ite integers of the f o r m ---(g any given integer > 1 , p prime) and

ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I: PRACE MATEMATYCZNE IX (1965)

Л.

(Warszawa)

On the composite Lehmer numbers with prime indices, I W. Sierpiński lias recently deduced from a certain conjecture of the present writer (cf. [3], p. 188) the existence of infinitely many compos-

t f - l

ite integers of the f o r m ---(g any given integer > 1 , p prime) and

g

2 * 4 1

also of the form — -— (cf. [4] and [5]).

av- b v

This suggests an analogous problem for —--- (a,b integers) and a — b

more generally for the so-called Lehmer numbers Ч (an- n i * - P , n odd,

[ (а» _ ^ ) / а2_ ^ ? n even,

where a, ft are roots of the trinomial z2- L ll2z f-M and L, M are ration­

al integers.

The aim of this paper is to deduce from the aforesaid conjecture H. I f •■■■>f к are irreducible polynomials with integral coeffi­

cients and the highest coefficients positive and such that f x(x)f2( x ) . . . / fc(<r) has no fixed factor greater than 1, then for infinitely many integers x , fi(x) (i = 1 k) are primes;

the existence of infinitely many composite integers Pp{a, ft) if a, ft are rational

0, а

ф ft or it a, ft are irrational and some additional restrictions hold. Since the same method of proof gives some weaker but unconditional results we shall prove simultaneously the following two theorems.

Theorem 1.

I f L M Ф

К = L —

ф

and none of the numbers

— K L , —3KL, —K M , —3 K M is a perfect square or each of the numbers K , L is a perfect square, then there exists an integer h > 0 such that for every integer D

0 one can find a prime q satisfying {l)

q I P«r- {KLW)fk and {q - { KL | q)), В j = 1.

P) (KL\q) is Jacobi’s symbol of quadratic character.

96 A. Sehinzel

Theorem

2. Under the assumptions of Theorem 1, Conjecture H im­

plies the existence of infinitely many primes p such that Pp(a, /5) is com­

posite.

1 think it very probable that Theorems 1 and 2 are true under the necessary assumption that a//? is not a root of unity, but the proof of that would require the use of reciprocity laws of degrees > 4. In our case we manage with the quadratic, cubic and biquadratic reciprocity laws only. We begin with two very elementary lemmas.

Lemma

1. I f a, b are two square-free integers, such that b ф —1, — 3 and ab ± — d2, — 3d2 then there exists an odd integer m > 0 such that (1) (a\m) = 1 and {^{m--(b\m)), 2ab} = 1 .

P r o o f. Let us remark that for every odd square-free modulus у Ф ± 1, ± 3 , there exists an integer v such that (v + 1, у) — 1 and (v\y) = —1. In fact, such a v — v(y) clearly exists if \y\ is a prime.

If у is not a prime, and q is its greatest prime factor, it suffices to take v(y) = v(q) modę,

1 mod у Jq.

How, put a = dax, b = dbx, where {ax, Ъг) = 1 and bx > 0 . Since a, b are square-free, we have (da1, b 1) = 1. We notice that if bx = 1 or 3, then ax Ф — 1, — 3 and d Ф —1, —3 .W e shall define m by a system of congruences different for each of the following 7 cases.

1. bx even:

2. bx odd ^ 1 , 3 :

m = lm od daxbx, 5mod8.

I lm o d idax, v(61)mod61.

3. bx = 1 or 3, ax > 0, d > 0:

m = — lm od ЫахЪх.

4. bx = 1 or 3, dax< 0 and min(a1, d) is even:

m = — lm od daxbx, 3m od8.

5. bx — 1 or 3, dax < 0 and min(a1, d) is odd:

lmod8&1max(a1, d),

г ( т т ( а ! , d))modmin(a1, d).

On the composite Lehmer numbers with prim e indices, 1 97

m ~

6. bx — 1 or 3, ax< 0, d < 0 and dax is even:

1 m odbxx even (ax, d) v(odd(a1? ^))mododd(a1, d), 5m od8,

where even(tt1?d) and odd(ax, d) denote the even and the odd one of the numbers ax and d.

7. bx = 1 or 3, ax and d are odd < 0:

I lmod4&1, r(d1)m oda1, r(d)m odd.

It follows from the Chinese Remainder Theorem that the above systems of congruences are solvable and from the quadratic reciprocity law that any of their positive solutions satisfies (1).

Lemma 2. I f у

is any integer >

and c any square-free integer, then there exists an odd integer m > 0 such that (2)

(2) 2v\\m — 1, (c\m) = —1 and (m — l , c ) = (2,c) except for

1 if с = 1 or 3, 7 [2 if c = ± 1 or ± 6 ,

у > 3 if c = ± 1 or ± 2 .

P r o o f. For

= 1, we put in Lemma 1: a = — с, b — 1 . Н с ф 1 , 3 , the conditions of that lemma are satisfied and there exists an odd inte­

ger m > 0 such that

( —c\m) = l , {%{m — 1), 2c) = 1.

Hence m = 3m od4, (e|m) = —1 and (m —1 ? c) = (2, c); so in this case conditions (2) are satisfied.

For у ^ 2 we define m by a system of congruences different for each of the following cases.

1. }c\ = 3mod4 or |c| == 2m od8,

= 2 or |c| = 6m od8,

^ 3:

— lm o d c, m = 2v+ l m o d 2 x+ . 2. |c| = lm o d 4 , с Ф ± 1 :

— r(c)m odc, 2y+ l m o d 2 y+1.

(2) 2y \\m— l means that 2y \m — I and 2y+1f m — l .

Prace Matematyczne IX. 1 7

98 ^{A .} Schinżel

3. |c| — 6m od8, у = 2, с Ф ± 6 or |с| = 2 m od8, у > 3, с Ф - ± 2 :

— r(c/2)m ode, m = 2V+ lm od 2 y+1.

It follows from the Chinese Remainder Theorem that the above systems of congruences are solvable and from the quadratic reciproci­

ty law that any of their positive solutions satisfies (2).

P r o o f o f T h e o re m s 1 and 2. In the course of this proof we can assume without loss of generality that (L , M) — 1, i.e. (a, (3) = 1.

We shall denote by k(n) the square-free kernel of any given integer n Ф 0.

Suppose first that none of the numbers —K L , —3 K L , — K M , — 3K M is a perfect square. Then it suffices to take к = 2. In fact, by Lemma 1 there exists an odd integer m > 0 such that

(Te{LM)\m) = 1 and ^ [ m - ( k { K L ) \ m ) ) , 2k{KL)lc{LM)} = 1.

Let us put

/,( * ) = ik(KZ) k( LM)cc + m, f,(x) = Ш x ) - \ { k ( K L ) \ m ).

It is clear that the polynomial f i ( x ) f 2(x) has no fixed factor > 1. Further, if /a (a?) = q is a prime, and q\ K L M , we have

{LM\q) = (k{LM)\m) = 1 and (KL\q) = (k{KL)\m),

because ( LM j r) and { KL | r) considered for r > 0 are characters with conductors dividing k KLM. Hence by the formula (cf. [2], p. 223)

(а/0)в/2- (Ы,|в)/2 = (LM\q)modq, we find

(3) ?|-f>Q'/2-(JS:i|ff)/2(«) P)-

Now the theorems follow easily. First, for every D Ф 0 there exists an integer x0 such that (/i(a?0) / 2(#o)> 4-K

CJLD) = 1. By Dirichlet’s theorem there exists a prime

q = fi(oo) = f x{x0)mo<\.LKLMD.

The prime q satisfies (3) and besides

l q - ł ( K L \ q ) ^ f 2(x0)mo&I), (£<?-\{KL\q), D) = 1,

which proves Theorem 1 in this case. As to Theorem 2, Conjecture H implies the existence of infinitely many integers x such that fi(x) = q and / 2(ж) = p are both primes. Taking x so large that q i K L M we have by (3)

(

4

q\Pp{°, /3).

On the composite Lehmer numbers with prim e indices, I 99

On the other hand, we have for К > 0 (6) |P„(a,

)I > (max(|a|, ^j))»-2(a + /i)

> / i £ >/2+ i X l / 2 p 2>

+ V

\v~2 2 )

and for К < 0, p > Ж(а, fi) by the fundamental lemma of [1]:

(5') \Pp(a, fi) I > |a|p- log325 > (|/2)p- log4

Thus for p large enough, \Pp(a, fi)\ > 2 p -\ -l — q and (4) implies that P p(a, fi) is composite.

It remains to consider the case where К , L and one of the numbers

— K M , ~ 3 K M is a perfect square. In this case the numbers a and fi are rational integers. For every pair L, M in question we shall give the

1 .

value of fc and construct two polynomials f 1{x, у ) , f 2{x, у) — — Пх( x , у ) — I

к '

of degree d < 2 satisfying the following conditions :

(i) polynomials f x(x, y)^fi(x, У) have integral coefficients, are prim­

itive, irreducible as polynomials in x for every integral value of у and (6) (/i(0 , 0 )/2(0, 0), (2d)!) = 1,

(ii) if (f1{x0, y0), y) = 1, there exist infinitely many primes of the form f x(x, у) = ) х{х0, y0)m odia,

(iii) if q = f x{x, y) is a prime -fiY, then

(7) j K a / ^ - ^ - l .

These conditions being satisfied, Theorems 1 and 2 follow easily.

First, by condition (i) and Gauss’ Lemma the polynomial f x(x, y)f2(x, y) . is primitive of degree 2d. Formula (6) implies that it has no fixed factor

> 1. Thus for every Р Ф 0, there exist integers x 0, yQ such that (8) (fi(x0, Уо)/а(®о>

kKLMD) = 1.

Now by condition (ii) there exist infinitely many primes of the form

/ i ( ® » У) = / i ( ® 0> Уо)

m°d k K L M D .

For every such prime we have by (iii) divisibility (7); thus

(9) q\P{a-i),tc(a,fi)

because q^KL.

On the other hand, q — 1 1

—— - = д;(/г(Я о,У о)-1) = / 2(%,?/o)niodD;

q

thus in view of (8), к I) = 1 .

100 A. Schinziel

Further, (i) implies the existence of an integer y 0 such that f i i x > У o) bas no fixed factor > 1 . Since again by (i) the polyno­

mials /i(a ?,y0) and / 2(ж, y Q) are irreducible, Conjecture H implies that for infinitely many integers x, q = / х(ж, y0) and p = / 2(ж, y0) are both primes. If we take x so large that

q^KLM and I— - — I > kp-\ -1, q — 1

we have (9) and since --- - = p , the number P p(a, j3) is composite.

к

We now proceed to the construction of polynomial / x (x, y) and de­

note by N the greatest odd factor of M.

If — 3 M is a perfect square, we put

к = 6, М х , у ) = 4(ЗЖж+1)2 + 27А2(2у + 1)2.

Condition (i) is satisfied trivially. Further, if (fi{x0, y0), y) = 1, then by Dirichlet’s theorem for the field K ( V —3), there exist infinitely many primes Q of that field satisfying the congruence

Q = 6Аж0 + 2 + З.Ж(2у0 + 1)1// — 3m odl2W /w.

A norm of any such prime Q is a rational prime q of the form f i ( x, у ) and satisfies the congruence q = / x {xQ, y0) mod у . Condition (ii) is satis­

fied. As to (iii), we notice that / х(ж, y) = lm od 6 and ( / х(ж, у), M) — 1;

thus if q = fi(oc,y) is a prime, we get

(10) (a p f~ 1)l2 == (M\q) = ( - 3 M \ q ) ( - 3 \ q ) = lm od q.

Further, it follows from the cubic reciprocity law that a and /5 are both cubic residues modgq thus

(11) (a//?)(3-1)/3 = lm o d g .

It follows from (10) and (11) that (a//?)(3-1)/6 = lm o d g , i.e. condition (iii) is satisfied.

If —I f a perfect square, we put

(12) ~ M = e2V~\

where у is an integer ^ 2 and e is a positive integer not being a perfect square (this is possible since — М Ф 0,1).

If neither у = 2, e = 6/2 nor у > 3, e = 2/2, then by Lemma 2 there exists an odd integer m > 0 such that

(13) 2y\\m~l, (k{e)\m) = - 1 , ( m - 1 , k{e)) = (2, k(e)).

W e put

k = 2y, f x(x, y) = /i(a ?) = 2у+1к(е)хф m.

On the composite Lehmer numbers with prim e indices, I 101

Condition (i) is satisfied trivially. Further if = 1 , then by Dirichlet’s theorem there exist infinitely many primes

q = f 1(x0)m.o&2v+1'k(e) ft.

These primes are clearly of the form fi(x), and thus condition (ii) is satisfied. As to (iii), if q = / i ( # ) > 0 and (q, e) — 1, we have by (13)

2 1 a - l ,

(e\q)

since (e|r) considered for r > 0 is a character with conductor dividing 4:1c (e). Hence if q is a prime t M , we get

(14) JLf«-Wsv = ( - 1 = _ e« - W s _ ( е |д) = i mod«.

On the other hand, it follows from (12),

= M and (a, ft) = 1 that (a2)(ff_1)/2V = (ft2)(Q~l'>l2V = lm od # ;

thus by (14)

(a lft)(-Q- 1V2V = == lm od q, which proves condition (iii).

If у = 2, в = 6/ 2, we put

' 1c = 12, f t (x, у) = (6Аж+ЗЖ + 2)2 + 108А2(2у + 1)2.

Condition (i) is satisfied trivially. Further, if (/i(# 0, yQ), /л) = 1, then by Dirichlet’s theorem for the field K{ V — 3) there exist infinitely many primes Q of this field satisfying the congruence

Q = 6Nx0+ 3 N + 2 + 6N(2y0 + l )V^-3mo&24Ny.

A norm of any such prime Q is a rational prime q of the form f x (x, у ) and satisfies the congruence q = / х(ж0, y0)m od/j. Condition (ii) is satis­

fied. As to (iii), if q = f t (x, y) is a prime f M, then by the cubic reci­

procity law

and ft are both cubic residues modg, and thus

(15) (а//?)(а_1)/3 = lm od g.

Further, since

= M = — 36/4 and (a, ft) = 1, each of the numbers

|a| and \ftI is a perfect square and we get

( a f f t f - W = Jf<3- 1)/4 = ( - l f - 1)l4&ą- 1)l2modq.

However, q = 13 mod 24, and thus

(16) ( a f f t f - W = _ 6 ( 3- 1)/2 = — (6|g) = lm o d q.

It follows from (15) and (16) that (а//?)((*~1)/12 = lm o d # , i.e. con­

dition (iii) is satisfied.

102 A. S c h i n z e l

If у > 3, e = 2/2 we put e = 2a</2, g odd.

By the Chinese Remainder Theorem the systems of congruences 4mod&(y), ■ 14mod3ft(</),

2y+1 + lm o d 2 y+2, V ~ \ 2v+1 + l - 161c{ gf mo&2v+2

are solvable for every у > 3. Denote by uY, vY any of their solutions and put Tc = 2y+1,

f ( x = [(2у+1Цд) х + иу)2 + 22у{1с(д)/3)ЦЗу + 1)2, if 3 1 Tc(g), {Х,У) ~ \(3-2y+1k ( g ) x + v y)2+Jc(gf(3-2Yy + 4:f, if 3< Щ) . Condition (i) is satisfied in view of the choice of uY, vY. Further, if (/j (x0, y Q), y) = 1, then by Dirichlet’s theorem for the field К (i) there exist infinitely many primes Q of that field satisfying the congruence

Q =

2y+1]c(g)x0 + uv + 2y (3y0 + 1 )i mod 2y+1 h {g), 3-2y+1Tc(g)x0 + vy+Jc(g)(3-2yy 0 + 4;)i m od3-2Y+lJc(g),

if 3| Tc{g), if 3fk{g).

A norm of any such prime Q is a rational prime q of the form f i { x, у ) and satisfies the congruence q ~ /г(х0, y0)mody. Condition (ii) is, there­

fore, satisfied. As to (iii), we consider the cases 3|k(g) and 3i~k{g) sepa­

rately.

If 3| Jc(g) and q is a prime of the form / j (x, y), then by the biquadra­

tic reciprocity law, 2 and Tt{g)j3 are both biquadratic residues modg;

thus 2 'У /9 = e[9 is also such a residue. If q^M we have, since q ~ 5 mod 12, e(a-m = (в/9)(®- 1)/4 9(*_ 1)/4 = 3(<z_1)/2 = (3| q ) = - l m o d ^ .

Since (q— l)/2 y+1 is odd, we get

М П -1)12Г ± '

_ = _„(g-D/4 s lm o d g .

Further, since aft = M = — -if ' and (a, fl) = 1, one of the numbers a2, fl2 must be a perfect 2y+1-th power; thus

(a//5 f - 1)'2V+1 = = lm od « , which proves condition (iii).

If 3\ Jc(g) and q is a prime of the form ^ (ж, y), then by the biquadra­

tic reciprocity law, lc{g) is a biquadratic residue m odq but 2 is not. If q f M, this gives

6(a-i)/4 s ^ 2(я-1)/4)*5^ - 1)/2 = — 1 mod q,

whence condition (iii) follows as before. This completes the proof.

On the composite Lehmer numbers with prim e indices, I 103

References

[1] A . S c h in z e l, The intrinsic divisors of Lehmer numbers in the case of negative discriminant, Ak. Mat. 4 (1962), pp. 413-41 6.

[2] — On primitive prime factors of Lehmer numbers I , Acta Arith. 8 (1963), pp. 213-22 3.

[3] — et W . S i e r p iń s k i, Sur certaines hypotheses concernant les nombres premiers, Acta Arith. 4 (1958), pp. 185-208.

[4] W . S ie r p iń s k i, Sur les nombres premiers dont tous les chiffres sont egaux a 1, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fiz. Mat. Nat. 31 (1961), pp. 3 4 7 -3 4 9 .

[5] — O liczbach złożonych postaci (2P + l ) / 3 , gdzie p jest liczbą pierwszą, Prace Mat. 7 (1962), pp. 169-172.