PRIME RECIPROCALS AND PRIMES IN ARITHMETIC PROGRESSION

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1 PRIME RECIPROCALS AND PRIMES IN ARITHMETIC PROGRESSION DANIEL LITT Abstract. This aer is a exository accout of some (very elemetary) argumets o sums of rime recirocals; though the statemets i Proositios 5 ad 6 are well kow, the argumets, to my kowledge, are origial. Dirichlet s theorem o rimes i arithmetic rogressio states that if a, b are relatively rime ositive itegers, the there are ifiitely may rimes satisfyig a mod b. We reset a well-kow methods of rovig the secial case a =, ad alter it to obtai a elemetary estimate o the sum X mod b, rime i some cases.. Itroductio Dirichlet s theorem o rimes i arithmetic rogressio states Theorem (Dirichlet). Let a, b be relatively rime itegers. satisfyig a mod b. The there are ifiitely may rimes Cosider the degeerate case a = 0, b = ; the this is simly the claim that there are ifiitely may rimes. To rove this, we might cosider the sum. rime If this sum diverges, the there are certaily ifiitely may rimes. Proositio. The sum diverges. rime Proof. The roof here is similar i sirit to that i Aostol [], albeit slightly altered. Assume to the cotrary that the sum coverges; the there exists N such that c := <. For a fixed rime, let Now cosider the exressio >N, rime G = =. ( + c + c 2 + c ) N, rime As c <, the exressio coverges absolutely; but by absolute covergece, we may (rearragig terms) write ( + c + c 2 + c ) G N, rime N which diverges by e.g. the itegral test. So we have a cotradictio. G.

2 This roositio suggests aother aroach to cosider the asymtotics of the sum. We may easily show the followig: rime,<n Proositio 2. There exists a costat c, ideedet of N, such that l l(n) + c. Proof. We have rime, <N ( ) π 2 6 ep <N = 2 N ( ) 2 N <N N x dx = l N. rime,<n rime,<n e ( + ) where the secod lie follows from the Talor series for e x. Takig logarithms o both sides gives the desired claim. How might we geeralize these roofs to the situatio (a, b) (0, )? Cosider the followig argumet: Proositio 3. There are ifiitely may rimes satisfyig for ay >. Before roceedig we eed a lemma: mod Lemma. Let f(x) Z[x] be a o-costat olyomial. Let The P f is ifiite. P f := { rime N s.t. f() 0}. Proof of Lemma. Assume the cotrary, ad let,..., k be a eumeratio of P f. Choose a iteger s so that f(s) = t 0; such a s exists as f is o-costat. Now ote that f(s + t k x) = f(s) + t k g(x) = t( + k g(x)) for some g(x) Z[x]; i articular, f(s + t k x) is divisible by t for ay x Z. Now cosider h(x) := t f(s + t k x) = + k g(x). But h is o-costat, so we may choose u Z with h(u). So h(u) mod k, ad thus h(u) is divisible by some rime i for i =,..., k. But the P f, which is a cotradictio. We ow rove the roositio. Proof of Proositio 3. Let Φ (x) Z[x] be the -th cyclotomic olyomial, that is, the miimal olyomial of a rimitive -th root of uity ζ over Q. Let a Z ad cosider rime with Φ (a) 0, where. Let m be the order of a mod ; we claim that = m. Ideed, Φ (x ), so a ad thus m. Assume m <. But the Φ (a), a m ; but both Φ (x), x m divide x, ad the two olyomials are relatively rime mod (ideed, the former is irreducible ad does ot divide the latter), so x has a double root mod at a. But the discrimiat of of x is, which is o-zero mod (as ), so this is a cotradictio. So we must have m =. 2

3 But ote that a mod, so, ad thus mod. So ay rime i P Φ(x) either divides or satisfies mod. But by Lemma, there are ifiitely may rimes i P Φ(x), ad oly fiitely may rimes divide, so there are ifiitely may rimes satisfyig mod. 2. Sums of Recirocals Ufortuately, Proositio 3 does ot aswer the followig questio: Does mod, rime coverge or diverge? Cosider the case = 4. Our aroach to Proositio 3 suggests lookig at the umber field Q[i], with rig of itegers Z[i]. Let N be the orm ma N : Z[i] Z give by a + bi a 2 + b 2, with a, b Z. Cosider the set N(Z[i]) Z. Let r 2 () = N () We claim the followig: Proositio 4. Let Z, = 2 s a a 3 where all the rime factors of a j satisfy j mod 4. The Proof. () N(Z[i]) oly if a 3 is a square. (2) Let b = q q2 2 q k k be the rime factorizatio of b. The r 2 () 2 q+q2+ +q k+2. () First, let be a odd rime, ad assume = x 2 + y 2 for x, y Z. Notig that the quadratic residues mod 4 are 0,, this imlies that 0,, 2 mod 4; as is odd we have mod 4. Now cosider N(Z[i]), that is, = x 2 + y 2. As Z[i] is a PID (ad thus a UFD), x + iy factors as (a + ib ) q (a k + ib k ) q k for some rimes a j + ib j. Note that N is multilicative, so x 2 + y 2 = (a 2 + b 2 ) (a 2 k + b2 k ). So we may reduce to the case where = x2 + y 2 with x + iy a rime. But the N(x + iy) = (x + iy)(x iy) = x 2 + y 2, so x + iy divides N(x + iy). Let k be the rime factorizatio of N(x + iy) i Z; as x + iy is a rime, it must divide oe of the j. But the x iy = x + iy must divide j = j. So N(x + iy) = (x + iy)(x iy) 2 j, so x2 + y 2 is either a rime or the square of a rime. But the by the first aragrah, we have that if x 2 + y 2 = a odd rime, mod 4. So by writig = x 2 + y 2, x + iy = (a + ib ) q (a k + ib k ) q k we must have that the odd squarefree art of is divisible oly by rimes mod 4, as desired. (2) Note that the uits of Z[i] (e.g. by aalysis of the orm) are {,, i, i}. Note that for a rime, 3 mod 4, we have r 2 () = 0 ad r 2 ( 2 ) = 4; that is, the reimages are {,, i, i} (as such a rime caot slit, by the aalysis above). For mod 4, we have r 2 () 8 as such a rime may slit ito at most two rimes (as Gal(Q[i]/Q) = Z/2Z) of the form x iy, x + iy. So the reimages of are the four uits multilied by these two rimes. Usig multilicativity of the orm ad the fact that Z[i] is a UFD, we may write a reimage of N as x = u( i) s a a 3 where u is a uit ad a, a 3 are the reimages of a, a 3 resectively. We have four choices for u ad oe choice for the reimage of 2; havig chose a uit already, each rime factor of a 3 gives us o choice. Fially, for each rime factor of a we may choose oe of the at most two rimes (u to a uit) lyig above that rime i Z[i]; let t = q + + q k be the total umber of rimes (with multilicity) dividig a. The this aalysis gives that r 2 () 4 2 t = 2 q+ +qk+2, as desired. Remark. Note that Proositio 4() is actually a if ad oly if; we omit the roof of the other directio, though it follows easily from a aalysis of the slittig of rimes i Z[i]. We may use this argumet to show: 3

4 Proositio 5. The sum diverges. rime, mod 4 Proof. For coveiece, let P,4 deote the set of rimes mod 4. Assume the theorem is false; the there exists N such that c := < 2. Let G be as i the roof of Proositio, let N P,4,>N G = = 2 ad cosider the exressio ( ) D := 4 2 G 2 ( + 2c + (2c) 2 + (2c) 3 + ) G. N, P,4 Note that this exressio coverges absolutely, by our choice of N. For = 2 s a a 3 as i Proositio 4, with a = q q k k, let s 2 () = j q j, ad let t() = a. The by absolute covergece of D, we may rearrage terms to achieve D 4 2 s2() = N,t()a square N,t()a square r 2 () N = N(z) = z Z[i] (x,y) Z 2 {(0,0)} 2 s2()+2 x 2 + y 2 where the iequality o the third lie comes from Proositio 4(2). But this last exressio diverges, as x 2 + y 2 (x + y) 2 = 2 = =, x0,y>0 N N (x,y) Z 2 {0,0} cotradictig the claim that D coverged absolutely. So we have the desired divergece. Remark 2. This argumet ca easily be exteded to the case = 3, relacig the Gaussia itegers with the Eisestei itegers; it roceeds essetially idetically. Ufortuately, we caot use a idetical argumet for rimes mod, > 5 as the estimate i Proositio 4(2) relies heavily o the fiiteess of the uit grou of Z[i]. The argumet goes through, however, by comarig a exressio aalogous to D above to the Dedekid Zeta fuctio of the umber field Q[ζ ], where ζ is a rimitive -th root of uity the Dedekid Zeta fuctio diverges at, givig the desired comariso, but the roof of this is o-elemetary ad thus we will ot exosit it here. We ca, however, fid a lower boud o the artial sums of, P,4 4

5 ad a aalogous argumet works for mod 3. The roof follows similarly to that of Proositio 2. Proositio 6. There exists a costat c, ideedet from N, such that l l(n) + c. 2 Proof. We have 2π 2 3 e2 P,4,<N ( ) P <N = 4 2 N ( ) 4 2 N 2 s2()+2 <N r 2 () <N 0<x 2 +y 2 <N rime,<n rime,<n x 2 + y 2 0<(x+y) 2 <N; x0,y>0 0<<N l N 2 e 2 ( + 2 ) (x + y) 2 where the secod lie follows from the Taylor series for e x ad the last lie follows by boudig by a itegral, as i the roof of Proositio 2. Takig logarithms o both sides gives the desired claim. 3. Further Remarks Ufortuately, it seems that geeralizig this argumet to large as i Remark would require estimates o the artial sums of the Dedekid Zeta fuctio for Q[i]; these estimates are far from elemetary. We might ask how likely such methods are to work for boudig the sum of recirocals of rimes a mod b with a mod b. There are several egative results i this directio: First, to have a hoe of usig orms from the rig of itegers of a umber field to aalyze rimes a mod b, we must be workig i a umber field with rime slittig cotrolled by some cogruece coditios. But by results of Murty i [2], such umber fields exist oly if a 2 mod b. Number fields whose rime slittig is cotrolled by cogruece coditios are Abelia extesios of Q, by Arti recirocity; by the Kroecker-Weber Theorem, such fields are subfields of cyclotomic fields. So a mod b will ofte slit, ad by our ow aalysis the rimes give by this slittig diverge, domiatig or obscurig divergece by rimes i other cogruece classes mod b. (This is of course heuristic.) Note also that Mertes Theorem imlies that the estimate i Proositio 2 is asymtotically shar. Cosider the followig quatitative form of Dirichlet s theorem: Theorem 2 (Dirichlet). Let P a,b be the set of rimes cogruet to a mod b, with (a, b) =. The teds to /φ(b) as teds to ifiity. P a,b {,..., } P 0, {,..., } 5

6 Together with Proositio 2, this imlies that the estimate i Proositio 6 is also asymtotically shar, as φ(4) = 2. Refereces [] T. Aostol, Itroductio to Aalytic Number Theory (Udergraduate Texts i Mathematics), Sriger (May 28, 998). Pgs [2] M.R. Murty, Primes i certai arithmetic rogressios, J. Madras Uiv. (988),