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On a Property of the Primitive Roots of Unity Leading to the Evaluation of Ramanujan’S Sums

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On a Property of the Primitive Roots of Unity Leading to the Evaluation of Ramanujan’S Sums On a Property of the Primitive Roots of Unity Leading to the Evaluation of Ramanujan's Sums Titu Andreescu and Marian Tetiva 1. Introduction. Ramanujan's sums are the sums of powers with the same exponent of the primitive roots of unity of some order. More specifically, let q and m be positive integers, and let x1; : : : ; xn (with n = '(q), where ' is Euler's totient function) be the roots of the qth cyclotomic polynomial Φq, that is, the primitive roots of unity of order q. We denote by X 2jmπ 2jmπ c (m) = xm + ··· + xm = cos + i sin q 1 n q q 1≤j≤q; (j;q)=1 and call this expression Ramanujan's sum. The formula X q c (m) = µ d q d dj(q;m) holds{see [2] for a proof. Here µ is the usual M¨obiusfunction defined on positive integers by µ(1) = 1, µ(t) = (−1)k when t is a product of k distinct prime factors, and µ(t) = 0 in any other case. The resemblance of the above identity with the well-known expression of ', X q '(q) = µ d d djq is, probably, not just a coincidence, as long as cq(m) = '(q) whenever q j m (it is easy to see that, P in that case, cq(m) = 1≤j≤q; (j;q)=1 1 = '(q)). Thus, Ramanujan's sums generalize Euler's totient function. They also represent a generalization of the M¨obiusfunction, because, for (q; m) = 1 we have cq(m) = cq(1) = x1 + ··· + xn = µ(q); according to a well-known result about the sum of primitive qth roots of unity, which we will mention again (and we will use) immediately. Ramanujan expanded the usual arithmetic functions in terms of his sums. Also, Ramanujan's sums were used in the proof of Vinogradov's theorem stating that every sufficiently large odd positive integer is the sum of three primes. 2. The main result. We have seen the formula X q c (m) = µ d: q d dj(q;m) Let us denote here D = (q; m), and let q0 be such that q = Dq0. Then let us rewrite the above equation as X D X µ(aq0) c (m) = µ q0 d = D : q d a djD ajD Because the values of the M¨obiusfunction for integers that are not square-free are zero, and due to its multiplicativity (µ(xy) = µ(x)µ(y) for relatively prime positive integers x and y) we further have X µ(a)µ(q0) X µ(a) c (m) = D = Dµ(q0) : q a a ajD; (a;q0)=1 ajD; (a;q0)=1 Mathematical Reflections 5 (2019) 1 Notice that only for a a product of distinct primes we get non-zero terms in this sum (and do not forget q = Dq0), so we have Y 1 c (m) = Dµ(q0) 1 − q p pjq; p6jq0 where the product is over all primes p that divide q, but do not divide q0. Finally we get Y 1 q 1 − p 0 pjq 0 '(q) cq(m) = µ(q ) = µ(q ) ; Y 1 '(q0) q0 1 − p pjq0 hence we have the following result (that gives a closed form expression for the Ramanujan sums): Proposition 1. For positive integers q and m we have '(q) q c (m) = µ : q q (m; q) ' (m; q) The interested reader can find another proof of proposition 1 (in a more general context) in [2]. 3. How do we prove the main result. It is time to see what is the purpose of this note. Namely, we intend to give yet another proof of proposition 1, which avoids the usual calculations with arithmetic functions (and is not very complicated, either). It relies on lemmas 1 and 3 below (lemma 2 helps proving lemma 3). Lemma 1. Let x1; : : : ; xn (with n = '(q))) be the qth primitive roots of unity. Then their sum is x1 + ··· + xn = µ(q). Proof. This is a well-known property of the primitive roots of unity. In order to prove it, let f(t) be the sum of the primitive roots of unity of order t (for any positive integer t). We then see that X f(d) djt is the sum of all roots of unity of order t, that is, X 1; t = 1 f(d) = 0; t > 1: djt However, it is well-known again that X 1; t = 1 µ(d) = 0; t > 1; djt therefore X X f(d) = µ(d); djt djt for any positive integer t. We also have f(1) = µ(1) = 1, thus f(t) = µ(t) follows for any t. Indeed, if the two functions f and µ differ, there exists a minimal t0 > 1 such that f(t0) 6= µ(t0). But then we would also have X X f(d) 6= µ(d); djt0 djt0 Mathematical Reflections 5 (2019) 2 (as f(d) = µ(d) for d j t0, d < t0, while f(t0) 6= µ(t0)), a contradiction. It remains that f and µ coincide, thus x1 + ··· + xn = f(q) = µ(q), as desired. One can note that the result of lemma 1 also represents a particular case of proposition 1 (for relatively prime q and m). Nevertheless, we will use it in our proof for proposition 1. Lemma 2. Let a, b, and c be positive integers such that a and b are relatively prime. For any positive integer u denote by Mu = fv 2 f0; 1; : : : ; u − 1g j (v; u) = 1g: Then the numbers at with t 2 Mbc leave when divided by b all possible remainders from Mb, and each such remainder is obtained the same number of times, that is, each remainder from Mb appears '(bc) '(b) times when at (mod b) (with t 2 Mbc) are listed. First proof. Since a is relatively prime to b, for two numbers t1; t2 2 Mbc (actually, for any integers t1 and t2) we have that at1 ≡ at2 (mod b) if and only if t1 ≡ t2 (mod b). Thus the remainders of the numbers at, t 2 Mbc appear with the same multiplicities as the remainders leaved by the numbers from Mbc themselves{thus it is enough to prove the statement for a = 1. That is, we need to prove that the remainders left by the numbers from Mbc at division by b can be organized into groups of '(bc) '(b) equal numbers, and they cover all the remainders from Mb. Any number from Mbc is relatively prime to bc, hence it is relatively prime to b, too, consequently the remainder it leaves when divided by b is also relatively prime to b, that is, it belongs to Mb. Also, we note that remainder 1 is surely obtained. Thus if we denote, for any r 2 Mb, by Nr = fx 2 Mbc j x ≡ r (mod b)g we have N1 6= ; (since 1 2 Mbc). Moreover, every Nr is nonempty. Indeed, let p1; : : : ; ps be the primes that divide c but do not divide b. By the Chinese remainder theorem, the system of congruences x ≡ r (mod b); x ≡ 1 (mod pi); 1 ≤ i ≤ s has a solution in the set f0; 1; : : : ; bp1 ··· ps − 1g. Of course bp1 ··· ps ≤ bc and the solution of the system of congruences is relatively prime to any of b; p1; : : : ; ps, hence to bc. Thus this solution is a number from Mbc which leaves remainder r when divided by b, and its existence shows that Nr is nonempty. Now let r be any element from Mb, and let y0 be one fixed element from Nr. We define the function f : N1 ! Nr by letting f(x) to be the remainder of y0x at division by bc. Since f(x) ≡ y0x (mod bc), we also have f(x) ≡ y0x (mod b), and x ≡ 1 (mod b), y0 ≡ r (mod b) imply f(x) ≡ r (mod b). Thus, indeed, f(x) 2 Nr and f is well defined. Also, it is clear that f is a bijection, and this shows that Nr and N1 have the same number of elements, that is every Nr has the same number of the elements. Also, the sets Nr, with r 2 Mb partition Mbc, hence jMbcj '(bc) jNrj = = jMbj '(b) for each r 2 Mb, as we intended to prove. Mathematical Reflections 5 (2019) 3 Second proof. We know now that it is enough to consider the case a = 1. Let r be any element from Mb, and let Nr = fx 2 Mbc j x ≡ r (mod b)g; which is obviously the same as Nr = fr + by j y 2 f0; 1; : : : ; c − 1g; (r + by; bc) = 1g: Let P be the set of primes that divide c, but do not divide b. For any prime p 2 P that divides c, let Ap = fy 2 f0; 1; : : : ; c − 1g j r + by is divisible by pg: We see that an element r + by, y 2 f0; 1; : : : ; c − 1g is relatively prime to bc (that is, it belongs to Nr) if and only if there is no prime p 2 P such that pjr + by. This means that Nr has the same number of elements as the set ! [ f0; 1; : : : ; c − 1g n Ap ; p2P that is, [ jNrj = c − Ap : p2P The cardinality of the union of the Aps is easily obtained by inclusion-exclusion principle. By noting that for any nonempty subset P 0 of P we have c \ Ap = Q 0 p p2P 0 p2P we immediately get Y 1 jN j = c 1 − ; r p p2P or 1 bc Q 1 − pjbc p '(bc) jN j = = ; r 1 '(b) b Q 1 − pjb p as desired.
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