Beitr Algebra GeomDOI 10.1007/s13366-012-0086-6

ORIGINAL PAPER

On the least odd quadratic non-residue

Wolfgang Knapp · Markus Köcher · Peter Schmid

Received: 29 September 2011 / Accepted: 21 December 2011© The Managing Editors 2012

Abstract Given a prime p > 3 let νp be the least positive odd integer which is anon-square mod p. Then νp = q is a prime less than p, and one knows that νp <√

p unless p ∈ {5, 7, 11, 13, 23, 59, 109, 131} (Brauer, Nagell, Rédei). Using ana-

lytical methods it has been shown that νp = O(p1

4√

e ) (Vinogradov, Burgess), andνp = O(log2 p) if one assumes the Extended Riemann Hypothesis (Ankeny). In thispaper we prove, with elementary algebraic means, that νr <

32 log r−1

2 for each primer ≡ p (mod 4q̂), with at most one exception. Here q̂ is the product of all odd primesless than or equal to q.

Keywords Quadratic non-residues · Primes · Arithmetic progressions

Mathematics Subject Classification (2010) 11A15 · 11N13

1 Introduction

Let p be an odd prime. We are interested in the least positive odd integer q = νp whichis a quadratic non-residue mod p, i.e., where the Legendre symbol ( q

p ) = −1. Thisνp is a prime (as the Legendre symbol is a character). We have ν3 = 5 and νp < pwhen p > 3, because there are (p − 1)/2 integers in {1, 2, . . . , p − 1} which are

W. Knapp (B) · M. Köcher · P. SchmidMathematisches Institut, Universität Tübingen,Auf der Morgenstelle 10, 72076 Tübingen, Germanye-mail: wolfgang.knapp@uni-tuebingen.de

M. Köchere-mail: markus.koecher@t-online.de

P. Schmide-mail: peter.schmid@uni.tuebingen.de

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quadratic non-residues (residues) mod p, and also (p − 1)/2 odd (even) integers inthis interval, and 4 is an (even) square in the interval when p > 3.

In his celebrated “Disquisitiones Arithmeticae” Gauss (1801) showed that νp <

1 + 2√

p when p ≡ 1 (mod 8), that is, when both −1 and 2 are squares mod p. Ithas been noticed by Brauer (1931), Nagell (1951), Rédei (1953), Tate (see Lam 1973,p. 179) and others that even νp <

√p in this case. For p �≡ 1 (mod 8) the square root

bound νp <√

p does not hold in general. By analytical methods Vinogradov (1927)

and Burgess (1957) have shown that νp = O(p1

4√

e ). If one assumes the Extended Rie-mann Hypothesis, then one can even prove that νp = O(log2 p) (cf. Ankeny 1952).

Motivated from Salié (1949), and using only elementary (algebraic) arguments, weestablish the following.

Theorem 1 Let νp = q for some prime p > 3, and let q̂ be the product of all oddprimes less than or equal to q. Then νr <

32 log r−1

2 for each prime r ≡ p (mod 4q̂ ),with one possible exception. The exception can only happen when the least positiveinteger in the arithmetic progression p + 4q̂ Z is a prime, say r , in which caseνr >

910 log r+3

2 .

The proof makes just use of quadratic reciprocity, as proposed by Euler, and ofelementary properties of the Chebyshev ϑ-function as derived by Felgner (1991). Theestimates can be slightly improved using the more advanced results of Rosser andSchoenfeld (1962) and Schoenfeld (1976).

For any odd prime q there exist primes p > q with νp = q, and then νr = qfor each prime r in the residue class p + 4q̂ Z (see Lemma 1). It seems that thereare always such residue classes mod 4q̂ where the least positive integer is a primebut that, for q > 7, there are also classes where this is not a prime (cf. Sect. 2).So the exception in Theorem 1 definitely can be excluded in certain cases. On theother hand, the much weaker square root bound is true whenever q > 17. IndeedNagell (1923), Nagell (1951) and Rédei (1953) have shown that νp <

√p unless p ∈

{3, 5, 7, 11, 13, 23, 59, 109, 131} (where νp = q equals 5, 3, 3, 7, 5, 5, 11, 11, 17,respectively). We give a slight refinement of this result by using a simple property oftriplets of primes in arithmetic progression (Lemma 2).

Theorem 2 Let the prime p > 23 and different from 59, 109 and 131. Then νp =√(p − 23)/2 for p ∈ {41, 73} and νp <

√(p − 31)/2 otherwise for p ≡ 1 (mod 8),

νp = √p − 12 for p ∈ {37, 61}, νp = √

p − 20 for p = 29, νp = √p − 22 for

p ∈ {31, 47, 71}, and νp <√

p − 33 in the remaining cases.

It is supposed that for any real number c > 14√

e, say, there exists an effective bound

α(c) such that νp < pc for all primes p > α(c). By the above we may take α(c) = 131for c ≥ 1

2 , and computations indicate that we possibly may take α(c) = 105 for c ≥ 13

(cf. the Table in Sect. 2). By the results of Brauer (1931) α(c) exists for c > 25 (the

bound being rather large when c is close to 25 ).

For an odd prime p let �p be the least odd prime ( �=p)which is a quadratic residuemod p. The study of this invariant is quite different from that of νp, and is closelyrelated to the class number of the quadratic number field Q(

√−p). In a subsequentpaper we shall show that �p <

√p except for eight or nine odd primes p.

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2 The set of primes N (q)

For an odd prime q let N (q) denote the set of primes p > 3 satisfying νp = q, thefibre over q of the map p �→ νp from the primes p > 3 to the odd primes (which is asurjection).

Lemma 1 Suppose q = qn is the nth odd prime in the natural ordering (q1 = 3,q2 = 5, etc.) and q̂ is the product of the first n odd primes (n ≥ 1). Then N (q)consists of all the primes in the ϕ(q̂ )/2n pairs of arithmetic progressions ±a + 4q̂ Z

with a odd and prime to q̂, q < a < 2ϕ(q̂ ), satisfying the Jacobi symbols ( qna ) = −1

and ( qia ) = 1 for i < n (if any). In particular, N (q) has density 1/2n.

Proof The lemma may be considered as a variant of an observation made by Salié(1949). (There is an analogous result for the set R(q) of all primes p > 7 satisfying�p = q.) For convenience we sketch a proof. Let X be the subset of G(q̂) = (Z/q̂ Z)∗mapping, via the Chinese remainder theorem, onto the non-squares in G(qn) ∼= F

∗qn

and the squares in G(qi ) for i < n. Then the cardinality |X | = ϕ(q̂)/2n . Let X∗ bethe subset of G(q̂) consisting of the classes −a + q̂ Z for integers a with a + q̂ Z ∈ X(so that |X∗| = |X |).

Let p > 3 be a prime. Recall that νp = q implies that p > q. Hence by quadraticreciprocity p ∈ N (q) if and only if p + 4q̂ Z in G(4q̂) ∼= {±1} × G(q̂) correspondsto (1, x) for some x ∈ X when p ≡ 1 (mod 4), and to (−1, x∗) for some x∗ ∈ X∗otherwise (cf. Proposition 1.10 in Cox 1989). By Dirichlet’s prime number theoremN (q) has (analytic, natural) density 2|X |/ϕ(4q̂) = 1/2n . Use finally the relevant(reciprocity) properties of the Jacobi symbol. ��

For example, N (3) consists of the primes in 5+12Z and 7+12Z = −5+12Z, andN (5) is the set of primes in the arithmetic progressions mod 60 of 13, 23, 37 = 60−27and 47 = 60−13. Similarly, N (7) is the set of all primes in the arithmetic progressionsmodulo 420 of 11, 61, 71, 179, 181, 191, 229, 239, 241, 349, 359, 409 (which are allprimes). For q = 11 we have 4q̂ = 4620 and p = 109 ∈ N (q), but 4q̂ − p = 13 ·347is not a prime. For q = 13 we have p = 709 ∈ N (q) but 4q̂ − p = 17 · 2903.

By a theorem of Linnik, as improved by Heath-Brown (1992), one knows thatp < (4q̂)6 if p ∈ N (q) and p is the least prime in its arithmetic progression mod 4q̂ .

Remark Let p(q) = min N (q) be the least prime in N (q). By the theorem of Nag-ell–Rédei we have p(q) > q2 for q > 17. There is some evidence that p(q) < 4q̂(so that the residue class of p(q) mod 4q̂ is exceptional in Theorem 1). We list p(q)for the first 13 odd primes q:

q 3 5 7 11 13 17 19 23 29 31 37 41 43p(q) 5 13 11 59 421 131 1,811 2,939 13,381 12,011 66,491 148,139 275,651

3 Proof of Theorem 1

We consider the Chebyshev ϑ-function, with ϑ(x) = ∑

r≤x log r (r prime, x > 0).By the prime number theorem ϑ(x) ∼ x for x → ∞ (e.g. see Theorem 9.5 inNathanson 2000). However, we need (sharp) effective estimates for ϑ(x). Already

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Chebyshev has shown that ϑ(x) < log 4 · x (see Nathanson 2000). Felgner (1991)proved (with elementary means) that one even has ϑ(x) < log 3 · x < 10

9 · x forintegers x ≥ 1.

Now fix a prime p > 3, and let νp = q. Thus p ∈ N (q) (and p > q). By Lemma 1we have νr = q for each prime r in the arithmetic progression p + 4q̂ Z (and for r in−p + 4q̂ Z).

Suppose first that the prime r is the least positive integer in the residue class p +4q̂ Z. Then r < 4q̂ and the primes in (4q̂ − r) + q̂ Z belong to N (q) by Lemma 1.It follows that 4q̂ − r > q. Consequently log(r + q) < log(4q̂) = log 2 + ϑ(q) andνr = q > 9

10 log r+32 by Felgner’s theorem.

Suppose next that the prime r ≡ p (mod 4q̂) but that it is not the least positiveinteger in p + 4q̂ Z. Then r > 4q̂ , even r > q + 4q̂ (Lemma 1). Thus log(r − q) >log 2 + ϑ(q). For n = 1(q = 3) we have r ≥ 5 + 12 = 17 and q = 3 < 3

2 log r−12 .

For n ≥ 2 (q ≥ 5) we have ϑ(q) ≥ 23 q. This is easily checked for 5 ≤ q ≤ 41, and

Felgner (1991) has proved (with elementary means) that ϑ(q) > log 2.1 · q > 0.7 · qfor q > 41. It follows that for q > 3 always

νr = q <3

2log

r − q

2.

Hence the result. ��Remark Using more advanced (analytical) methods it has been proved by Rosserand Schoenfeld that ϑ(x) < 1.001102x for all x > 0 (see in particular Schoenfeld1976, Theorem 6∗). In fact, it is conjectured that ϑ(x) < x for all x > 0. One alsoknows that ϑ(x) ≥ 0.84x for x > 101, and that one gets better estimates for larger x(see Theorem 10 in Rosser and Schoenfeld 1962). Of course, this could be used inorder to improve the estimates in Theorem 1.

4 Triplets of primes in arithmetic progression

It is a well known interesting (and still open) question whether there are infinitely manytwin primes q and q + 2. However, if q, q + 2 and q + 4 are primes, then necessarilyq = 3. This is immediate from the following special case of a more general result onprimes in arithmetic progression (see Ribenboim 1996, p. 284).

Lemma 2 Let d ≥ 2 be an integer. Assume that q, q +d, q +2d is a triplet of rationalprimes. Then either q = 3 or d is divisible by 3.

Proof Let us recall the simple argument. It is clear that q must be odd and d even.Suppose d is not divisible by 3. Then q, q +d, q +2d represent the three distinct resi-due classes modulo 3. Hence exactly one of these primes is divisible by 3, so equals 3.This forces that q = 3 (and q + 3d is no prime in this case). ��Consider in particular the situation that d = 2s for some integer s ≥ 1. Fors = 1, 2, 3, 6, 7, 10, 11 the string 3, 3 + 2s, 3 + 2s+1 indeed consists of primes,and one might ask whether this is a chain of primes for infinitely many s. Of course,

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3 + 2s is divisible by 5 whenever s ≡ 1 (mod 4). One knows that there are infinitelymany triplets of primes in arithmetic progressions (see Ribenboim 1996, p. 285).

So far the longest (explicitly) known string of primes in arithmetic progression con-tains 25 terms. However, one knows that there exist strings of arbitrarily large length(Erdös–Green–Tao); see Green and Tao (2008).

5 Proof of Theorem 2

Let the prime p > 5 in what follows, and let ν = νp. We know that ν < p (henceν ≤ p − 2 as ν is odd). There are unique integers k ≥ 1 and r �= 0 such that

p = 2kν + r and |r| < ν

(dividing p by 2ν). Here r is odd and so |r | ≤ ν−2. We distinguish the obvious cases.

Case 1 p ≡ 1 (mod 8)Then ( r

p ) = (|r |p ) = 1 (by the very definition of ν). Using that (−1

p ) = 1 we see

that ( 2kp ) = ( νp ) = −1. Since ( 2

p ) = 1, this implies that (the odd part of) ( kp ) = −1

and so k ≥ ν (by definition of ν). Hence p ≥ 2ν2 + r ≥ 2ν2 − ν + 2. Arguing likeNagell (1951) we show that p ≥ 2ν2 − 1. There is an integer a ≥ 0 such that

p = 2ν2 − ν + 2 + 2a = 2(ν − 1)(ν − 1 − a)+ (2a + 3)ν.

Clearly ν − 1 is a quadratic residue mod p. If a ≥ ν − 1, then p ≥ 2ν2 + ν.If a < ν− 1, then ν− 1 − a is a quadratic residue and 2a + 3 a quadratic non-residuemod p, whence 2a + 3 ≥ ν and so p ≥ 2ν2 − ν + 2 + (ν − 3) = 2ν2 − 1. Hence inboth cases p ≥ 2ν2 − 1.

We next show that if p = 2ν2 − 1 then necessarily p = 17. Assume that ν > 3 butp = 2ν2 − 1. Then we have the decompositions

p = 2ν(ν − 1)+ 2ν − 1 = 2ν(ν + 1)− (2ν + 1).

Since −1, 2, ν− 1 and ν+ 1 = 2(ν+ 1)/2 are squares mod p, we see that 2ν− 1 and2ν+1 are quadratic non-residues mod p. Both integers are odd, and ν < 2ν−1 < 2ν,2ν < 2ν + 1 < 3ν. It follows that 2ν − 1 and 2ν + 1 are primes. For otherwise each(proper) prime divisor is less than ν (as ν > 3), and there is one which is a quadraticnon-residue mod p (as the Legendre symbol is a character). This would contradict thevery definition of ν = νp. In particular, 2ν−1 and 2ν+1 are not divisible by 3. But ifν ≡ 2 (mod 3) then 2ν−1 ≡ 0 (mod 3), and if ν ≡ 1 (mod 3) then 2ν+1 ≡ 0 (mod 3).Hence we must have ν ≡ 0 (mod 3). Thus ν = 3, against our assumption.

Consequently p > 2ν2 − 1(p > 17). We cannot have p = 2ν2 + 7, because thencertainly ν �= 3 and, therefore, 2ν2 + 7 ≡ 0 (mod 3). We have to exclude the possibil-ity that p = 2ν2 + 15 (noting that p ≡ 1 (mod 8)). Here ν �= 3 (as 2 · 32 + 15 = 33),ν �= 5 (as 75 is no prime), ν �= 7 (as ν113 = 3), ν �= 11 (as ν257 = 3) and ν �= 13(as 353 is no prime). Hence ν > 13. Write p = 2ν2 + 15 in the following ways:

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p = 2ν(ν + 3)− 3(2ν − 5) = 2ν(ν − 3)+ 3(2ν + 5) = 2ν(ν − 1)+ 2ν + 15.

Using that −1, 2, 3, ν ± 3 and ν − 1 are squares mod p we obtain that ( 2ν−5p ) =

( 2ν+5p ) = ( 2ν+15

p ) = −1. Now q = 2ν − 5, q + 10 = 2ν + 5 and q + 20 = 2ν + 15are odd integers in the interval (ν, 3ν) (ν ≥ 17). It follows that q, q + 10, q + 20 isa triplet of primes in arithmetic progression, which contradicts Lemma 2 since q �= 3and d = 10 is not divisible by 3.

Therefore p ≥ 2ν2 + 23. We cannot have p = 2ν2 + 31 since then ν �= 3 andso 2ν2 + 31 ≡ 0 (mod 3). It remains to show that νp = √

(p − 23)/2 if and only ifp ∈ {41, 73}. Otherwise p = 2ν2 + 23 and ν > 7. Now 2ν2 + 23 is not a prime forν = 11, 13, 19, 23, and 2 · 172 + 23 = 601 but ν601 = 7. Hence ν > 23. From

p = 2ν(ν − 1)+ (2ν + 23) = 2ν(ν + 1)− (2ν − 23)

we infer that 2ν + 23 and 2ν− 23 are quadratic non-residues mod p, because −1, 2,ν−1 and ν+1 = 2(ν+1)/2 are squares mod p. It follows that 2ν−1 and 2ν+1 areprimes (as they are odd and in the interval (ν, 3ν), and ν > 3). In particular both 2ν−1and 2ν+ 1 are not divisible by 3. But if ν ≡ 1 (mod 3) then 2ν− 23 ≡ 0 (mod 3), andif ν ≡ 2 (mod 3) then 2ν + 1 ≡ 0 (mod 3). Hence we must have ν ≡ 0 (mod 3). Thusν = 3, against our assumption.

Case 2 p ≡ 7 (mod 8)Now ( r

p ) = 1 if r > 0 and ( rp ) = −1 if r < 0, and ( 2

p ) = 1. If r < 0,

then ( kp ) = −1 and k ≥ ν, hence p ≥ 2ν2 + r ≥ 2ν2 − ν + 2. If r > 0, then

2 ≤ ν − r ≤ ν − 1 and ( ν−rp ) = (

(ν−r)/2p ) = 1. From p = (2k + 1)ν − (ν − r) we

get that ( 2k+1p ) = −1 and 2k + 1 ≥ ν (by definition of ν), hence p ≥ ν2 − ν + 1 in

this latter (worse) case. At any rate we have

p = ν2 − ν + 1 + a = (ν − 2)(ν + 1)+ (a + 3)

for some (even) integer a ≥ 0. Since ( ν+1p ) = (

(ν+1)/2p ) = 1 = ( ν−2

p ) and (−1p )=−1,

a + 3 must be a quadratic non-residue mod p. Therefore a + 3 ≥ ν and p ≥ ν2 − 2.We assert that if p = ν2 − 2 = (ν + 3)(ν − 3)+ 7, then p = 7 or p = 23. Assumethat this is false. Then ν �= 3, 5. Since ν ± 3 [and (ν ± 3)/2] are quadratic residuesmod p and p > 23, we must have ( 7

p ) = (−1p ) = −1 and so ν = 7. It follows that

p = ν2 − 2 = 47. But ν47 = 5. Hence the assertion. We cannot have p = ν2 + 6,because otherwise ν �= 3 and (−1

p ) = (−6p ) = ( ν

2

p ) = 1. If p = ν2 + 14 then p = 23

when ν = 3, but ν23 = 5, and ν2 + 14 ≡ 0 (mod 3) when ν �= 3. Thus p ≥ ν2 + 22.Suppose that p = ν2 + 22. We claim that then p = 31, 47 or 71, that is, ν ≤ 7.

Assume ν > 7. Since 112 +22 = 143 is no prime, since 132 +22 = 191 but ν191 = 7,since 172 + 22 = 211 but ν211 = 3, and since 192 + 22 = 383 but ν383 = 5, we thenhave ν ≥ 23. From

p = ν2 + 22 = ν(ν + 1)− (ν − 22)

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we may conclude that ( ν−22p ) = −1 and ν − 22 ≥ ν, which is absurd.

We finally have to exclude the possibility that p = ν2 + 30. In this case we musthave ν ≥ 31 (ν �= 7 as ν79 = 3, ν �= 11 as ν151 = 3, and otherwise ν2 + 30 is noprime for ν ≤ 29). Now from p = ν(ν+1)− (ν−30)we can infer that ( ν−30

p ) = −1

and so ν − 30 ≥ ν, which again is absurd. Thus p ≥ ν2 + 38.

Case 3 p ≡ 5 (mod 8)Now (−1

p ) = 1 and ( 2p ) = −1. We could show as before that p > ν2 (using

that p �∈ {13, 109}; cf. Nagell 1951); the approach given by Rédei (1953) is some-what different (studying the largest integer e such that e2 < p). We assume thatp > ν2(ν <

√p) in what follows. We first show that then even p ≥ ν2 + 12, with

equality if and only if p ∈ {37, 61}. Of course ν2 ≡ 1 (mod 8) (as ν is odd) andp = ν2 + s with s ≡ 4 (mod 8). Assume that p = ν2 + 4 (s = 4). Then ν �= 3 (asν13 = 5), ν �= 5 (as ν29 = 3, ν �= 7 (as ν53 = 3) and ν �= 13 (as ν173 = 3). Considerthe following decompositions of p = ν2 + 4:

p = (ν + 2)(ν − 2)+ 8 = (ν + 10)(ν − 10)+ 104.

Using that −1, ν − 2, ν − 10 are squares mod p and that 8 and 104 = 8 · 13 are non-squares we deduce that both ν + 2 and ν + 10 are non-squares mod p in the interval(ν, 2ν) and so must be primes. If ν ≡ 1 (mod 3) then ν + 2 ≡ 0 (mod 3), which isimpossible. If ν ≡ 2 (mod 3) then ν+10 ≡ 0 (mod 3), which again is impossible. Butthis forces that ν = 3, which has been already ruled out.

We next show that p = ν2 + 12 if and only if p = 31 (ν = 5) or p = 61 (ν = 7).Clearly then ν �= 3. Assume that ν > 7 and consider the decompositions

p = ν2 + 12 = (ν + 4)(ν − 2)− 2(ν − 10) = (ν + 4)(ν − 4)+ 4 · 7.

Since −1, ν − 2, ν − 10 and 4, ν − 4 are squares mod p, we get ( ν+4p ) = −1 and,

hence, that also ( 7p ) = −1, against our assumption ν > 7.

We assert that p = ν2 + 20 if and only if p = 29 (and ν = 3). Assume that ν �= 3here. Then one checks that we must have ν ≥ 17 (at least), and from

p = (ν + 4)(ν − 2)− 2(ν − 14) = (ν + 8)(ν − 2)− 6(ν − 6)

one infers that ( ν+4p ) = −1 = ( ν+8

p ). It follows that ν, ν + 4, ν + 8 is a triplet ofprimes, and Lemma 2 forces that ν = 3, against our assumption.

It remains to rule out the possibility p = ν2 + 28. Using that ν37 = 5, ν53 = 3,ν149 = ν197 = 3 and that 72 + 28 is no prime we get that here ν ≥ 19 (at least). From

p = ν2 + 28 = (ν + 4)(ν − 2)− 2(ν − 18)

we deduce that ( ν+4p ) = −1. Applying this to p = (ν + 4)(ν − 4) + 44 we see that

44 = 4 · 11, hence 11, is a non-square mod p, against ν ≥ 19.

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Case 4 p ≡ 3 (mod 8)Now (−1

p ) = −1 = ( 2p ). By the Nagell–Rédei theorem we know that p > ν2,

because p �∈ {11, 59, 131} by hypothesis. Having also excluded p = 19 we havep ≥ 43. Since ν43 = 3 we therefore may assume that ν ≥ 5 in what follows. Wecannot have p = ν2 + 2, because ν2 + 2 ≡ 0 (mod 3) (as ν �= 3). Assume thatp = ν2 + 10. Then ν ≥ 23, because ν2 + 10 is no prime for ν = 5, 17, 19 and ν �= 7as ν59 = 11, ν �= 11 as ν131 = 17 and ν �= 13 as ν179 = 7. Consider the followingdecompositions of p = ν2 + 10:

p = ν(ν − 4)+ 2(2ν + 5) = ν(ν − 8)+ 2(4ν + 5) = ν(ν − 12)+ 2(6ν + 5).

Since −2, ν − 4, ν − 8 and ν − 12 are squares mod p, we see that 2ν + 5, 4ν + 5 and6ν + 5 are non-squares mod p. By Lemma 2 this cannot be a triplet of primes. Since2ν + 5 and 6ν + 5 are primes, 4ν + 5 is no prime and hence divisible by 3. It followsthat ν ≡ 1 (mod 3). Now consider the decomposition p = ν(ν − 20) + 10(2ν + 1)and use that −10 and ν − 20 are squares mod p. This yields that ( 2ν+1

p ) = −1 andthat 2ν + 1 is a prime, which is impossible since ν ≡ 1 (mod 3).

Assume next that p = ν2 +18. Then ν ≥ 37, because if ν ≤ 31 then either ν2 +18is no prime (ν = 3, 13, 31) or p = ν2 + 18 is a prime with νp = 3. From

p = (ν + 2)2 − 2(2ν − 7) = ν(ν − 36)+ 18(2ν + 1)

we infer that 2ν−7 and 2ν+1 are non-squares mod p, hence are primes. Since ν �= 3and 2ν − 7 �≡ 0 (mod 3), we see that we must have ν ≡ 1 (mod 3). Similarly, from2ν + 1 �≡ 0 (mod 3) it follows that ν ≡ 2 (mod 3). This is the desired contradiction.

We finally note that we cannot have p = ν2 + 26, because ν2 + 26 ≡ 0 (mod 3)(as ν �= 3). Thus p ≥ ν2 + 34. ��

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