primes of the form aq2+1

Approaches to primes of the form n2 + 1Primes of the form aq2 + 1

Primes of the form aq2 + 1

Kaisa Matomäki

Department of Mathematics

Royal Holloway, University of London

March 6, 2008

Kaisa Matomäki Primes of the form aq2 + 1


Outline

1 Approaches to primes of the form n2 + 1

2 Primes of the form aq2 + 1



Primes of the Form n2 + 1

A long-standing conjecture:

Conjecture

There are in�nitely many primes of the form n2 + 1.

The �rst twelve instances of such primes are

2 = 12 + 1 101 = 102 + 1 577 = 242 + 1

5 = 22 + 1 197 = 142 + 1 677 = 262 + 1

17 = 42 + 1 257 = 162 + 1 1297 = 362 + 1

37 = 62 + 1 401 = 202 + 1 1601 = 402 + 1



Primes Represented by Polynomials

The case of primes of the form n2 + 1 = f (n) is a special case of amore general conjecture.

Conjecture

Any reasonable polynomial f (n) ∈ Z[x ] takes prime values in�nitely

often.

The linear case f (n) = an + b (where reasonable meansgcd(a, b) = 1) was proved by Dirichlet in 1837.

No instance of the conjecture in higher degree is known.

In 1922, Hardy and Littlewood gave a conjectural asymptoticformula for the number of primes of the form f (n).



Prime factors of n2 + 1

Theorem (Iwaniec 1978)

There are in�nitely many integers n such that n2 + 1 has at most

two prime factors.

Write P(m) for the greatest prime factor of m. Then of course

m ∈ P ⇐⇒ P(m) = m.

The previous theorem implies P(n2 + 1) > n in�nitely often.

Theorem (Deshouillers and Iwaniec 1982)

There are in�nitely many integers n such that P(n2 + 1) > n6/5.

In the proof the authors use their deep results on averages ofKloosterman sums.



Approximations with Polynomials in Two Variables

Theorem (Harman and Lewis 2001)

There exists in�nitely many primes p with

p = n2 + m2, m, n ∈ Z,m < p0.119.

The proof uses a generalization of Harman's sieve method toGaussian integers (complex numbers m + ni with m, n ∈ Z).

Theorem (Friedlander and Iwaniec 1998)

There are in�nitely many primes of the form n2 + m4.

The proof is almost one hundred pages long!



Approximations Using Fractional Parts

For x ∈ R, write {x} for the fractional part of x .

So {x} ∈ [0, 1) and x − {x} ∈ Z.If 0 < {x1/2} < x−1/2, then for some integer n,

n < x1/2 < n + x−1/2 =⇒ n2 < x < n2 + 2.

Thus if we could show that {p1/2} < p−1/2 in�nitely often,our conjecture would follow. But we only know

Theorem (Balog 1983, Harman 1983)

For any ε > 0, there are in�nitely many solutions to

{p1/2} < p−1/4+ε.



Approximating with aq2 + 1

Still another way to approach primes of the form n2 + 1 is toconsider primes of the form p = aq2 + 1 with a as small aspossible.

The case a = 1 is of course the conjecture itself.

Baier and Zhao 2006: a ≤ p5/9+ε for any ε > 0.

Theorem (Matomäki)

Let ε > 0. There are in�nitely many primes of the form

p = aq2 + 1, where a ≤ p1/2+ε and q is a prime.



Re�ning the Task

Let X be a large number. We write, for q2 ∼ Q = X 1/2−ε,

A(q) = {aq2+1 | aq2+1 ∼ X} = {n ∼ X | n ≡ 1 (mod q2)}.

We need that for in�nitely many q holds∑

p∈A(q) 1 > 0.

No reason for the residue class 1 (mod q2) to be special amongthose coprime to q2 and thus de�ning

B(q) = {n ∼ X | gcd(n, q2) = 1}

we would expect that∑p∈A(q)

1 =1

ϕ(q2)

∑p∈B(q)

1 + smaller errorPNT=

X (1 + o(1))

ϕ(q2) logX.

This holds assuming the generalized Riemann hypothesis.



Sums over Primes

Sums over primes di�cult to handle.

But, it is possible to decompose them into easier sums.

We want to split∑

p∈A(q) 1 ≈1

ϕ(q2)

∑p∈B(q) into showing

that for type I sums∑mn∈A(q)m∼M

am =1

ϕ(q2)

∑mn∈B(q)m∼M

am + error

and type II sums∑mn∈A(q)m∼M

ambn =1

ϕ(q2)

∑mn∈B(q)m∼M

ambn + error ,

where M lays in certain ranges.



Type I Sums

For a set E ⊂ N, we write Ed = {m | dm ∈ E}. Then

|A(q)d | = |{m ∼ X/d | dm = aq2 + 1}|= |{a ∼ X 1/2+ε | aq2 ≡ −1 (mod d)}|

=

{X 1/2+ε

d+ O(1) if gcd(d , q2) = 1,

0 else,

=1

ϕ(q2)

q2∑k=1

(k,q2)=1

|{a ∼ X 1/2+ε | aq2 ≡ −k (mod d)}|+ O(1)

=|B(q)d |ϕ(q2)

+ O(1).



Type I Sums Continue

Since

|A(q)d | =|B(q)d |ϕ(q2)

+ O(1),

we have∑mn∈A(q)m∼M

am =∑m∼M

am|A(q)m| =∑m∼M

am

(|B(q)m|ϕ(q2)

+ O(1)

)

=1

ϕ(q2)

∑mn∈B(q)m∼M

am + O(M)

=1

ϕ(q2)

∑mn∈B(q)m∼M

am + O

(X 1−ε/4

Q

)

for M ≤ X 1/2.Kaisa Matomäki Primes of the form aq2 + 1


Type II Sums

For type II sums we need deeper arguments.

A large sieve result by Baier and Zhao implies that

∑q∈Pq2∼Q

∣∣∣∣∣∣∣∣∑

mn∈A(q)m∼M

ambn −1

φ(q2)

∑mn∈B(q)m∼M

ambn

∣∣∣∣∣∣∣∣ = O

(X 1−ε/2

Q1/2

)

for M ∈ [X 3/8+2ε,X 5/8−2ε].

Thus for most prime squares q2 ∼ Q (all but O(Q1/2X−ε/4))∣∣∣∣∣∣∣∣∑

mn∈A(q)m∼M

ambn −1

φ(q2)

∑mn∈B(q)m∼M

ambn

∣∣∣∣∣∣∣∣ = O

(X 1−ε/4

Q

).



Arithmetic information

By previous slides, we have type I information

∑mn∈A(q)m∼M

am =1

ϕ(q2)

∑mn∈B(q)m∼M

am + O

(X 1−ε/4

Q

)

for M ≤ X 1/2 and type II information

∑mn∈A(q)m∼M

ambn =1

ϕ(q2)

∑mn∈B(q)m∼M

ambn + O

(X 1−ε/4

Q

).

for most q with M ∈ [X 3/8+2ε,X 5/8−2ε].

Next task is to split∑

p∈A(q) 1 into type I and type II sums forwhich we use Harman's sieve method.



Sieve Notation

We write

S(E , z) = |{m ∈ E | p | m =⇒ p > z}|.

Then A(q) ∩ P = S(A(q), 3X 1/2).

Buchstab's identity states that for z > w ≥ 1,

S(E , z) = S(E ,w)−∑

w≤p<z

S(Ep, p).

This holds since by de�nitions

S(Ep, p) = |{pn ∈ E | p0 | n =⇒ p0 > p}|= |{n ∈ E | smallest prime factor of n equals p}|.



Decomposing A(q) ∩ P

So we want information about A(q) ∩ P = S(A(q), 3X 1/2).

We use Buchstab's identity with w = X 1/4−4ε in order todecompose

S(A(q), 3X 1/2) = S(A(q),w)−∑

w<p<3X 1/2

S(A(q)p, p)

and further

S(A(q),w) =∑

n∈A(q)

1−∑p1<w

∑p1n∈A(q)

1 +∑

p1<p2<w

∑p1p2n∈A(q)

1− . . .



Finding type I and type II sums

So, we ended up with sums of the form∑p1<···<pj<w

∑p1...pjn∈A(q)

1.

If p1 . . . pj ≤ X 1/2, we have a type I sums with

am =

{1 if m = p1 · · · pj with p1 < · · · < pj < w ,

0 otherwise.

If p1 . . . pj > X 1/2, then for some k ≤ j , the productp1 . . . pk ∈ [X 3/8+2ε,X 5/8−2ε] since pi < w = X 1/4−4ε. Thuswe have a type II sum.

Hence

S(A(q),w) = type I sums + type II sums.



Further decomposing

Since

S(A(q), 3X 1/2) = S(A(q),w)−∑

w<p<3X 1/2

S(A(q)p, p),

we still need to handle ∑w<p<3X 1/2

S(A(q)p, p)

If p > X 3/8+2ε, we have type II sum. Also if summand isreplaced by S(A(q)p,w) we have type I and II sums as before.

Hence we can apply Buchstab's identity and �nd that

S(A(q), 3X 1/2) = type I/II sums +∑

w<p2<p1<X 3/8+2ε

S(A(q)p1p2 , p2).



Final steps of the proof

Now we have

S(A(q), 3X 1/2) ≥ S(A(q), 3X 1/2)−∑

w<p2<p1<X 3/8+2ε

S(A(q)p1p2 , p2)

= type I sums + type II sums

=1

ϕ(q2)

S(B(q), 3X 1/2)−∑

w<p2<p1<X 3/8+2ε

S(B(q)p1p2 , p2)

+ O

(X 1−ε/4

Q

)≥ X

2Q logX> 0

for most q: Using the prime number theorem it is easy to see that

S(B(q), 3X 1/2)−∑

w<p2<p1<X 3/8+2ε

S(B(q)p1p2 , p2) ≥2X

3 logX.



Further thoughts

We have actually proved that for most prime squaresq2 ∼ X 1/2−ε, there are at least X/(φ(q2)2 logX ) primes withp ∼ X and p ≡ 1 (mod q2).

The residue class 1 is not special here, and the result could beshown even uniformly for any k such that gcd(k, q2) = 1.

Getting over barrier 1/2 needs new ideas - type II informationdisappears at this point.



Summary

Di�erent approximations to p = n2 + 1 showed well howinnocent-looking number theoretic problems lead to very deepmethods.

Actually it does not seem likely that any of the approaches can�nally settle the conjecture.

Our approach was to show that there are primes p such thatp − 1 possesses a large square factor.



Thank you

Any questions?


primes of the form aq2+1

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