsequence converging to Then d ≥ 0 and Since proving that ν p (f (n, m)) is periodic and minimal period is a divisor of p d+1 .On the other hand, if f (x, y) has a zero α(y) inZ p [y], Then v p (f ) ≥ v p (x − α), and V p (f ) is unbounded.
Theorem 4.2.Let v be a vertex at the k-th level of the valuation tree of f (x, y) labelled with * for k > 1.Then v splits into four vertices such that either all are non-terminating or two of them are non-terminating.Proof.We are given that f (x, y) = ax 2 + by 2 + cxy + dx + ey + g.Let (b k−1 , c k−1 ) be associated with the vertex v at the k-th level of the valuation tree.So we have f Here i 0 , i 1 , . . ., i k , j 0 , j 1 , . . ., j k ∈ {0, 1} and (b 0 , c 0 ) = (i 0 , j 0 ).We want to find (i k , j k ) such that f (b k , c k ) ≡ 0 (mod 2 k+1 ).
On putting the expression for (b k , c k ) in the above equation we get The following Table 1 gives the all possible cases for (i k , j k ) when α ≡ 0 (mod 2).
Hence the theorem is proved for α ≡ 0 (mod 2).For the case α ≡ 1 (mod 2), we can find the appropriate label by interchanging the * and k in the last column of Table 1.Hence the theorem is proved in this case as well.
Remark.If the label of k-th level is given then Theorem 4.2 describes only about the label of the (k + 1)-th level of the valuation tree but Theorem 4.1 describes the label of the next k levels of the valuation tree of given f (x, y).
To construct the whole tree one starts with the root vertex and recursively apply the following algorithm to every non-terminal vertex: Algorithm 1.To find the label of the children of given non-terminating vertex v at k-th level, k > 1: 2 : Use Table 1, (i 0 , j 0 ), α to find the label of the children of given vertex where (i 0 , j 0 ) is corresponding to the parent vertex of the given non-terminating vertex v.

polynomial in two variables
To study the 2-adic valuation tree of the higher degree polynomials effectively we need some advanced techniques.One of the technique, we would like to highlight is the generalized Hensel's lemma [9,10].We study the 2-adic valuation tree of the polynomial x 2 y + 5 shown in Figure 7.
On putting the expression for (b k , c k ) in the above equation we get Therefore there are two vertices labelled with k descending from v with j k ̸ ≡ a (mod 2) and the other two vertices are non-terminating labelled with * .
We can prove Theorem 5.1 by using the generalized Hensel's lemma [9].

Conjecture 1 . 1 .
Let k ∈ N be fixed.Then we conjecture that (a) there exists a level m 0 (k) and an integer µ(k) such that for any m ≥ m 0 (k), the number of non-terminal classes of level m is µ(k), independently of m;(b) moreover, for each m ≥ m 0 (k), each of the µ(k) non-terminal classes splits into one terminal and one non-terminal subclass.The latter generates the next level set.

Figure 3 .
Figure 3.The first three levels of the 2-adic valuation tree of x 2 + y 2

Figure 4 .
Figure 4.The first few levels of the 2-adic valuation tree of x 2 + y 2 + xy + x + y

Now a n
and b n satisfy: 0 ≤ a n , b n ≤ p n and a n ≡ a n+1 (mod p n ) , b n ≡ b n+1 (mod p n ).Hence sequences {a n } and {b n } are convergent to some element in the fieldQ p .Let {(a n , b n )} converges to (x, y) for x, y ∈ Q p .Since the polynomial f (x, y) is continuous so f (a n , b n ) converges to f (x, y).Now by Lemma 3.1, we know that v p (f (a n , b n )) tends to ∞ as n tends to ∞ so f (a n , b n )tends to 0 when n tends to ∞. Hence f (x, y) = 0.So any infinite branch in the tree associated to polynomial f (x, y) corresponds to a root of f (x, y) = 0 in Q p × Q p .A valuation tree having an infinite branch cannot have a closed-form.Since the polynomials in examples 2.1, 2.2 and 2.3 have zeros in Q p × Q p , they do not admit closed-form for 2-adic valuation.

Theorem 5 . 1 .Figure 7 .
Figure 7.The first two levels of the p-adic valuation tree of x 2 y + 5