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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
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
Outline
1 Approaches to primes of the form n2 + 1
2 Primes of the form aq2 + 1
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes 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
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 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).
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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!
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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).
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 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
Approaches to primes of the form n2 + 1Primes 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
).
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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}|.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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− . . .
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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).
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
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.
Kaisa Matomäki Primes of the form aq2 + 1
Approaches to primes of the form n2 + 1Primes of the form aq2 + 1
Thank you
Any questions?
Kaisa Matomäki Primes of the form aq2 + 1