Restricted sums of four squares

0.24 MB·

by  Zhi-Wei Sun

Preprint, arXiv:1701.05868
RESTRICTED SUMS OF FOUR SQUARES
Zhi-Wei Sun
7
1
0 Department of Mathematics, Nanjing University
2 Nanjing 210093, People’s Republic of China

r [email protected]
a
M http://math.nju.edu.cn/∼zwsun

8
2 Abstract. We refine Lagrange’s four-square theorem in new ways by imposing
somerestrictionsinvolvingpowersof two(including1). Forexample,weshowthat
T] eachn∈N={0,1,2,...} can be writtenas x2+y2+z2+w2 (x,y,z,w ∈N) with
N |2x−y|∈{4k : k ∈N} (or x+y−z ∈{±8k : k ∈N}∪{0} ⊆{t3 : t∈Z}), and
that any positive integer can be written as x2+y2+z2+w2 (x,y,z,w ∈Z) with
.
h x+y+z (or x+y+2z, or x+2y+2z) a power of four. We also pose some open
t
a conjectures; for example,we conjecture that any positive integercan be written as
m x2+y2+z2+w2 with x,y,z,w ∈N such that x+2y−2z is a power of four, and
[ that each n = 0,1,2,... can be written as x2 +y2 +z2 +w2 with x,y,z,w ∈ N

such that both x and x+24y are squares.
5
v
8
6
8 1. Introduction
5
0
The celebrated four-square theorem (cf. [N, pp.5-7]) proved by J. L. Lagrange
.
1
in 1770 states that any n ∈ N = {0,1,2,...} can be written as x2 +y2 +z2 +w2
0
7 with x,y,z,w ∈ N. Recently, the author [S17b] found that this can be refined
1
in various ways by requiring additionally that P(x,y,z,w) is a square, where
:
v P(x,y,z,w) is a suitable polynomial with integer coefficients. (For example, we
i
X may take P(x,y,z,w) = x2y2 + y2z2 + z2x2.) Here is a challenging conjecture
r posed by the author.
a
1-3-5 Conjecture ([S17b, Conjecture 4.3(i)]). Any n ∈ N can be written as
x2 +y2 +z2 +w2 with x,y,z,w ∈ N such that x+3y +5z is a square.
In this paper we aim to refine Lagrange’s four-square theorem in a new direc-
tion by imposing restrictions involving power of two (including 20 = 1). [S17b,
Theorem 1.1] asserts that for any a ∈ {1,4} and m ∈ {4,5,6} we can write each
n ∈ N as axm+y2+z2+w2 with x,y,z,w ∈ N. Actually the proof in [S17b] works
for a stronger result which requires additionally that x ∈ {2k : k ∈ N} ∪ {0}.
Similarly, by modifying the proof of [S17b, Theorem 1.2(i)]slightly we get that for
2010 Mathematics Subject Classification. Primary 11E25; Secondary 11D85, 11E20, 11P05.
Keywords. Representations of integers, sums of four squares, powers of two.
Supported by the National Natural Science Foundation (Grant No. 11571162) of China.
1
2 ZHI-WEI SUN
any a ∈ {1,2} each n ∈ N can be written as x2 +y2+z2 +w2 with x,y,z,w ∈ N
and a(x−y) ∈ {4k : k ∈ N}∪{0}.
Now we state our first theorem.
Theorem 1.1. (i) Any n ∈ Z+ = {1,2,3,...} can be written as x2+y2+z2+8k
with x,y,z ∈ N and k ∈ {0,1,2}.
(ii) Any n ∈ Z+ can be written as x2 +y2 +z2 + w2 with x,y,z,w ∈ N such
that x−y = 2⌊ord2(n)/2⌋, where ord (n) is the 2-adic order of n.
2
(iii) Each n ∈ N not of the form 26k+3×7 (k ∈ N) can be written as x2+y2+z2+
w2 with x,y,z,w ∈ N and x−y ∈ {8k : k ∈ N}∪{0}. Consequently, we can write
any n ∈ N as x2+y2+z2+w2 with x,y,z,w ∈ Z and x+y ∈ {8k : k ∈ N}∪{0}.
(iv) Any n ∈ Z+ can be written as x2 +y2 +z2 +w2 with x,y,z,w ∈ Z such
that x+y +z +w = 2⌊(ord2(n)+1)/2⌋.
Remark 1.1. For integers y and z of the same parity, clearly
2 2
y +z y −z
y2 +z2 = 2 +2 .
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
So, Theorem 1.1(i) implies that any n ∈ N can be written as x2 +2y2 +2z2 +8k
with x,y,z ∈ N and k ∈ {0,1,2}. As {8k : k ∈ N} ∪ {0} ⊆ {m3 : m ∈ N},
Theorem 1.1(iii) implies a conjecture stated in [S17b, Remark 1.2]. Theorem
1.1(iv) with ord (n) = 1 was first realized by Euler in a letter to Goldbach dated
2
June 9, 1750 (cf. Question 37278 in MathOverFlow). See [S, A281494] for the
number of ways to represent n ∈ Z+ as x2 +y2 +z2 +w2 with x,y,z,w ∈ Z and
|x| 6 |y| 6 |z| 6 |w| such that x+y +z +w = 2⌊(ord2(n)+1)/2⌋. For example,
14 = 02 +12 +(−2)2 +32 with 0+1+(−2)+3 = 2 = 2⌊(ord2(14)+1)/2⌋
and
107 = (−1)2+(−3)2+(−4)2+92 with(−1)+(−3)+(−4)+9 = 1 = 2⌊(ord2(107)+1)/2⌋.
It was proved in [SS] that any n ∈ N can be written as x2 + y2 + z2 + w2
(x,y,z,w ∈ Z) with x+y +z +w a square.
The author (cf. [S17b, Conjecture 4.1]) conjectured that for each ε ∈ {±1} any
n ∈ N can be written as x2 +y2 +z2 +w2 (x,y,z,w ∈ N) with 2x+εy a square.
Y.-C. Sun and the author [SS] confirmed this for ε = 1, but the case ε = −1
remains unsolved. Our next theorem provides an advance in this direction.
Theorem 1.2. (i) Any n ∈ Z+ can be written as x2+y2+z2+w2 with x,y,z,w ∈
N and |2x−y| ∈ {4k : k ∈ N}.
(ii) Let a ∈ {1,2}. Then any n ∈ N can be written as x2 +y2 +z2 +w2 with
x,y,z,w ∈ N and 2x−y ∈ {±a8k : k ∈ N}∪{0} ⊆ {at3 : t ∈ Z}.
Remark 1.2. Motivated by Theorem 1.2(i), we conjecture that any n ∈ Z+ can be
written as x2 +y2 +z2 +w2 (x,y,z,w ∈ Z) with 3x+y ∈ {4k : k ∈ N}.
The author (cf. [S17b, Conjecture 4.3(iii)-(iv)]) conjectured that each n ∈ N
can be written as x2 +y2 +z2 +w2 (x,y,z,w ∈ N) with x+y−z (or x−y −z)
a square. In contrast, we have the following result.
RESTRICTED SUMS OF FOUR SQUARES 3
Theorem 1.3. (i) Any n ∈ Z+ can be written as x2+y2+z2+w2 with x,y,z,w ∈
Z and x + y + z ∈ {4k : k ∈ N}. Also, for any a ∈ {0,2} we can write each
n ∈ Z+ as x2 +y2 +z2 +w2 with x,y,z,w ∈ N and |x+y −z| ∈ {ak,4k}.
k∈N
(ii) Let a ∈ {1,2}. Then any n ∈ N can be written as x2 +y2 +Sz2 +w2 with
x,y,z,w ∈ N and x+y −z ∈ {±a8k : k ∈ N}∪{0} ⊆ {at3 : t ∈ Z}.
Theauthor[S17b]conjectured thatanyn ∈ Ncanbewrittenasx2+y2+z2+w2
with x,y,z,w ∈ N such that P(x,y,z) is a square, where P(x,y,z) may be any
of the polynomials
x+y −2z, 2x+y −z, 2x−y −z, x+2y −2z, 2x−y −2z.
Here we give the following result.
Theorem 1.4. (i) Let c ∈ {1,2}. Then each n ∈ Z+ can be written as x2+y2+
z2 +w2 with x,y,z,w ∈ Z and x+y +2z ∈ {c4k : k ∈ N}.
(ii) Any n ∈ Z+ can be written as x2 + y2 + z2 + w2 with x,y,z,w ∈ Z and
x+2y +2z ∈ {4k : k ∈ N}.
(iii) If n ∈ Z+ does not belong to {24k,24k+3}, then we can write n as
k∈N
x2 + y2 + z2 + w2 with x,y,z,w ∈ ZSand x + 2y + 2z ∈ {3 × 4k : k ∈ N}.
Consequently, any integer n > 1 can be written as x2+y2+z2+w2 with x,y,z,w ∈
Z and x+2y +2z ∈ {3×2k : k ∈ N}.
Remark 1.3. Y.-C. Sun and the author [SS, Theorem 1.2(ii)] showed that for each
d = 1,2,3andm = 2,3, we canwriteany n ∈ N asx2+y2+z2+w2 (x,y,z,w ∈ Z)
with x+2y +2z ∈ {dtm : t ∈ N}.
In contrast with the 1-3-5 Conjecture, we have the following curious conjecture
motivated by Theorem 1.4.
Conjecture 1.1. (i) Any n ∈ Z+ can be written as x2 + y2 + z2 + w2 with
x,y,z,w ∈ N such that x+2(y −z) is a power of four (including 40 = 1).
(ii) Let c ∈ {1,2,4}. Then each n ∈ N can be written as x2+y2+z2+w2 with
x,y,z,w ∈ N and y 6 z such that 2x+y −z ∈ {c8k : k ∈ N}∪{0}.
Remark 1.4. (i) We have verified the first part for all n = 1,... ,2 × 107, and
Qing-Hu Hou extended the verification for n up to 109. See [S, A279612 and
A279616] for related data. For example,
111 = 92 +12 +52 +22 with 9+2×1−2×5 = 40.
(ii) We have verified part (ii) of Conjecture 1.1 for all n = 0,... ,2×106. See
[S, A284343] for related data. For example,
2976 = 202 +162 +482 +42 with 16 < 48 and 2×20+16−48 = 8.
In 1917 Ramanujan [R] listed 55 possible quadruples (a,b,c,d) of positive inte-
gerswitha 6 b 6 c 6 dsuch thatany n ∈ Ncanbewrittenasax2+by2+cz2+dw2
with x,y,z,w ∈ Z, and 54 of them were later confirmed by Dickson [D27] with
the remaining one wrong (see also [W]).
4 ZHI-WEI SUN
Theorem 1.5. (i) Any n ∈ Z+ can be written as x2+y2+z2+2w2 with x,y,z,w ∈
N and x−y = 1. Also, any n ∈ Z+ can be written as x2 +2y2 +2z2 +2w2 with
x,y,z,w ∈ Z and x+y +z = 1.
(ii) Any n ∈ Z+ can be written as x2 + y2 + z2 + 2w2 (x,y,z,w ∈ Z) with
x+y+2z = 1. Also, any n ∈ Z+ can be written as x2+4y2+z2+2w2 (x,y,z,w ∈
Z) with x+2y+2z = 1, and any n ∈ Z+ can be written as x2 +2y2+2z2 +2w2
(x,y,z,w ∈ Z) with x+y +3z = 1.
(iii) Any n ∈ Z+ can be written as x2 + y2 + z2 + 2w2 (x,y,z,w ∈ Z) with
y+z+2w = 1. Also, any n ∈ Z+ can be written as x2+y2+4z2+2w2 (x,y,z,w ∈
Z) with y+2z+2w = 1, and any n ∈ Z+ can be written as x2+2y2+2z2+2w2
(x,y,z,w ∈ Z) with x+y +z +2w = 1.
(iv) Any n ∈ Z+ can be written as x2 + y2 + z2 + 2w2 (x,y,z,w ∈ Z) with
y +z +w = 1.
(v) Any n ∈ Z+ can be written as x2 +y2 +2z2 +5w2 with x,y,z,w ∈ Z and
y+w = 1. If any positive integer congruent to 1 modulo 7 can be written as x2+
7y2+14z2 with x,y,z ∈ Z, then any n ∈ Z+ can be written as x2+y2+2z2+6w2
with y +w = 1.
(vi) Any n ∈ Z+ can be written as x2 + y2 + z2 + 3w2 (x,y,z,w ∈ Z) with
x+y +2z = 2.
(vii) Any integer n > 4 can be written as x2+y2+z2+2w2 with x,y,z,w ∈ Z
such that x+y +z = t2 for some t = 1,2.
(viii) Any integer n > 7 can be written as x2 +y2 +z2 +2w2 (x,y,z,w ∈ Z)
with x+y +2z = 2t2 for some t = 1,2.
Remark 1.5. We can also prove several other results similar to the first asser-
tion in Theorem 1.3(v). Concerning the second assertion in Theorem 1.3(iii),
we conjecture that the condition always holds, i.e., for any n ∈ N we can write
7n+1 = x2 +7y2 +14z2 with x,y,z ∈ Z. I. Kaplansky [K] reported that 2, 74
and 506 are the only positive integers n 6 105 with n ≡ 1,2,4 (mod 7) which
cannot be represented by x2 +7y2 +14z2 with x,y,z ∈ Z.
Theorem 1.6. (i) Any n ∈ N can be written as x2+y2+z2+2w2 (x,y,z,w ∈ Z)
with x+y +z +w a square (or a cube).
(ii) Any n ∈ N can be written as x2 + y2 + z2 + 2w2 (x,y,z,w ∈ Z) with
x+2y +2z a square (or a cube).
Remark 1.5. Our proof of Theorem 1.6 depends heavily on a new identity similar
to Euler’s four-square identity. We even conjecture that any n ∈ N can be written
as x2 +y2 +z2 +2w2 (x,y,z ∈ N and w ∈ Z) with x+y −z +w ∈ {0,1}.
We will prove Theorems 1.1-1.4 and 1.5-1.6 in Sections 2 and 3 respectively,
and pose some related conjectures in Section 4.
2. Proofs of Theorems 1.1-1.4
It is known that the set
E(a,b,c) := {n ∈ N : n 6= ax2 +by2 +cz2 for all x,y,z ∈ Z} (2.1)
RESTRICTED SUMS OF FOUR SQUARES 5
is infinite for any a,b,c ∈ Z+.
Lemma 2.1 (The Gauss-Legendre Theorem). We have
E(1,1,1)= E := {4k(8l+7) : k,l ∈ N}. (2.2)
0
Remark 2.1. This is a well-known result on sums of three squares, see, e.g., [N96,
p.23] or [MW, p.42]).
Proof of Theorem 1.1. (i) If n − 1 6∈ E(1,1,1), then n = 80 + x2 + y2 + z2 for
some x,y,z ∈ N.
Below we assume that n − 1 ∈ E(1,1,1). Then n − 1 = 4k(8l + 7) for some
k,l ∈ N.
If k > 0, then n−8 ≡ n ≡ 1 (mod 4) and hence n−8 = x2+y2+z2 for some
x,y,z ∈ N.
Now we consider the case k = 0. In this case n = 8l +8. If l is odd, then by
Lemma 2.1 we can write n−8 = 23l as x2 +y2 +z2 with x,y,z ∈ N. Clearly,
8 = 02+02+02+8, 3×8 = 42+02+02+8, 5×8 = 42+42+02+8, 7×8 = 42+42+42+8.
If l is even and at least 8, then n−64 = 8(l+1)−64 = 23(l−7) 6∈ E(1,1,1) and
hence n = 82 +x2 +y2 +z2 for some x,y,z ∈ N. This proves Theorem 1.1(i).
(ii) Write n = 2am with a ∈ N, m ∈ Z+ and 2 ∤ m. Let n = n/4⌊a/2⌋ =
0
2a0m, where a is 0 or 1 according as a is even or odd. As 4 ∤ n , we have
0 0
2n − 1 6≡ 7 (mod 8). By Lemma 2.1, 2n − 1 = u2 + v2 + (2y + 1)2 for some
0 0
u,v,y ∈ N with u ≡ v (mod 2). Let z = (u + v)/2 and w = |u − v|/2. Then
2n = 4y2 + 4y + (z + w)2 + (z − w)2 and hence n = x2 + y2 + z2 + w2 with
0 0
x = y +1. It follows that
2 2 2 2
n = 2⌊a/2⌋x + 2⌊a/2⌋y + 2⌊a/2⌋z + 2⌊a/2⌋w
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
with
2⌊a/2⌋x−2⌊a/2⌋y = 2⌊a/2⌋.
(iii) Obviously, for any k ∈ N we have
26k+3 ×7 = (8k ×6)2 +(8k ×4)2 +(8k ×2)2 +02 with 8k ×6+8k ×2 = 8k+1.
So it suffices to prove the first assertion in Theorem 1.1(iii) by induction.
For each n ∈ {0,1,... ,63}\{56}, we can verify via a computer that n can be
written as x2 +y2 +z2 +w2 with x,y,z,w ∈ N and x−y ∈ {8k : k ∈ N}∪{0}.
Now let n > 64 be an integer not of the form 26k+3 ×7 = 64k × 56 (k ∈ N),
and assume that any m ∈ {0,1,... ,n−1} not of the form 26k+3×7 (k ∈ N) can
be written as x2+y2+z2+w2 with x,y,z,w ∈ N and x−y ∈ {8k : k ∈ N}∪{0}.
6 ZHI-WEI SUN
If64 | n, then bythe inductionhypothesiswecan writen/64asx2+y2+z2+w2
with x,y,z,w ∈ N and x−y ∈ {8k : k ∈ N}∪{0}, and hence
n = (8x)2+(8y)2+(8z)2+(8w)2 with 8x−8y = 8(x−y) ∈ {8k : k ∈ N}∪{0}.
Below we suppose that 64 ∤ n.
Case 1. n 6∈ 4k(16l+14) for any k,l ∈ N.
In this case, we have 2n 6∈ E(1,1,1) by (2.2), and hence 2n = (2y)2 +z2 +w2
for some y,z,w ∈ N with z ≡ w (mod 2). Thus
2 2
z +w z −w
n = y2 +y2 + + with y −y = 0.
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
Case 2. n = 16l+14 for some l ∈ N.
In this case, 2n−1 ≡ 3 (mod 8) and hence by (2.2) we can write n as (2y +
1)2 +z2 +w2 with y,z,w ∈ N and z ≡ w (mod 2). It follows that
2 2
z +w z −w
n = (y +1)2 +y2 + + with (y +1)−y = 0.
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
Case 3. n = 4k(16l+14) for some k ∈ {1,2} and l ∈ N.
In this case, 2n−64 = 4k+1(8l+3k). In light of (2.2), 8l+3k = x2+y2+z2 for
some integers x > y > z > 0. If k = 1, then 8l+3k = (2n−64)/16 > 64/16 = 4.
If k = 2, then 8l +3k > 3k = 6. So, x > 2 and hence 2k+1x = 2v +8 for some
v ∈ N. Therefore
2n−64 = (2k+1x)2+(2k+1y)2+(2k+1z)2 = (2v+8)2+2(2ky+2kz)2+2(2ky−2kz)2
and hence
n = (v+8)2 +v2 +(2k(y +z))2 +(2k(y −z))2 with (v +8)−v = 81.
The induction proof of Theorem 1.1(iii) is now complete.
(iv) We distinguish two cases.
Case 1. 4 ∤ n.
Let
1 if 2 ∤ n,
δ = 1−ord (n) =
n 2
(cid:26) 0 if 2kn.
Then
3 (mod 8) if 2 ∤ n,
4δnn−1 ≡
(cid:26) 1 (mod 4) if 2kn.
By Lemma 2.1, there are u,v,w ∈ Z such that
4δnn−1 = 2δnu−1 2 + 2δnv −1 2 + 2δnw−1 2.
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
RESTRICTED SUMS OF FOUR SQUARES 7
If 2kn, then δ = 0 and u + v + w ≡ n ≡ 0 (mod 2). In the case 2 ∤ n, since
n
(2u−1)2 = (2(1−u)−1)2, without loss of generality we may also assume that
u+v +w ≡ 0 (mod 2). Set
u+v −w u−v +w −u+v +w
x = , y = , z = .
2 2 2
Then u = x+y, v = x+z and w = y +z. Therefore
4δnn−1 = 2δn(x+y)−1 2 + 2δn(x+z)−1 2 + 2δn(y +z)−1 2
=(cid:0)2δn(x+y +z)(cid:1)−2 (cid:0)2 + 2δnx 2 + (cid:1)2δny(cid:0)2 + 2δnz 2 −1(cid:1)
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
and hence
n = (x+y +z)−21−δn 2 +x2 +y2 +z2 = x2 +y2 +z2 + 21−δn −x−y −z 2
(cid:0) (cid:1) (cid:0) (cid:1)
with x+y +z +(21−δn −x−y −z) = 2⌊(ord2(n)+1)/2⌋.
Case 2. 4 | n.
Write n = 4kn with k,n ∈ Z+ and 4 ∤ n . By the above, there are
0 0 0
x ,y ,z ,w ∈ Z such that
0 0 0 0
n = x2 +y2 +z2 +w2 with x +y +z +w = 2⌊(ord2(n0)+1)/2⌋.
0 0 0 0 0 0 0 0 0
It follows that
n = 2kx 2 + 2ky 2 + 2kz 2 + 2kw 2
0 0 0 0
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
with
2kx +2ky +2kz +2kw = 2k+⌊(ord2(n0)+1)/2⌋ = 2⌊(ord2(n)+1)/2⌋.
0 0 0 0
Combining the above, we have proved Theorem 1.1. (cid:3)
Lemma 2.2. For any integer n > 16 with 16 ∤ n, we have {n−1,n−16} 6∈ E ,
0
where E is the set defined in (2.2).
0
Proof. Clearly one of n − 1 or n − 16 is odd. If n − 1 ≡ 7 (mod 8), then
n − 16 ≡ n ≡ 8 (mod 16). If n − 16 ≡ 7 (mod 8), then n − 1 ≡ 7 + 15 ≡ 6
(mod 8). Thus {n−1,n−16} 6∈ E as desired. (cid:3)
0
Remark 2.2. It follows from Lemmas 2.1 and 2.2 that any n ∈ Z+ can be written
as x2 + y2 + z2 + w2 with x a power of two and y,z,w ∈ N. A stronger result
given in [S17b, Theorem 1.2(v)] states that any positive integer can be written as
4k(1+4x2 +y2)+z2 with k,x,y,z ∈ N.
8 ZHI-WEI SUN
Lemma 2.3 ([S17a, Lemma 2.1]). . Let u and v be integers with u2+v2 a positive
multiple of 5. Then u2 +v2 = x2 +y2 for some x,y ∈ Z with 5 ∤ xy.
Lemma 2.4. Let n > 4 be an integer not divisible by 64.
(i) We have {n,n−4} 6⊆ E .
0
(ii) If 16 ∤ n, then {n,n−1} 6⊆ E . If 16 | n, then {n,n−1,n−64} 6⊆ E .
0 0
Proof. (i) If n = 4k(8l +7) for some k ∈ N and l ∈ N, then k < 3 as 64 ∤ n, and
hence n−4 6∈ E since
0
8l+7−4 = 8l+3 if k = 0,
n−4 =  4(8l+7)−4 = 4(8l+6) if k = 1,


42(8l+7)−4 = 4(8(4l+3)+3) if k = 2.


(ii) If n = 8l+7 for some l ∈ N, then n−1 = 8l+6 6∈ E . If n = 4(8l+7) for
0
some l ∈ N, then n −1 = 32l +27 = 8(4l +3)+ 3 6∈ E . So {n,n− 1} 6⊆ E if
0 0
16 ∤ n.
Now we consider the case 16 | n. As 64 ∤ n, if n 6∈ E , then n = 42(8l+7) for
0
some l ∈ N, and hence n−64 = 42(8l+3) 6∈ E .
0
The proof of Lemma 2.4 is now complete. (cid:3)
Proof of Theorem 1.2. (i) For n = 1,... ,15 we can easily verify the desired result.
Now fix an integer n > 16 and assume that each m = 1,... ,n − 1 can be
written as x2 +y2 +z2 +w2 (x,y,z,w ∈ N) with |2x−y| ∈ {4k : k ∈ N}.
Let’s first consider the case 16 | n. By the the induction hypothesis, there are
x,y,z,w ∈ N for which n/16 = x2 +y2 +z2 +w2 with |2x−y| ∈ {4k : k ∈ N},
and hence |2(4x)−4y| = 4|2x−y| ∈ {4k : k ∈ N}.
Now we suppose that 16 ∤ n. Then {5n − 1,5n − 16} 6⊆ E by Lemma 2.2.
0
Let δ = 0 if 5n − 1 ∈ E , and δ = 1 otherwise. In view of Lemma 2.1, we can
0
write 5n − 16δ as x2 + y2 + z2 with x,y,z ∈ Z. Since a square is congruent to
one of 0,1,−1 modulo 5. one of x2,y2,z2 must be congruent to −1 modulo 5.
Without loss of generality, we may assume that x2+1 ≡ y2+z2 ≡ 0 (mod 5). If
y2+z2 6= 0, then by Lemma 2.3 we can write y2+z2 = y2+z2 with y ,z ∈ Z and
1 1 1 1
5 ∤ y z . Without loss of generality, we simply assume that x ≡ −2×4δ (mod 5)
1 1
(otherwise we use −x instead of x) and that either y = z = 0 or y ≡ 2z ≡ −2×4δ
(mod 5).
Clearly, r = (x+2×4δ)/5, s = (2x−4δ)/5, u = (2y+z)/5 and v = (2z−y)/5
are all integers. Observe that
(4δ)2 +x2 y2 +z2
r2 +s2 +u2 +v2 = + = n with 2r −s = 4δ.
5 5
If x > −2, then x > 2 since x ≡ (−1)δ−12 (mod 5), and hence r,s ∈ N. When
x 6 −2, clearly s < 0, and
r > 0 ⇐⇒ δ = 1 & x = −3.
RESTRICTED SUMS OF FOUR SQUARES 9
If r 6 0 and s 6 0, then
|2|r|−|s|| = |2(−r)−(−s)| = |2r−s| = 4δ.
Now it remains to consider the case δ = 1 and x = −3. Note that
y2 +z2
= u2 +v2 = n−r2 −s2 = n−12 −(−2)2 > 0
5
and hence y ≡ 2z ≡ −2 ×4δ (mod 5). When y 6= −3, in the spirit of the above
arguments, all the four numbers
y +2×4δ 2y −4δ 2x+z 2z −x
r¯= , s¯= , u¯ = , v¯ =
5 5 5 5
are integral, and
(4δ)2 +y2 x2 +z2
r¯2 +s¯2 +u¯2 +v¯2 = + = n
5 5
with |2|r¯|−|s¯|| = |2r¯−s¯| = 4δ. If y = −3, then
5n−16 = x2 +y2 +z2 = (−3)2 +(−3)2 +z2 ≡ z2 +2 6≡ 0,1 (mod 4)
thus 5n−1 6≡ 0,3 (mod 4) and hence 5n−1 6∈ E which contradicts δ = 1. This
0
concludes our induction proof of Theorem 1.2(i).
(ii) For every n = 0,1,... ,63, we can verify the desired result directly.
Now fix an integer n > 64 and assume that the desired result holds for all
smaller values of n.
If64 | n, then bythe inductionhypothesiswecan writen/64asx2+y2+z2+w2
with x,y,z,w ∈ N and 2x−y ∈ {±a8k : k ∈ N}∪{0}, and hence
n = (8x)2 +(8y)2 +(8z)2 +(8w)2 with 2(8x)−(8y) ∈ {±a8k : k ∈ N}∪{0}.
Below we suppose that 64 ∤ n.
By Lemma 2.4, 5n−(aδ)2 6∈ E for some δ ∈ {0,1,8} satisfying
0
δ = 8 =⇒ a = 1 & 16 | n. (2.3)
In view of (2.2), 5n−(aδ)2 = x2 +y2 +z2 for some x,y,z ∈ Z. Since any square
is congruent to one of {0,±1} modulo 5, one of x2,y2,z2, say x2, is congruent to
−(aδ)2 modulo 5. Without loss of generality, we simply suppose that x ≡ −2aδ
(mod 5). As y2 ≡ (2z)2 (mod 5), we may also suppose that y ≡ 2z (mod 5).
Thus all the numbers
x+2aδ 2x−aδ 2y +z y −2z
r = , s = , u = , v =
5 5 5 5
10 ZHI-WEI SUN
are integral. Note that
(aδ)2 +x2 y2 +z2
n = + = r2+s2+u2+v2 with2r−s = aδ ∈ {a8k : k ∈ N}∪{0}.
5 5
If x > aδ/2, then r > 0 and s > 0. If x 6 −2aδ, then r 6 0 and s 6 −aδ 6 0.
Now we handle the remaining case −2aδ < x < aδ/2. Clearly, r > 0 > s,
aδ = 2r −s > 2, hence δ = 8, and a = 1 and 16 | n by (2.3). As 2r −s = 8, we
must have
(r,s) ∈ {(1,−6), (2,−4), (3,−2)}.
Note that 2×2−4 = 0 and 2×2−3 = 1. If (r,s) = (1,−6), then
n = 12 +(−6)2 +u2 +v2 6≡ 0 (mod 4),
which contradicts 16 | n. This ends our proof of Theorem 1.2(ii). (cid:3)
Lemma 2.5. We have
E(1,3,6)= {3q+2 : q ∈ N}∪{4k(16l+14) : k,l ∈ N} (2.4)
and
E(2,3,6) = {3q+1 : q ∈ N}∪E . (2.5)
0
Remark 2.3. (2.4) and (2.5) are known results, see, e.g., L. Dickson [D39, pp.112-
113].
Proof of Theorem 1.3. (i) We can easily verify the desired result for n = 1,... ,15.
Now let n > 16 and a ∈ {0,2}. Assume that each m = 1,2,... ,n−1 can be
writtenasx2+y2+z2+w2 (x,y,z,w ∈ Z)withx+y+z ∈ {4k : k ∈ N}, andwecan
also write m as x2+y2+z2+w2 with x,y,z,w ∈ N and |x+y−z| ∈ {ak,4k}.
k∈N
We first suppose that 16 | n. By the induction hypothesis, there aSre x,y,z,w ∈
Z and k ∈ N for which n/16 = x2 + y2 + z2 + w2 with x + y + z = 4k, and
hence n = (4x)2 +(4y)2 +(4z)2 +(4w)2 with 4x+4y +4z = 4k+1. Also, by the
induction hypothesis, there are x,y,z,w ∈ N for which n/16 = x2 +y2 +z2 +w2
and |x+y−z| ∈ {ak,4k}, and hence n = (4x)2+(4y)2+(4z)2+(4w)2 with
k∈N
|4x+4y −(4z)| =S4|x+y −z| ∈ {ak,4k}.
k∈N
Now we suppose that 16 ∤ n. BSy Lemmas 2.2 and 2.5, for some δ ∈ {0,1} we
have 3n−16δ 6∈ E(2,3,6). So, there are x,y ∈ N and z ∈ Z such that
3n−16δ = 3x2 +6y2 +2(3z −4δ)2 = 3(x2 +2y2 +2(3z2 −2×4δz))+2×16δ
and hence
n = x2 +2y2 +6z2 −4δ+1z +16δ = x2 +(y +z)2 +(z −y)2 +(4δ −2z)2