A pdf version is Dirichlet hyperbola method.

**1. Introduction **

Remark 1I thought this problem initial 5 years ago, cost me several days to find a answer, I definitely get something without the argument of Dirchlet hyperbola method and which is weaker but morally the same camparable with the result get by Dirichlet hyperbola method.

Remark 2How to get the formula:

In fact,

Which is the integer lattices under or lying on the hyperbola .

Remark 3By trivial argument, we can bound the quantity as following way,

The error term is , which is too big. But fortunately we can use the symmetry of hyperbola to improve the error term.

*Proof:*

Given a natural number k, use the hyperbola method together

with induction and partial summation to show thatwhere denotes a polynomial of degree with leading term .

Remark 4is the residue of at .

*Proof:*

We can establish the dimension 3 case directly, which is the following asymptotic formula,

The approach is following, we first observe that

The problem transform to get a asymptotic formula for the lattices under 3 dimension hyperbola. The first key point is, morally is the central point under the hyperbola.

Then we can divide the range into 3 parts, and try to get a asymptotic formula for each part then add them together. Assume we have:

- .
- .
- .

Then the task transform to get a asymptotic formula,

But we can do the same thing for and then integral it. This end the proof. For general , the story is the same, by induction.

Induction on and use the Fubini theorem to calculate .

There is a major unsolved problem called Dirichlet divisor problem.

What is the error term? The conjecture is the error term is , it is known that is not right.

Remark 5To beats this problem, need some tools in algebraic geometry.

**2. Several problems **

, is there a asymptotic formula for ?

, is a polynomial with degree , is there a asymptotic formula for ?

, is a polynomial with degree , is there a asymptotic formula for ?

*Proof:*

Now we try to estimate

In fact, we have,

Where , .

So by 1 we know,

So we have,

Remark 6In fact we can get , by combining the theorem 3 and 1.

**3. Lattice points in ball **

Gauss use the cube packing circle get a rough estimate,

In the same way one can obtain,

Remark 7Where is the volume of the unit ball in dimension.

Dirchlet’s hyperbola method works nicely for the lattic points in a ball of dimension . Langrange proved that every natural number can be represented as the sum of four squares, i.e. , and Jacobi established the exact formula for the number of representations

Hence we derive,

This result extend easily for any , write as the additive convolution of and , i.e.

Apply the above result for and execute the summation over the remaining squares by integration.

Remark 8Notice that this improve the formula 12 which was obtained by the method of packing with a unit square. The exponent in 16 is the best possible because the individual terms of summation can be as large as the error term (apart from ), indeed for we have if is odd by the Jacobi formula. The only case of the lattice point problem for a ball which is not yet solved (i.e. the best possible error terms are not yet established) are for the circle() and the sphere ().

Theorem 4

**4. Application in finite fields **

Suppose is a irreducible polynomial. And for each prime , let

By Langrange theorem we know . Is there a asymptotic formula for

A general version, we can naturally generated it to algebraic variety.

Is there a asymptotic formula for

Example 1We give an example to observe what is involved. . We know is solvable iff or . One side is easy, just by Fermat little theorem, the other hand need Fermat descent procedure, which of course could be done by Willson theorem. In this case,

Which is a special case of Dirichlet prime theorem.

Let be an algebraic number field, i.e. the finite field extension of rational numbers, let

Theorem 5

- is a ring, we call it the ring of integer of .
- He showed further every non-zero ideal of could write as the product of prime ideal in uniquely.
- the index of every non-zero ideal in is finite, i.e. , and we can define the norm induce by index.
Then the norm is a multiplication function in the space of ideal, i.e. .

- Now he construct the Dedekind Riemann zeta function,

Now we consider the analog of the prime number theorem. Let , does the exist a asymptotic formula,

Given a prime , we may consider the prime ideal

Where is different prime ideal in . But the question is how to find these ? For the question, there is a satisfied answer.

Lemma 6 (existence of primitive element)There always exist a primetive elements in , such that,

Where is some algebraic number, which’s minor polynomial .

Theorem 7 (Dedekind recipe)Take the polynomial , factorize it in the polynomial ring ,

Consider . Then apart from finite many primes, we have,

Where .

Remark 9The apart primes are those divide the discriminant.

Now we can argue that 4 is morally the same as counting the ideals whose norm is divide by in a certain algebraic number theory.

And we have following, which is just the version in algebraic number fields of 2.

Theorem 8 (Weber)of ideals of with norm equal to,