How many lattice points are there on a circle or a sphere?
In this chapter we study the distribution of points with integral coordinates on spheres.
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How many lattice points are there on a circle or a sphere?
A point in Rn with integral coordinates is called a lattice point. In this chapter we study the distribution of lattice points on circles and spheres in Rn . We start by finding a formula for the number r (n) of points with integral coordinates on the circle x 2 + y 2 = n for a natural number n. We then prove a famous theorem of Gauss that gives an expression for the sum kn=1 r (n). We then state similar theorems for the number of points on higher dimensional spheres. At the end we state and prove a theorem of Jarnik (Theorem 9.9), and a recent generalization due to Cilleruelo and Córdoba (Theorem 9.10), about integral points on arcs. In the Note, we discuss the error term in Gauss’ theorem mentioned above.
9.1 The case of two squares For a natural number n, we let r (n) be the number of representations of n as a sum of two integral squares, i.e., the number of integral points on the circle x 2 + y 2 = n. By Theorem 5.7 we know that if we write pβ p n = m · 2α p≡1 mod 4
with m a product of primes of the form 4k + 3, then r (n) = 0 unless m is a square. Theorem 9.1. If m is a square, r (n) = 4
(1 + β p ).
p
Proof. If n = x 2 + y 2 , then n = N (x + i y). So we need to determine the number of Gaussian integers z such that n = N (z). By Theorem 5.10 any such z is a product e f p p pp . z = uk(1 + i)a p≡1 mod 4
© Springer Nature Switzerland AG 2018 R. Takloo-Bighash, A Pythagorean Introduction to Number Theory, Undergraduate Texts in Mathematics, https://doi.org/10.1007/978-3-030-02604-2_9
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9 How many lattice points are there on a circle or a sphere?
Here u is one of the four units in Z[i]; k ∈ N is a product of primes of the form 4k +3; and all but finitely many of the non-negative integers e p , f p are zero, meaning the product is finite. Then we have N ( p )e p N ( p ) f p N (z) = N (k)N (1 + i)a p≡1 mod 4
= k 2 2a
p e p p f p = k 2 2a
p≡1 mod 4
pe p + f p .
p≡1 mod 4
Consequently, since N (z) = n we get βj p j = k 2 2a m2α
pe p + f p .
p≡1 mod 4
j
This implies m = k 2 , a = α, and for each p of the form 4k + 1, e p + f p = β p . The number of such e p , f p is 1 + β p . Since there are four possibilities for the unit u, i.e., ±1, ±i, we get a total of 4 (1 + β p ) p
choices for z. This finishes the proof.
For example, we have 180 = 32 · 22 · (2 + i) · (2 − i). So we get the following numbers as the list of numbers z that have the property that N (z) = 180: u · 3 · (1 + i)2 · (2 + i) = u(−6 + 12i) and u · 3 · (1 + i)2 · (2 − i) = u(6 + 12i) for u ∈ {+1, −1, i, −i}. This means that the possibilities for the ordered pairs (a, b) such that 180 = a 2 + b2 are: (±6, ±12), (±12, ±6), a total of eight possibilities. Now that we have a formula for r (n) one could ask natural statistical questions about it. For example, one could ask what the average behavior of r (n) is like. Let us make this notion precise. Definition 9.2. Suppose f : N → C is a function. We say f has average value equal to c if the
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