On the composite Lehiner numbers with prime indices, II

J.

On the composite Lehiner numbers with prime indices, II

A. Schinzel has deduced from his Conjecture H (cf. [2], p. 188) a theorem about the so-called Lehmer numbers

where a, ft are roots of the trinomial: zi - L x,iz-\-M and L, M are rational integers.

More exactly, he proved the following two theorems:

I. I f LM Ф О, К = L — 4cM Ф 0 and none of the numbers —K L ,

~ 3 K L , — 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 Тс > 0 such that for every integer D Ф 0 one can find a prime q satisfying

(1) q\P s and ( s , D ) = 1,

II. Under the assumptions of I, Conjecture H implies the existence of infinitely many primes p such that P p(a,

is composite.

The aforesaid conjecture reads as follows:

H. I f f l , f 2, . . . , f k are irreducible polynomials with integral coeffi­

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

The aim of this paper is to prove the following theorems:

Th e o r e m

1. I f L M Ф 0,

L , M) = 1, (\L\, \M\} Ф

1>, К =

= L — IM Ф 0, —K L is a perfect square and either 1Q\L or фЪМ and

± 5 3M are not perfect biquadrates, then there exists an integer ~k > 0 such that for every integer D ф 0 one can find a prime q satisfying

Pn(a, /9) = (an— j3n)[a — fi, if n is odd, (an—(3n) /a2 — /92, if n is even

where Jacobi's symbols of quadratic

character.

10G J. W ó j c i k

Theorem

2. I f L M Ф 0 , (L , M) — 1, <|L|, \M\) Ф ( 1 , 1 ) , <3, 1 ), К — L — 1M Ф 0, —3KL is a perfect square and either 27 f L or ± 7 M and ± 75M are not perfect sixth powers, then there exists an integer Тс > 0 such that for every integer T) Ф 0 one can find a prime q satisfying:

q\P(ą-i)lk and

q - i = i .

Theorem 3.

Under the assumption of Theorem 1 or 2, Conjecture H implies the existence of infinitely many primes p such that P p{a, fi) is com­

posite.

The assumption (L , M) = 1 does not essentialy affect generality and simplifies a little the formulation of the theorems. Let us remark that Theorems 1 and 2 can be proved with g - 1

Те replaced throughout by g + 1

Те The proof is based on the arithmetic of Gauss’s field K( i) and of Eisensteins field K (

). The only laws of reciprocity used are quadratic, cubic and biquadratic. As to the notation, we denote by A (a) the absolute norm of an element a of K( i) or K (

). An integer a is called primitive if it is not divisible by a rational integer > 1, primary if a

~ 1 m od(l + ^)3 for aeK( i ) , a = —lm od 3 for a e l ( o ) (cf. [1], p. 152 and 187). are Jac°b i’s symbols of cubic and biquadratic char­

acter, respectively. Tc2(a)

and &4(a) denote the square- free, the cube-free and the biquadrate-free kernel* of a, respectively

( а Ф

0). If a is primitive and

is its conjugate, then clearly (1) Nke{a) = ke(a)lce{a') = JceN{a) (e = 2, 3 or 4).

We begin with four elementary lemmas.

Lemma

1. For any biquadrate-free, primary, primitive integer aeK(i ) and for any rational integer g there exists a rational integer v(a, g) such that

(2) ((~ ~ a~~)) = i0 and И а ’ =

except for

g = 0m od4, A (a ) = 5 or 53;

(3) g = 2m od4, N (a) = 1;

g = 1 or 3m od4, A (a) is a perfect square.

P r o o f. The lemma is true if N{a) is a prime. Indeed, if Ж (a) > 5 , then

(a) —1)/4 > 1; thus for every value of

there exist at least two residue classes rmodiV^a) satisfying ((—]) = ia.

\\a//

Since N (a) is prime, at least one of them satisfies (r — 1, N {a)) = 1.

If Ж(a) = 5 and g Ф 0m od4, the existence of v(a, g) is obvious.

Suppose that Ж (a) is not a perfect square. If there exists a prime factory of Ж (a) such that

(4) <* = р1<*ц Р = Я ( р 1) > 5 , 1 = 1 or 3, then we define v(a,g) by the system of congruences

v{a, g)

— lmodJVr(a1),

v

\P

,9£ +

Ж (cq)— 1

m od p .

If there exists no prime factor p satisfying (4), then we have

—

«i

{px) = 5, £ = 1 or 3

and either g ф 0mod4 or ax = q\a2, where q = Ж(qx) is a prime.

We put in the first case

( - I m o d N K ) , via, g) = {

I

#£)mod5, an d. in the second case

j — l m o d 5 N ( a 2),

via, q) — I

I v{qx, l)m od g.

Now, let Ж (a) be a perfect square and Ж (а) ф 1. We have to con­

sider only the case g = 0 or 2 mod 4.

We put a = p\ax-, p = Ж(px) a prime,

— lmodJ\r(a1), v(Pi, g/2)modp.

Finally, if Ж (a) = 1 and g = 0mod4 we put v(a, g) = 1. It follows from the Chinese Remainder Theorem that the above systems of congruences are solvable and clearly any of their solutions satisfies (2).

Lemma 2.

I f a is any biquadrate-free, primary, primitive integer of K ( i ) , g arbitrary rational integer, and none of the exceptions (3) holds, then for any integer у > 1 there exists a positive integer m such that (5) m = 2y-f- lm o d 2 y+2,

and (m— 1, Ж {a)) — 1.

a

m

108 J. W ó jc i k

P r o o f. If we put

m = l - f 2 vmod2y+2, v(a, g)mo6.N(a),

it follows from the biquadratic reciprocity law that m satisfies (5).

Lemma 3.

For any square-free, primary, primitive integer aeK( o) and for arbitrary rational integers g, Ti there exists an integer v( a, g) such that

(6⁾

[

9, h)

a

] ^{Q°t y(g? N( a)}

^{( - I f}

and {v(a, g, h) — 1, N (a)) = 1 except for

# = 0 m od3, h = 0m od2, N (a) = 7, g ф 0mod3 or h ф 0m od2, N (a) = 1.

P r o o f. The lemma is true if N (a) is a prime.

Indeed, if N ( a ) > 7 , then (N(a) — l ) /6 > 1; thus for each value of g and h there exist at least two residue classes r m o d l(a ) satisfying

r

~N (a) ( - I f .

Since N (a) is a prime, at least one of them satisfies [г— 1,Ж (а)) — 1.

If N (a) — 7 and g =£ 0mod3 or h Ф 0mod2 the existence of v(a, g, h) is obvious. If N(a) is composite, and p — Ж( рг) is its greatest prime factor, we put

— lm o d N(a)/p, v(a, g, h)

v

2

m od p .

Finally, if N (a) — 1, g = 0m od3, h = 0mod2 we put v(a, g, h) = 1.

Lemma 4.

Let a, (3 be any primary, primitive integers of the field K (

) such that a is cube-free and is square-free. Let g , h be any rational integers and у, 3 any positive integers. There exists a positive integer m such that

m = 2<5-b lm o d 2 <5+1, m = 3y-f lm o d 3 y+1,

[ — 1 = ^ , Ш Ш = ( - 1)л and (m — 1, N (ap)\ — 1;

L

J

except for

g = 0m od3, h = 2<5-1mod2, N(a) = 7 or 72, N (ft) = 7;

7 i^ 0 m o d 2 , N((3) — 1.

# ^ 0 m o d 3 , N(a) = l ,

P r o o f. Let

x, where

—

—

(oq),

(/Si)) = 1. Further, let

=

c^),

=

a2 =

x;

<dx, d 2,

2 , 1 ) , if

* ^4? S у Щ ) —

<<*а, dlf 1 , 2>, if N{ dx) < Sid, ).

Clearly N( ds) > N( d4) ‘, a = d\dla2, N ф) — N( d3) N(d4)N(fix) and by the assumptions about a and /3, the numbers N( d3), N(d4), N (a2) and N ({3X) are relatively prime in pairs. Moreover, a2 is cube-free and pri­

mitive, and thus it can be represented in the form a2 = a\ a4, where (Na3, J a 4) = 1 and a3, a4 are square-free. We put

<«5, аб, С

'd') — <«3 J a4 ? ^ , 1^>,

<a4 ? аз > 1 ^ ?

if JV(as) ^ ^ ( a 4), if W(a3) < N (a4).

Clearly ЗУ(а5) ^ N( a6), a2 = agCtg and (3Va5, JVa6) = 1 .

We shall define w by a system of congruences different for each of the following 6 cases.

1. N( d3) > 7 or N( d3) = 7, g ф 0m od3:

— 1 mod N (d4 a2 fix),

v(d3, eg, h + 2s^ { N { f ) ) - l ) + - (dt^?) ~-1- )modJV-(d,), 2<3 + lm o d 2 <5+1,

3y+ l m o d 3 y+1;

2. N (d3) — 7, g = 0mod3 and either Ж(| 31) > 1 or W(j31) = l ,

== 2<5_1 + lm o d 2 :

--lm odTV (a),

»(/?!, 0, h + 2 s- i { N(§1) + l) + l\ m oiN (ti1),

tyyi -— ~ ' '

2й+ 1 mod 2d+1, 3y+ ln a o d 3y+1;

3. 3V(c73) — 7, g = 0m od3, N (fix) = 1; h = 2d_1mod 2:

— lm o d W(ag),

у(а5,2 ^ !С ; 0)mod N( as), m ^ 2 mod 7,

2d + lm o d 2d+1,

3y+ lm o d 3y+1, where

п о J. W ó jc ik

4. N(dz) = 1, N( a 2) > 1 , and either N ((Зг) > 1 or h == 0m od2:

where

m =

— lm o d JV'(aJ),

»( f t , *, Ъ + 2"-! (J (ft) - 1 ) ) mod N (ft), r(a5, £#, l)m od (a5),

2a + lm o d 2a+1, 3y+ lm o d 3y+I,

5. N( d3) = N (a2)

= Omod 2:

(O, if ДГ(0 ) = 1 , (1 , if N ( P ) > 1;

= 1; g = 0m od3 and either N {f$x) > 1 or

where

m =

V ( f t , ® , Ь+ а а- 2 ( У ( f t ) - 1 ) ) m o d # ( f t ) ,

2a + lm o d 2a+1, 3y+ lm od 3y+1,

(0 , if N(P) = 1, j l , if

A =

It follows from the Chinese Remainder Theorem that the above systems of congruences are solvable, and from the quadratic and cubic reciprocity law and from Lemma 3 that any of their positive solutions satisfies (7).

P r o o f o f T h e o re m s 1, 2 and 3. For every pair L , M in question we shall give the value of Tc and construct two linear functions f 1(co),f2{x)

= — (fi(oc)~ l) satisfying the following conditions:

К '

(i) f i ( x )i f i { x ) have integral coefficients, are primitive, their leading coefficient divide 12TcKLM2 and

(8)

(ii) if q = /, (x) is a prime - f l l , then (9)

These conditions being satisfied, Theorems 1, 2 and 3 easily follow.

First, by condition (i) and Gauss’ s Lemma the polynomial f x(x)f2(x) is primitive of degree 2. Formula (8) implies that it has no fixed factor

> 1 .

Thus for every D ф 0, there exists an integer such that

(10) (/i(®o)/s(*o)> 12 ШЪМгП) _{= 1}

Now by Dirichlet’s theorem, there exist infinitely many primes q == / 1(a?0)mod 12kKLM2l). Clearly, every snch prime q is of the form fx(x). For every such prime we have by (ii) divisibility (9); thus

because qi KL.

On the other hand,

4—~~ = y ( / i ( ® o ) - l ) = /a(® 0)modl>,

thus in view of (10) , 7>j == 1. Since by (8) f 1(x)f2(oc) has no fixed factor > 1 , Conjecture H implies that for infinitely many integers x, 0. = f i i x ) and V = h ( x ) are both primes. On the other hand, for p > N (a, ft) we have by inequality (5') of paper [3]

Thus, for p large enough, \Pp(a, /8)| > к р ф 1 = q and (11) implies that P p{a, fi) is composite.

1. Assume first that —K L is a perfect square. Since K L M ф 0 and (L , M) = 1 we have

= в

V2 S A ,

e ~ ~ W ~

where A is a primary, primitive integer of K( i ) , в4 = 1, t — 0 or 1, s = 0 or 1. Since <|£|, \М\)Ф < 2 ,1>, А ф 1. We put A = U2V~2, where у ^ 2, U is primary; у = 2, U — A , i i t = s = 0. U is not a perfect square if t — 1 or s — 1 (this is possible since A # 0 , 1 ) . Let e = N( U) . We put in Lemma 2: g = 2t — s, a = k4{U). Under the assumptions of Theorem 1 the conditions of this lemma are satisfied and by (1) there exists a positive integer m such that

(12) m — 2v + lm o d 2V+2, ^(*7) \\ _

m Jf ^amd 'm - 1 , fc4(e)) = 1.

We put Jc = 2у, / г(а?) = 2y+2k4l(e)x-\-m. Condition (i) is satisfied trivially. As to (ii) if q = f^ x ) and (q, e) = 1, we have by (12)

(13) g - 1

2y = 1 mod 4 and || —

since 11 — ) J, considered for r > 0 , is a character with conductor dividing u

4&4(e). We notice that if q — qxq2 is the decomposition of q into prime fac-

112 J. W ó j c i k

tors in К (i), we have (14) (UIU')iQ- m

Analogously

(15) ( UIU')(9~1)!i = ( ( y j j mod q2.

It follows from (14) and (15) that

mod qx.

(Z7/Z7')(a-1)/4 mod q.

Hence if q is prime we get by (13)

(16) (a/,? )« -Ч'2У = i(a+*)(«-4/*>'( = j«+» = i mod g , and conditions (ii) is satisfied. This completes the proof of Theorem 1 and of the first part of Theorem 3.

2. Assume now that — 3K L is a perfect square. Since K L M Ф 0 and (L , M) = 1, we have (if we neglect the order of a, ft):

a — &

\ 1 e)5

A , ft

( 1 -

2)

Vb*

A

where A is a primary, primitive integer of K (

), = 1, t = 0 or l , s = 0 or 1. Since <|.L|, |Ж|> ^ < 1 ,1>, < 3 ,1>; А

—1. We put A = Z72<3-1 = 7 зУ-1, where <5, у > 0, U, V are primary;

ó = l , U = A , if s = 0,

U is not a perfect square if s = 1 (this is possible since А

0 ,1 ), у = 1, V — A , if t = s,

V is not a perfect cube if t ф s (this is possible since А

0, ± 1 ) . Let e = N( U) , f = N{ V) . W e put in Lemma 4: g = r-f-2s, h = s, a — Jc3(V), ft = Tc(TJ). Under the assumptions of Theorem 2 the conditions of this lemma are satisfied and by (1) there exists a positive integer m such that

m = 2ó-f-1 mod 2a+1, m = 3y+ l mod 3y+1, (17)

Г М А 1 = ( M ) = ( _ ! ) • and ( т ~ 1 , Ц е Ш П ) = 1.

L m J \ m /

We put

Tc = 2a3y, f t (cc) = 2a+13y+Ifc(e)fc3( /) r + m .

Condition (i) is satisfied trivially. As to (ii), if q = f i ( o c ) > 0 and (q,fe) = 1? we have by (17)

0 — 1 0 — 1

(18) ^ - = l m o d 2 , - = lm o d 3 ■ И- ^{ą A}

^{\q ( - 1 ) %}

XVl I e\

since I — I, I —I considered for r > 0 are cubic or quadratic characters with conductor dividing Nh3(V) — fc3(/) or 4&(e), respectively.

We notice that if q = qxq2 is the decomposition of q into prime factors in K (

), we have

™ r a ■ [ а з - н а - ш —

Analogously

(20) ( UI U ' f - 1)/3 = j ^{mod q2.}

It follows from (19) and (20) that

( Z7/f7')(a_1)/3 = mod q.

Hence, if q is a prime ~/~M, we get by (18)

(21) = ( - 1 f Q~mS { s - t f ~ 1)l2d {U I U 'f ~ 1)l2

= ( — 1) j —J = 1 mod q and

(22) («/£)<«-1)/3V = ( ~ l f q- m %is- ł){q~m \ v / V T ~ 1)lb = е8~*[—] s l m o d ^ . It follows from (21) and (22) that

(а//?)(а~1)/2<5зУ = lm o d q,

i.e. condition (ii) is satisfied. This completes the proof.

References

[1] P. B a c h m a n .n , Die Lehre von der Kreistheilung und ihre Beziehungen zur Zahlentheorie, Leipzig 1872.

[2] A . S c h in z e l et W . S ie r p iń s k i, Sur certaines hypotheses concernant les nombres premiers, A cta Arith. 4 (1958), pp. 185-208.

[3] A . S c h in z e l, On the composite Lehmer numbers with prime indices, I , Prace Mat. this volume, pp. 95-103.

Prace Matematyczne IX. l