Representation of Integers of the Form x^{2} + my^{2} − z^{2}

Supawadee Prugsapitak and Nattaporn Thongngam Algebra and Applications Research Unit Division of Computational Science Faculty of Science Prince of Songkla University Hatyai, Songkhla 90110 Thailand

                     supawadee.p@psu.ac.th
                     6310220031@psu.ac.th

Abstract Let m be a positive integer. A positive integer k is called m-special if every integer n can be expressed as n = x^{2} + my^{2} − kz^{2} for some nonzero integers x, y and z. In this article, we show that 1 is m-special if and only if m is not divisible by 4.

1 Introduction

Representations of integers as sums of squares have a long history. Ramanujan [4] proved that there exist 54 quadruples (a, b, c, d) with 1 ≤ a ≤ b ≤ c ≤ d such that every natural number n is representable in the form aw^{2} +bx^{2} +cy^{2} +dz^{2}. Panaitopol [3] proved that there exists no triple (a, b, c) with 1 ≤ a ≤ b ≤ c such that every natural number is representable in the form ax^{2} + by^{2} + cz^{2}. However, if we allow c in the representation ax^{2} + by^{2} + cz^{2} to be negative, then the representation is possible. Nowicki [2] showed that if all natural numbers are representable in the form x^{2} + y^{2} − cz^{2}, then c is of the form q or 2q, where either q = 1 or q is a product of primes of the form 4m + 1. Lam [1] proved its sufficiency. In what follows, we are concerned with the problem of representing natural numbers n in the form n = x^{2} + my^{2} − z^{2} for a given positive integer m, where xyz ≠ 0. We provide necessary and sufficient conditions for representing all integers n in the form x^{2} + my^{2} − z^{2} where x, y and z are nonzero.

Definition 1. A positive integer k is m-special if for every integer n there exist nonzero integers x, y, and z such that n = x^{2} + my^{2} − kz^{2}.

Nowicki [2] showed that 1 is 1-special, that is, for every integer n there exist nonzero integers x, y, and z such that n = x^{2} + y^{2} − z^{2}. Nowicki [2] and Lam [1] showed that k is 1-special if and only if k is of the form q or 2q where either q = 1 or q is a product of primes of the form 4m + 1.

2 Main Results

Theorem 2. Let m be a positive integer. If m is divisible by 4, then 1 is not m-special.

Proof. Let m be a positive integer divisible by 4. Assume that 1 is m-special. Then there exist nonzero integers x, y, z such that x^{2} + my^{2} − z^{2} = 2. So we have x^{2} − z^{2} ≡ 2 (mod 4). Since quadratic residues modulo 4 are 0 and 1, it follows that x^{2} − z^{2} ≡ 0, 1, 3 (mod 4). This is a contradiction. Hence 1 is not m-special.

In 2015, Nowicki [2] showed that 1 is 1-special.

Lemma 3. [2] 1 is 1-special.

We present an elementary result. The following well-known lemma plays an important role in our main result. The proof is given in [5, Theorem 3.18]:

Lemma 4. Let m be a rational number. Then m is an integer if and only if m^{2} is an integer.

Theorem 5. Let m > 1 be a positive integer . If m is not divisible by 4, then 1 is m-special.

Proof. Let m > 1 be a positive integer not divisible by 4. We show that for any integers n there exist nonzero integers x, y and z such that n = x^{2} + my^{2} − z^{2}, i.e., x^{2} − z^{2} = (x − z)(x + z) = n − my^{2}. In each case, we provide a choice of nonzero integers x, y and z.

Case 1: Suppose n ≡ 0 (mod 4). Thus n = 4l for some integer l. If l ≠ 0, then we choose

y = 4l + 2, x = 1 + l − m(2l + 1)^{2}, z = −1 + l − m(2l + 1)^{2}.

Obviously, y ≠ 0. We next show that x and z are nonzero. If x = 0 or z = 0, then m(2l + 1)^{2} = l + 1 or m(2l + 1)^{2} = l − 1 respectively. Since m > 1, we have l > 0 or l > 3, respectively. Both cases imply that m is not an integer, which is a contradiction.

Case 2: Suppose n ≡ 1 (mod 4). Thus n = 4l + 1 for some integer l. In this case, we choose

y = 2(4l + 1), x = 1 + 2l − 2m(4l + 1), z = −2l − 2m(4l + 1).

Again, we have that y ≠ 0. If x = 0, then 1 is divisible by 2, which is a contradiction. If z = 0, then m^{2}(4l + 1)^{2} = l^{2}. Thus m^{2} is not an integer for all l ≠ 0. Then by Lemma 4, we get that m is not an integer. Thus x and z are nonzero.

Case 3: Suppose n ≡ 2 (mod 4). Thus n = 4l + 2 for some integer l.

Case 3.1: Suppose m ≡ 2 (mod 4). Hence m = 4k + 2 for some non-negative integer k. We choose

y = 2l + 1, x = l − k − 2kl + 1, z = −3l − k − 2kl − 1.

It is easy to see that y ≠ 0. If x = 0 or z = 0, then k = \frac{l+1}{2l+1} or k = \frac{−(3l+1)}{2l+1} = −1 − \frac{l}{2l+1} respectively. Since k is an integer, we have l = −1 or 0. Therefore x or z can be zeros only if (m, n) = (2, −2), (6, 2). To handle these cases, we can use the representations

2 = x^{2} + 6y^{2} − z^{2} where (x, y, z) = (12, 3, 14)
−2 = x^{2} + 2y^{2} − z^{2} where (x, y, z) = (4, 3, 6).

Case 3.2: Suppose m ≡ 1 (mod 2). Therefore m = 2k + 1 for some positive integer k. Here we choose

y = 2l + 1, x = 1 − k − 2kl, z = −k − 2l − 2kl.

Since y is odd, we have that y is nonzero. If x = 0, then l = 0 and k = 1. We obtain m = 3 and n = 2. For this exceptional case, we use the representation 2 = 12^{2} + 3(3)^{2} − 13^{2}. If z = 0, then k^{2}(1 + 2l)^{2} = (2l)^{2}. Thus k^{2} is not an integer for l ≠ 0. Thus again by Lemma 4, k is not an integer for l ≠ 0.

Case 4: Let n ≡ 3 (mod 4). Thus n = 4l + 3 for some integer l. We choose

y = 2(4l + 3), x = 2l − 6m − 8ml + 2, z = −2l − 6m − 8ml − 1.

It is easy to see that y and z are nonzero. If x = 0, then m(4l+3) = l+1. Since m = \frac{l+1}{4l+3} > 1, we obtain that l is not an integer which is a contradiction.

In conclusion, we have proved the following theorem.

Theorem 6. Let m be a positive integer. Then 1 is a m-special if and only if m is not divisible by 4.

3 Acknowledgments

We thank the referee for valuable comments and suggestions.

References

[1] P. C. H. Lam, Representation of integers using a^{2} + b^{2} − dc^{2}, J. Integer Sequences 18 (2015), Article 15.8.6.

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

[3] L. Panaitopol, On the representation of natural numbers as sums of squares. Amer. Math. Monthly, 112 (2005), 168–171.

[4] 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.

[5] K. H. Rosen, Elementary Number Theory and its Applications, 5th edition, Pearson Education Inc. Press, 2005.

2010 Mathematics Subject Classification: Primary 11E25. Keywords: sum of squares.

Received March 18 2021; revised versions received April 6 2021; July 10 2021. Published in Journal of Integer Sequences, July 19 2021.

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