Counting Invertible Sums of Squares Modulo $ N $ and a New

Home , Multiplicative function

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ρ (n) := card (x ,...,x ) (Z/nZ)k : x2 + . . . + x2 λ (mod n) (2) k,λ { 1 k ∈ 1 k ≡ } which count the number of points on hyperspheres in (Z/nZ)k and, in particular, in the case gcd(λ, n) = 1. These functions were already studied in the case when n is an odd prime by V. H. Lebesgue in 1837. In particular he proved the following result ([4, Chapter X]).

Proposition 1. Let p be an odd prime and let k, λ be positive integers with p ∤ λ. Put t = ( 1)(p−1)(k−1)/4p(k−1)/2 and ℓ = ( 1)k(p−1)/4p(k−2)/2. Then − − pk−1 + t, if k is odd and λ is a quadratic residue modulo p; k−1 ρk,λ(p)= p t, if k is odd and λ is a not quadratic residue modulo p;  − pk−1 ℓ, if k is even. −  The paper is organized as follows. First of all, in Section 2 we study the values of ρk,λ(n) in the case gcd(λ, n) = 1, thus generalizing Lebesgue’s work. In Section 3 we study the functions Φk, in particular we prove that they are multiplicative and we give a closed formula for Φk(n) in terms of the prime-power decomposition of n. Section 4 is devoted to deduce an asymptotic formula for n≤x Φk(n). Finally, we close our work suggesting some ideas that leave the door open for future work. P 2 Counting points on hyperspheres (mod n)

Due to the Chinese Remainder Theorem, the function ρk,λ, deﬁned by (2) is multiplicative; i.e., if n = pr1 prm , then ρ (n) = ρ (pr1 ) ρ (prm ). Hence, we can restrict ourselves 1 · · · m k,λ k,λ 1 · · · k,λ m to the case when n = ps is a prime-power. Moreover, since in this paper we focus on the case gcd(λ, n) = 1, we will always assume that p ∤ λ. The following result will allow us to extend Lebesgue’s work to the odd prime-power case.

Lemma 1. Let p be an odd prime and let s N. If p ∤ λ, then ∈ s (s−1)(k−1) ρk,λ(p )= p ρk,λ(p).

2 2 s+1 Proof. It is easily seen that any solution to the congruence x1 + . . . + xk λ (mod p ) must s s ≡ be of the form (a1 + t1p , . . . , ak + tkp ), where 0 t1,...,tk p 1, for some (a1, . . . , ak) such 2 2 s s≤2 ≤ − s 2 s+1 that a1 + . . . + ak λ (mod p ). Now, (a1 + t1p ) + . . . + (ak + tkp ) λ (mod p ) if and ≡ 2 ≡ 2 s only if 2a1t1 + . . . + 2aktk K (mod p), where K is such that a1 + . . . + ak = Kp + λ. Since ≡− k−1 ai 0 (mod p) for some i 1,...,k , it follows that there are exactly p possibilities for 6≡ ∈ { s+1 } k−1 s (t1,...,tk). We obtain that ρk,λ(p )= p ρk,λ(p ), and the result follows inductively. If p = 2 we have a similar result.

Lemma 2. Let s 3 and let λ N be odd. Then, ≥ ∈ s (s−3)(k−1) ρk,λ(2 ) = 2 ρk,λ(8).

2 Proof. In the case p = 2 the proof of Lemma 1 does not work since 2a t + . . . + 2a t K 1 1 k k ≡ − (mod 2) holds only if K is even. Therefore, we use that every solution of the congruence x2 +. . .+x2 λ (mod 2s+1) is of the form (a +t 2s−1, . . . , a +t 2s−1), where 0 t ,...,t 3, 1 k ≡ 1 1 k k ≤ 1 k ≤ for some (a , . . . , a ) satisfying a2 + . . . + a2 λ (mod 2s−1), that is a2 + . . . + a2 = L2s−1 + λ 1 k 1 k ≡ 1 k with an integer L. Now, taking into account that s 3, (a + t 2s−1)2 + . . . + (a + t 2s−1)2 λ ≥ 1 1 k k ≡ (mod 2s+1) if and only if

2(a t + . . . + a t ) L (mod 4). (3) 1 1 k k ≡− 2 2 s Here the condition (3) holds true if and only if L is even, i.e., a1 + . . . + ak λ (mod 2 ). s ≡ Hence we need the solutions (a1, . . . , ak) of the congruence (mod 2 ), but only those satisfying

0 a , . . . , a < 2s−1 (4) ≤ 1 k s k It is easy to see that their number is ρk,λ(2 )/2 , since all solutions of the congruence (mod 2s) are (a + u 2s−1, . . . , a + u 2s−1) with (a , . . . , a ) verifying (4) and 0 u , . . . , u 1. 1 1 k k 1 k ≤ 1 k ≤ Since a must be odd for some i 1,...,k , for a ﬁxed even L, (3) has 2 4k−1 solutions i ∈ { } · (t ,...,t ). We deduce that ρ (2s+1) = 2 4k−1ρ (2s)/2k = 2k−1ρ (2s). Now the result 1 k k,λ · k,λ k,λ follows inductively on s.

3 As we have just seen, unlike when p is an odd prime, the recurrence is now based on ρk,λ(2 ). Hence, the cases s = 1, 2, 3; i.e., n = 2, 4, 8, must be studied separately. In order to do so, the following general result will be useful. Lemma 3. Let k, λ and n be positive integers. Then

n−1 ρk,λ(n)= ρ1,ℓ(n)ρk−1,λ−ℓ(n). Xℓ=0 Proof. Let (x ,...,x ) (Z/nZ)k be such that x2 + + x2 λ (mod n). Then, for some 1 k ∈ 1 · · · k ≡ ℓ 0,...,n 1 we have that x2 ℓ (mod n) and x2 + . . . + x2 λ ℓ (mod n) and hence ∈ { − } 1 ≡ 2 k ≡ − the result.

Now, given k,n N let us deﬁne the matrix M(n) = (ρ (n)) . If we consider ∈ 1,i−j 0≤i,j≤n−1 the column vector Rk(n) = (ρk,i(n))0≤i≤n−1, then Lemma 3 leads to the following recurrence relation: R (n)= M(n) R (n). k · k−1 s In the following proposition we use this recurrence relation to compute ρk,λ(2 ) for s = 1, 2, 3 and odd λ. Lemma 4. Let k be a positive integer. Then

k−1 i) ρk,1(2) = 2 ,

3k k−1 2 −1 πk ii) ρk,1(4) = 4 + 2 sin 4 , 3k iii) ρ (4) = 4k−1 2 2 −1 sin πk , k,3 − 4 3 k iv) ρ (8) = 22k−3 2k + 2 2 +1 sin πk + 2sin 1 π(k + 1) 2 cos 1 π(3k + 1) , k,1 4 4 − 4 k v) ρ (8) = 22k−3 2k 2 2 +1 sin πk 2 cos 1 π(k + 1) + cos 3 π(k + 1) , k,3 − 4 − 4 4 k vi) ρ (8) = 22k−3 2k + 2 2 +1 sin πk 2sin 1 π(k + 1) + 2 cos 1 π(3k + 1) , k,5 4 − 4 4 k vii) ρ (8) = 22k−3 2k 2 2 +1 sin πk 2sin 1 (3πk + π) + 2 cos 1 π(k + 1) . k,7 − 4 − 4 4 Proof. First of all, observe that 20002004 42000200 2 00 2 04200020 1 1 2 20 0 00420002 M(2) = , M(4) =   , M(8) =   . 1 1 0 22 0 20042000   0 02 2 02004200       00200420   00020042     Let us compute ii). We know that R (4) = M(4) R (4). Hence, since the eigenvalues of k · k−1 M(4) are 4, 2 + 2i, 2 2i, 0 , we know that { − } ρ (4) = C 4k + C (2 + 2i)k + C (2 2i)k. k,1 1 2 3 − In order to compute C1, C2 and C3 it is enough to observe that ρ1,1(4) = 2, ρ2,1(4) = 8 and ρ3,1(4) = 24. Hence 4C + (2 + 2i)C + (2 2i)C = 2, 1 2 − 3 16C + 8iC 8iC = 8, 1 2 − 3 64C (16 16i)C (16 + 16i)C = 24. 1 − − 2 − 3 We deduce 1 3k πk ρ (4) = 4k i(2 + 2i)k + i(2 2i)k = 22k−2 + 2 2 −1 sin , k,1 4 − − 4 as claimed. To compute the other cases note that the eigenvalues of M(2) are 0, 2 while the eigenvalues { } of M(8) are

8, 4 + 4i, 4 4i, √2( 2 2i), √2(2 + 2i), √2( 2 + 2i), √2(2 2i), 0 . − − − − − n o Thus, in each case we only need to compute the corresponding initial conditions and constants. The ﬁnal results have been obtained with the help of Mathematica “ComplexExpand” command.

Note that a diﬀerent approach to compute the values ρk,λ(n), using the Gauss quadratic sum was given in [20].

4 3 Counting invertible sums of squares (mod n)

Given positive integers k,n, this section is devoted to computing Φk(n), deﬁned by (1). Let A(k, λ, n) denote the set of solutions (x ,...,x ) (Z/nZ)k of the congruence x2 + . . . + x2 λ 1 k ∈ 1 k ≡ (mod n). First of all, let us deﬁne the set

(n) := A(k, λ, n). Ak 1≤λ≤n gcd([λ,n)=1

Hence, Φ (n) = card (n) and, since the union is clearly disjoint, it follows that k Ak

Φk(n)= ρk,λ(n). 1≤λ≤n gcd(Xλ,n)=1

The following result shows the multiplicativity of Φk for every positive k.

Proposition 2. Let k be a positive integer. Then Φk is multiplicative; i.e., Φk(mn)=Φk(m)Φk(n) for every m,n N such that gcd(m,n) = 1. ∈ Proof. Let us deﬁne a map F : (m) (n) (mn) by Ak ×Ak −→ Ak

F ((a1, . . . , ak), (b1, . . . , bk)) = (na1 + mb1, . . . , nak + mbk).

Note that if (a , . . . , a ) (m), then a2 + . . . + a2 λ (mod m) for some λ with 1 k ∈ Ak 1 k ≡ 1 1 gcd(λ ,m) = 1. In the same way, if (b , . . . , b ) (n), then b2 + . . . + b2 λ (mod n) for 1 1 k ∈ Ak 1 k ≡ 2 some λ2 with gcd(λ2,n) = 1. Consequently,

2 2 2 2 2 2 2 2 (na1 + mb1) + . . . + (nak + mbk) = n (a1 + . . . + ak)+ m (b1 + . . . + bk) + 2mn(b a + . . . + b a ) 1 1 k k ≡ n2λ + m2λ (mod mn). ≡ 1 2 Since it is clear that gcd(n2λ + m2λ ,mn) = 1, it follows that (na + mb , . . . , na + mb ) 1 2 1 1 k k ∈ (mn) and thus F is well-deﬁned. Ak Now, let (c ,...,c ) (mn). Then c2 + . . . + c2 λ (mod mn) for some λ such that 1 k ∈ Ak 1 k ≡ gcd(λ, mn) = 1. Letusdeﬁne a c (mod m) and b c (mod n) for every i = 1,...,k. It fol- i ≡ i i ≡ i lows that (a , . . . , a ) (m), (b , . . . , b ) (n) and, moreover, F ((a , . . . , a ), (b , . . . , b )) = 1 k ∈Ak 1 k ∈Ak 1 k 1 k (c1,...,ck). Hence, F is surjective. Finally, assume that

(na + mb , . . . , na + mb ) (nα + mβ ,...,nα + mβ ) (mod mn) 1 1 k k ≡ 1 1 k k for some (a , . . . , a ), (α ,...,α ) (m) and for some (b , . . . , b ), (β ,...,β ) (n). 1 k 1 k ∈ Ak 1 k 1 k ∈ Ak Then, for every i = 1,...,k we have that na + mb nα + mβ (mod mn). From this, it i i ≡ i i follows that a α (mod m) and that b β (mod n) for every i and hence F is injective. i ≡ i i ≡ i Thus, we have proved that F is bijective and the result follows.

5 Since we know that Φk is multiplicative, we just need to compute its values over prime- powers. We do so in the following result.

Proposition 3. Let k, r be positive integers. Then

r kr kr−1 i) Φk(2 )= ϕ(2 ) = 2 . ii) If p is an odd prime,

k Φ (pr)= ϕ(pkr) ( 1)k(p−1)/4ϕ(pkr−k/2)= pkr− 2 −1(p 1) pk/2 ( 1)k(p−1)/4 . k − − − − − Proof.

i) If r = 1, 2, 3 the result readily follows from Lemma 4 by a simple computation. Now, if r> 3 we can apply Lemma 2 to obtain that

r r (r−3)(k−1) Φk(2 )= ρk,i(2 ) = 2 ρk,i(8) 1≤i≤2r 1≤i≤2r X2∤i X2∤i 2r−3−1 (r−3)(k−1) = 2 ρk,i(8) = j=0 8j+1≤i≤8(j+1)−1 X X2∤i (r−3)(k−1) r−3 = 2 2 ρk,i(8) = 1≤i≤7 X2∤i = 2(r−3)(k−1)2r−323k−1 = 2rk−1 = ϕ(2kr).

ii) Due to Lemma 1 it can be seen, as is the previous case, that

p−1 r k(r−1) Φk(p )= p ρk,i(p). Xi=1 Thus, it is enough to apply Proposition 1.

Finally, we summarize the previous work in the following result.

Theorem 1. Let k be a positive integer. Then the function Φk is multiplicative and for every n N, ∈ nk−1ϕ(n), if k is odd;  ( 1)k(p−1)/4 Φk(n)= nk−1ϕ(n) 1 − , if k is even.  − pk/2  p|n ! p>Y2   6 Written more explicitly, we deduce that

nk−1ϕ(n), if k is odd; 1 nk−1ϕ(n) 1 , if k 0 (mod 4);  − pk/2 ≡  p|n Φ (n)=  Y k  p>2  1 1 nk−1ϕ(n) 1 1+ , if k 2 (mod 4). − pk/2 pk/2 ≡  Yp|n Yp|n  p≡1(mod 4) p≡−1(mod 4)   When k is a multiple of 4, Φk is closely related to Jk/2. The following result, which follows from Theorem 1 and the deﬁnition of Jordan’s totient function Jk makes this relation explicit. Corollary 1. Let k N be a multiple of 4 and let n N. Then, ∈ ∈ 2k/2 Φ (n)= nk/2−1J (n)ϕ(n) . k k/2 2k/2 1+ n(mod 2) − Moreover, if k/4 is odd, then we have

Φ (n) 2k/2 k = nk/4J (n) . Φ (n) k/2 2k/2 1+ n(mod 2) k/4 − Recall that in the case k = 4, Φ4(n) is the number of units in the ring H(Z/nZ). If, in addition, n is odd then Φ4(n)= nJ2(n)ϕ(n) which is the well-known formula for the number of regular matrices in the ring M2(Z/nZ). Of course, this is not a surprise since it is known that for an odd n the rings H(Z/nZ) and M2(Z/nZ) are isomorphic ([7]). Some elementary properties of Φk, well known for Euler’s function (the case k = 1) follow at once by Theorem 1. For example, we have Corollary 2. Let k N be ﬁxed. ∈ i) If m,n N such that n m, then Φ (n) Φ (m). ∈ | k | k ii) Let m,n N and let d = gcd(m,n). Then Φ (mn)Φ (d)= dkΦ (m)Φ (m). ∈ k k k k iii) If n,m N, then Φ (nm)= nkm−kΦ (n). ∈ k k

4 The average order of Φk(n)

The average order of ϕ(n) is well-known. Namely, 1 3 ϕ(n) x (x ), x ∼ π2 →∞ nX≤x see, for example, [10, Th. 330]. In fact, the best known asymptotic formula is due to Walﬁsz [22]: 3 ϕ(n)= x2 + O(x(log x)2/3(log log x)4/3). (5) π2 nX≤x We now generalize this result.

7 Theorem 2. Let k N be any fixed integer. Then ∈ C Φ (n)= k xk+1 + O(xkR (x)), k k + 1 k nX≤x where 6 C = ,R = (log x)2/3(log log x)4/3, if k is odd; k π2 k 3 1 ( 1)k(p−1)/4(p 1) Ck = 1 2 − − ,Rk(x) = log x, if k is even. 4 − p − pk/2+2 ! p>Y2 Proof. If k is odd, then this result follows easily by partial summation from the fact that Φk(n)= nk−1ϕ(n) using Walfisz’ formula (5). Assume now that k N is even. Since the function φ is multiplicative, we deduce by the ∈ k Euler product formula that ∞ Φ (n) k = ζ(s k)G (s), ns − k n=1 X where 1 1 ( 1)k(p−1)/4(p 1) G (s)= 1 1 − − k − 2s−k+1 − ps−k+1 − ps−k/2+1 p>2 ! Y is absolutely convergent for s > k. This shows that Φ = id g in terms of the Dirichlet ℜ k k ∗ k convolution, where id (n)= nk (n N) and the multiplicative function g is defined by k ∈ k 2k−1, if p = 2, r = 1; r − k−1 k(p−1)/4 k/2−1 gk(p )= p ( 1) p (p 1), if p> 2, r = 1; − − − − 0, otherwise.

We obtain  (x/d)k+1 Φ (n)= g (d)ek = g (d) ek = g (d) + O((x/d)k) k k k k k + 1 nX≤x deX≤x Xd≤x Xe≤x Xd≤x

k+1 x k+1 gk(d) k gk(d) = Gk(k +1)+ O x | k+1 | + O x | k | . (6) k + 1 d !  d  Xd>x Xd≤x Here for every k 4 we have   ≥ ∞ r k−1 k/2−1 k/2 gk(d) gk(p ) gk(p) p + p + p | k | | kr | = 1+ | k | 1+ k d ≤ p p ≪ p ! Xd≤x pY≤x Xr=0 pY≤x pY≤x ∞ 1 1 −1 < = 1 log x pr − p ≪ r=0 pY≤x X pY≤x 8 by Mertens’s theorem. In the case k = 2 this gives

g (d) 2 1 1 | 2 | 1+ 1+ d2 ≪ p − p2 p2 d≤x p≤x p≤x X p≡1(modY 4) p≡−1(modY 4)

1 2 1+ log x, ≪ p ≪ p≤x p≡1(modY 4) using that 1 1 c(log x)−1/2, − p ∼ p≤x p≡1(modY 4) with a certain constant c, cf. [21]. Hence the last last error term of (6) is O(xk log x) for every k 2 even. ≥ Furthermore, note that for every k 2, g (n) nk/2σ (n) (n N), where σ (n) = ≥ | k | ≤ k/2−1 ∈ t dt. Using that σ (n) < ζ(t)nt for t > 1 we conclude that h (n) nk−1 for k 4. d|n t | k | ≪ ≥ Therefore, P g (d) 1 1 | k | . dk+1 ≪ d2 ≪ x Xd>x Xd>x In the case k = 2, using that σ (n) n log n (this suﬃces) we have h (n) n3 log n and 1 ≪ 2 ≪ g (d) log d log x | 2 | . d3 ≪ d2 ≪ x Xd>x Xd>x Hence the ﬁrst error term of (6) is O(xk) for k 4 and it is O(x2 log x) for k = 2. This ≥ completes the proof.

Corollary 3. (k = 2, 4)

x3 2 1 1 Φ (n)= 1 + 1 + O(x2 log x), 2 4 − p2 p3 − p3 nX≤x p≡1(modY 4) p≡−1(modY 4) 3x5 1 1 1 Φ (n)= 1 + + O(x4 log x). 4 20 − p2 − p3 p4 p>2 nX≤x Y 5 Conclusions and further work

The generalization of ϕ that we have presented in this paper is possibly one of the closest to the original idea which consists of counting units in a ring. In addition, both the elementary and asymptotic properties of Φk extend those of ϕ in a very natural way. There are many other

9 results regarding ϕ that have not been considered here but that, nevertheless, may have their extension to Φk. For instance, in 1965 P. Kesava Menon [14] proved the following identity:

gcd(j 1,n)= ϕ(n)d(n), − 1≤j≤n gcd(Xj,n)=1 valid for every n N, where d(n) denotes the number of divisors of n. This identity has been ∈ generalized in several ways. See, for example [11, 12, 17, 19]. Also,

gcd(j2 1,n)= ϕ(n)h(n), − 1≤j≤n gcd(Xj,n)=1 where h is a multiplicative function given explicitly in [19, Cor. 15]. Our work suggests the following generalization:

gcd(x2 + . . . + x2 1,n)=Φ (n)Ψ (n), 1 k − k k 1≤x1,...,xk≤n 2 X 2 gcd(x1+...+xk,n)=1 where Ψk is a multiplicative function to be found. Another question is on minimal order. As well known ([10, Th. 328]), the minimal order of ϕ(n) is e−γn(log log n)−1, where γ is Euler’s constant, that is

ϕ(n) log log n lim inf = e−γ. n→∞ n It turns out by Theorem 1 that for every k N odd, ∈ Φ (n) log log n lim inf k = e−γ . n→∞ nk

Find the minimal order of Φk(n) in the case when k is even.

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C. Calderón Departamento de Matemáticas, Universidad del Pa´ıs Vasco Facultad de Ciencia y Tecnolog´ıa Barrio Sarriena, s/n, 48940 Leioa, Spain [email protected] J. M. Grau Departamento de Matemáticas, Universidad de Oviedo Avda. Calvo Sotelo, s/n, 33007 Oviedo, Spain [email protected] A. M. Oller-Marcén Centro Universitario de la Defensa Ctra. de Huesca, s/n, 50090 Zaragoza, Spain [email protected] L. Tóth Department of Mathematics, University of Pécs Ifjúság u. 6, H-7624 Pécs, Hungary [email protected]