Representation of Integers Using a^{2} + b^{2} − dc^{2}

Peter Cho-Ho Lam Department of Mathematics Simon Fraser University Burnaby, BC V5A1S6 Canada

                         chohol@sfu.ca

Abstract

A positive integer d is called special if every integer m can be expressed as a^{2} +b^{2} − dc^{2} for some nonzero integers a, b, c. A necessary condition for special numbers was recently given by Nowicki, and in this paper we prove its sufficiency. Thus, we give a complete characterization for special numbers.

1 Introduction

Many problems in number theory are concerned with the representation of integers by multivariate polynomials with integral coefficients and variables. For example, the well-known theorem of Lagrange asserts that every positive integer is the sum of four squares. Ramanujan [3] gave a complete list of general quadratic forms with four variables,

Q(x, y, z, w) = ax^{2} + by^{2} + cz^{2} + dw^{2},

that represent all positive integers, where a, b, c, d ∈ ℕ. Note that it is not possible to represent all positive integers if we reduce one variable in Q(x, y, z, w); in fact, it cannot even represent all integers from 1 to 10 by a very elementary argument. However, three variables are sufficient if we use indefinite quadratic forms. For example, the form Q(x, y, z) = x^{2} + y^{2} − z^{2} can represent all integers with integral x, y, z since x^{2} − z^{2} represents all odd integers and one can pick y = 0, 1. To generalize this, Nowicki [2] defined special numbers and proved a necessary condition for them, which is stated in Theorem 2:

Definition 1. A positive integer d is special if for every integer m there exist nonzero integers a, b, c such that m = a^{2} + b^{2} − dc^{2}.

Theorem 2. Every special number d is of the form q or 2q, where either q = 1 or q is a product of primes of the form 4m + 1.

Nowicki [2] further verified that the converse is true when d ≤ 50 through various identities. For example, when d = 13, we have

\begin{aligned}(2k − 4)^{2} + (3k − 10)^{2} − 13(k − 3)^{2} = (2k − 30)^{2} + (3k − 36)^{2} − 13(k − 13)^{2} & = 2k − 1, \\ (2k − 3)^{2} + (3k − 2)^{2} − 13(k − 1)^{2} = (2k − 29)^{2} + (3k − 54)^{2} − 13(k − 17)^{2} & = 2k. \end{aligned}

We need two identities for each parity since we require a, b, c to be nonzero. Similarly, when d = 34, we have

\begin{aligned}(3k − 7)^{2} + (5k − 16)^{2} − 34(k − 3)^{2} = (3k − 24)^{2} + (5k − 33)^{2} − 34(k − 7)^{2} & = 2k − 1, \\ (3k − 11)^{2} + (5k − 27)^{2} − 34(k − 5)^{2} = (3k − 45)^{2} + (5k − 61)^{2} − 34(k − 13)^{2} & = 4k, \\ (3k − 1)^{2} + (5k + 1)^{2} − 34(k)^{2} = (3k − 69)^{2} + (5k − 135)^{2} − 34(k − 26)^{2} & = 4k + 2. \end{aligned}

In this paper, we prove the converse of Theorem 2, and hence give a complete characterization of special numbers:

Theorem 3. If d is of the form q or 2q, where either q = 1 or q is a product of primes of the form 4m + 1, then d is special.

2 Proof of Theorem 3

First, we invoke the following well-known lemma, where the proof is given in [1, Theorem 3.20]:

Lemma 4. A positive integer n can be expressed as the form q or 2q where q is a product of primes of the form 4m + 1 if and only if n can be expressed as the form n = x^{2} + y^{2} where x, y ∈ ℕ and gcd(x, y) = 1.

Proof of Theorem 3. In what follows, we assume d > 1, since d = 1 is already known to be special.

Suppose d is odd. Then all prime factors of d are of the form 4m + 1. By Lemma 4, we can write d = x^{2}+y^{2} where gcd(x, y) = 1, and x ≢ y (mod 2). Now let a = xk+α, b = yk+β and c = k, where α and β are integers which will be chosen later. It follows that

\begin{aligned}a^{2} + b^{2} − dc^{2} & = (xk + α)^{2} + (yk + β)^{2} − (x^{2} + y^{2})(k)^{2} \\ & = 2(xα + yβ)k + α^{2} + β^{2}. \end{aligned}  (1)

We consider the solution pairs (α, β) to the equation

xα + yβ = 1.  (2)

It suffices to show that α^{2} + β^{2} cover both parities. Note that (2) must have an integral solution (α_{0}, β_{0}) since gcd(x, y) = 1. If we define α_{1} = α_{0} + y and β_{1} = β_{0} − x, then (α_{1}, β_{1}) is another solution of (2). Now observe

α_{1}^{2} + β_{1}^{2} = (α_{0} + y)^{2} + (β_{0} − x)^{2}
= (α_{0}^{2} + β_{0}^{2}) + x^{2} + y^{2} + 2(α_{0}y + β_{0}x)
≡ (α_{0}^{2} + β_{0}^{2}) + x^{2} + y^{2} (mod 2).

Since x^{2} + y^{2} = d is odd, α_{0}^{2} + β_{0}^{2} ≢ α_{1}^{2} + β_{1}^{2} (mod 2). The two identities given by

(xk + α_{i})^{2} + (yk + β_{i})^{2} − (x^{2} + y^{2})(k)^{2} = 2k + α_{i}^{2} + β_{i}^{2},

where i = 0, 1, cover both odd and even integers, and hence every integer can be expressed as the form a^{2} + b^{2} − dc^{2} for some integers a, b, c.

However, one of the variables a, b, c becomes zero in the representations of

m = α_{i}^{2} + β_{i}^{2}, −\frac{2α_{i}}{x} + α_{i}^{2} + β_{i}^{2}, −\frac{2β_{i}}{y} + α_{i}^{2} + β_{i}^{2}  (3)

for i = 0, 1. To fix this problem, we can simply set α_{n} = α_{0}+ny and β_{n} = β_{0}−nx to generate more identities, where n ∈ ℕ. As n → ∞, the absolute values of α_{n} and β_{n} approach infinity. Thus for sufficiently large n, the new exceptional cases do not overlap with the original ones, and the values in (3) can be represented using the new identities.

Now suppose d is even. Then d = 2q where q is a product of primes of the form 4m + 1. Again by Lemma 4, we can write d = x^{2} + y^{2} where gcd(x, y) = 1, but this time x ≡ y ≡ 1 (mod 2). We have a similar expansion as (1), and if xα + yβ = 1, then α ≢ β (mod 2) and α^{2} + β^{2} ≡ 1 (mod 2). Therefore we have an identity that generates all odd integers.

But in this case, shifting the solution (α, β) of (2) does not produce an identity for even integers. Therefore in (1) we consider the linear equation

xα + yβ = 2.  (4)

Now we pick a pair of solution (α_{0}, β_{0}), and construct the second solution pair (α_{1}, β_{1}) in a similar manner, and then

α_{1}^{2} + β_{1}^{2} = (α_{0} + y)^{2} + (β_{0} − x)^{2}
= (α_{0}^{2} + β_{0}^{2}) + x^{2} + y^{2} + 2(α_{0}y + β_{0}x)
≡ (α_{0}^{2} + β_{0}^{2}) + 2 + 2(α_{0}y + β_{0}x) (mod 4).

Since x and y are odd, and the right hand side of (4) is even, we deduce that α_{0} ≡ β_{0} (mod 2). Therefore 2 | (α_{0}y + β_{0}x) and

α_{1}^{2} + β_{1}^{2} ≡ α_{0}^{2} + β_{0}^{2} + 2 (mod 4).

Also note that α_{0}^{2} + β_{0}^{2} ≡ 0 (mod 2). Thus, the two identities given by

(xk + α_{i})^{2} + (yk + β_{i})^{2} − (x^{2} + y^{2})(k)^{2} = 2k + α_{i}^{2} + β_{i}^{2},

where i = 0, 1, cover both integers of the form 4m and 4m + 2, and hence every integer can be expressed as the form a^{2} + b^{2} − dc^{2} for some integers a, b, c. The exceptional cases can be handled similarly as in the d = q case.

References

[1] I. Niven, H. S. Zuckerman, and H. L. Montgomery, An Introduction to the Theory of Numbers, John Wiley & Sons, Inc., 1991.

[2] A. Nowicki, The numbers a^{2} + b^{2} − dc^{2}, J. Integer Seq. 18 (2015), Article 15.2.3.

[3] S. Ramanujan, On the expression of a number in the form ax^{2} + by^{2} + cz^{2} + dw^{2}, Proc. Cambridge Philos. Soc. 19 (1917), 11–21. Reprinted in Collected Papers of Srinivasa Ramanujan, AMS Chelsea Publishing, 2000, pp. 169–178.

2010 Mathematics Subject Classification: Primary 11D09. Keywords: sum of squares, sum of two coprime squares.

Received July 14 2015; revised version received July 29 2015. Published in Journal of Integer Sequences, July 29 2015.

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