ON THE GREATEST PRIME FACTORS OF n2 + 1
PING XI
Abstract. We study the work of Deshouillers-Iwaniec on the greatest prime factors of quadratic
polynomials: There are infinitely many n, such that P + (n2 + 1) > n1.202 . The proof started from
the Chebyshev-Hooley method, and succeeded by combining linear sieve and the estimate for
sums of Kloosterman sums from theory of automorphic forms. This note is prepared for the 2016
Analytic Number Theory Seminar in Academia Sinica, Beijing.
1. Introduction
It is widely believed that any fixed irreducible polynomial, which has no fixed prime factors,
can capture infinitely many prime values. The linear case is known due to the classical theorem of Dirichlet on the equidistributions of primes in arithmetic progressions. However, the
current machinery seems far from successful to solve the quadratic case, which was (at least) originally conjectured by Gauß. Before focusing our attention to the case of quadratic polynomials,
we would like to mention that there are certain generalizations of the above conjecture, saying
Schinzel’s hypothesis H, Bateman-Horn conjecture and so on.
In this note, we are interested in the prime values of the special polynomial n2 + 1, which is
irreducible in R (a fortiori in Z) and has no fixed prime factors. Most of the following statements
and arguments, with certain modifications, also apply to other general irreducible quadratic
polynomials. Let us state the conjecture explicitly in the case of n2 + 1.
Conjecture 1. There are infinitely many n, such that n2 + 1 takes prime values. More precisely, we have
X
(1)
Λ(n2 + 1) = SX (1 + o(1))
n6X
as X → +∞, where Λ denotes the von Mangoldt function and
Y
ρ(p) – 1 S=
1–
p–1
p
with %(p) = #{a (mod p) : a2 + 1 ≡ 0 (mod p)}.
The above quantitative form was first predicted by Hardy and Littlewood using their circle
method.
There are at least two ways as approximations to Conjecture 1.
Problem 1. Denote by P + (n) the greatest prime factor of n. Find the value of ϑ as large as possible, such that
there exist infinitely many n with
P + (n2 + 1) > nϑ .
(2)
Date: October 21, 2016.
1
2
PING XI
Problem 2. Denote by Pr a positive integer with at most r distinct prime factors. Find the value of r as small
as possible, such that there exist infinitely many n with
n2 + 1 = Pr .
(3)
Regarding Problem 2, Iwaniec [Iw] proved that one may take r = 2, from which we conclude
trivially that ϑ = 1 works in Problem 1.
Theorem 1 (Deshoulliers-Iwaniec). For ϑ = 1.202, (1) is solvable for infinitely many n.
Hooley [Ho1] is the first who was able to go beyond ϑ = 1 in Problem 1, for which one can
take ϑ = 1.1. In fact, the exponent ϑ = 1 follows trivially from Poisson summation, and Hooley
was able to control the resultant exponential sums successfully by appealing to the theory of
quadratic forms, based on which he arrived at the certain averages of Kloosterman sums. In
his times, Hooley can only use Weil’s bound for individual Kloosterman sums. Thanks to their
systematic study on spectral theory of automorphic forms, Deshouillers and Iwaniec [DI2] were
able to control cancellations among Kloosterman sums, which lead to Theorem 1 amongst other
things.
In the sequel, we put
η = 10–2016 .
Given a positive integer q, define
X
%(h, q) =
e
ha a (mod q)
a2 +1≡0 (mod q)
q
for each h ∈ Z and write %(0, q) = %(q).
2. Chebyshev-Hooley method
Let g be a non-negative smooth function with compact support in [1, 2]. The Fourier transform of g is defined by
Z
bg (λ) =
g(x)e(–λx)dx.
R
From integration by parts, we derive that
bg (λ) (1 + |λ|)–A
(4)
holds for any fixed A > 0 with an implied constant depending on A.
The proof of Theorem 1 starts from the asymptotic evaluation of the weighted sum
X n
(5)
H(X ) =
g
log(n2 + 1).
X
n>1
On one hand, we have
H(X ) = 2 log X (1 + o(1))
X n
g
= 2bg (0)X log X (1 + o(1)).
X
n>1
On the other hand, by virtue of the identity
log m =
X
l|m
Λ(l),
ON THE GREATEST PRIME FACTORS OF n2 + 1
3
we obtain
H(X ) =
X
Λ(l)
X
g
n
n2 +1≡0 (mod l)
l
X
.
The sum over l is in fact restricted to l X 2 due to the support of g. We now split the sum over
l to three segments and define
n
X
X
H1 (X ) =
Λ(l)
g
,
X
l6X 1–η
n2 +1≡0 (mod l)
n
X
X
H2 (X ) =
Λ(l)
g
,
X
n2 +1≡0 (mod l)
X 1–η <l6X ϑ
n
X
X
H3 (X ) =
Λ(l)
g
,
X
2
ϑ
n +1≡0 (mod l)
l>X
so that
H(X ) = H1 (X ) + H2 (X ) + H3 (X ).
Here ϑ > 1 – η is a constant to be chosen later. Note that
n
X
X
H3 (X ) =
log p
g
+ O(X 1–ϑ+ε ).
X
2
ϑ
n +1≡0 (mod p)
p>X
Thus, Theorem 1 follows if one can prove
H(X ) > H1 (X ) + H2 (X )
with ϑ = 1.202. The tasks now turn to evaluate/estimate H1 (X ) and H2 (X ) from above.
We will prove the following two propositions.
Proposition 1. For X large enough, we have
H1 (X ) = (1 – η)bg (0)X log X (1 + o(1)).
Proposition 2. For X large enough, we have
Z ϑ
H2 (X ) 6 2bg (0)
1–η
θdθ
X log X (1 + o(1)).
1 – 12 θ – 7η
With the help of Mathematica 9, one may check that
Z ϑ
θdθ
2>1–η+2
1
1–η 1 – 2 θ – 7η
with ϑ = 1.202. This proves Theorem 1.
3. Proof of Proposition 1
Lemma 1 (Poisson summation formula). Let q be a positive number. Suppose that F , Fb ∈ L1 (R) and
have bounded variations. For a ∈ Z, we have
X
1 X b h ha F (n) =
F
e
.
q
q
q
n≡a (mod q)
h∈Z
4
PING XI
From Poisson summation formula, we obtain
n
X
X
g
=
X
2
X
g
n
a (mod l)
n≡a (mod l)
a2 +1≡0 (mod l)
n +1≡0 (mod l)
(6)
=
X
X X hX bg
%(h, l).
l
l
h∈Z
From (4), we find
hX |h|X –A
bg
1+
(1 + |h|X η )–A
l
l
for l 6 X 1–η and any A > 0. Taking A = 3/η and noting that %(h, l) l ε , we have
n X
1 X
g
= bg (0) + O
,
X
l
lX 2
2
n +1≡0 (mod l)
from which we conclude
H1 (X ) = bg (0)X
X Λ(l)
+ O(1) = (1 – η)bg (0)X log X (1 + o(1))
l
1–η
l6X
as claimed.
4. Proof of Proposition 2
4.1. Initial transformation. From the definition of Λ, we have
H2 (X ) = H2∗ (X ) + O(X η+ε )
with
H2∗ (X ) =
X
log p
X
n2 +1≡0 (mod p)
X 1–η <p6X ϑ
g
n
X
.
Before applying Poisson summation to the n-sum, we would like to transform the sum over primes
to sum over integers by appealing to the linear sieve of Rosser-Iwaniec.
For the sake of later analysis, we would like to attach a smooth weight to the sum over p. The
following lemma is a kind of smooth partition of unity, which works well in our situation (see [Fo,
Lemme 2] for instance).
Lemma 2. For every ∆ > 1, there exists a sequence (W`,∆ )`>0 of smooth functions with support included in
[∆`–1 , ∆`+1 ], such that
X
W`,∆ (t) = 1
`>0
for all t > 1, and
(ν)
W`,∆ (t) ν
for all t > 1 and ν > 0.
∆ ν
t(∆ – 1)
ON THE GREATEST PRIME FACTORS OF n2 + 1
5
In what follows, we take
∆ = 1 + (log X )–B
for some large parameter B > 0.
By virtue of lemma 2, we may write
X X
H2∗ (X ) =
X
W`,∆ (p) log p
g
n
n2 +1≡0 (mod p)
`>0 X 1–η <p6X ϑ
X
.
Define
Ψ(W ) =
X
X
W (p) log p
p
g
n
n2 +1≡0 (mod p)
X
,
so that
X
H2∗ (X ) =
Ψ(W ) + O(X (log X )–B/2 ),
W
where the summation is over all smooth functions W of the shape W`,∆ with
1–η
ϑ
log X + 1 < ` 6
log X – 1.
1+∆
1+∆
The following arguments apply to an individual Ψ(W ) given a smooth function W with support in [L, L(1 + ∆)2 ], where X 1–η L X ϑ .
We now introduce the sieve method. Let λ = (λd ) be an linear upper bound sieve of level D,
so that
n
X
X
X
Ψ(W ) 6
λd
W (l) log l
g
.
X
2
l≡0 (mod d)
d6D
n +1≡0 (mod l)
From (6), it follows that
Ψ(W ) 6 X
X
l≡0 (mod d)
d6D
Put H = LX
X
λd
W (l) log l X hX bg
%(h, l).
l
l
h∈Z
–1+η+ε
. Hence
X
Ψ(W ) 6 X
λd
d6D
X
l≡0 (mod d)
W (l) log l
l
X
06|h|6H
hX bg
%(h, l) + O(1).
l
Define
Ψ1 (W ) = bg (0)X
X
λd
d6D
X
l≡0 (mod d)
W (l) log l
%(l),
l
and
Ψ2 (W ) = X
X
d6D
λd
X
l≡0 (mod d)
W (l) log l X hX bg
%(h, l),
l
l
16h6H
so that
Ψ(W ) 6 Ψ1 (W ) + Ψ2 (W ) + Ψ2 (W ) + O(1).
6
PING XI
4.2. Estimate for Ψ2 (W ). Here comes the most novel part of the proof. Let us recall one
classical result of Gauß on the representation of numbers by binary quadratic forms. We refer
to Disquisitiones Arithmeticae of or Smith’s report [Sm] for a very clear description of this theory in
a form suitable for our purpose.
Lemma 3. Let l > 1. If
a2 + 1 ≡ 0 (mod l)
(7)
is solvable for a (mod l), then l can be represented properly as a sum of two squares
l = r 2 + s2 ,
(8)
(r, s) = 1,
r > 0,
s > 0.
There is a one to one correspondence between the incongruent solutions a (mod l) to (7) and the solutions (r, s) to
(8) given by
a s
s
= – 2 2 .
l
r r(r + s )
By virtue of Lemma 3, we obtain
%(h, l) =
X ε Hs X X hs
hs X X hs e
– 2 2 =
e
+O
,
r r(r + s )
r
r(r 2 + s2 )
2 2
2 2
r +s =l
r,s>0,(r,s)=1
r +s =l
r,s>0,(r,s)=1
from which we get
Ψ2 (W ) = X
XX
d6D h6H
XX
λd
2
Φ(r 2 + s2 , h)e
2
r +s ≡0 (mod d)
r,s>0
(r,s)=1
hs r
+ O(LX –1+2η+ε ),
where
W (n) log n hX bg
.
n
n
Note that (r, d) = 1. Applying Poisson summation to the s-sum, we get
hs ha X
X
X
Φ(r 2 + s2 , h)e
=
e
Φ(r 2 + s2 , h)
r
r
2 2
Φ(n, h) =
a (mod rd)
r 2 +a2 ≡0 (mod d)
(a,r)=1
r +s ≡0 (mod d)
(r,s)=1
=
1 Xb
Φ(k; r, h, d)
rd
k∈Z
s≡a (mod rd)
X
a (mod rd)
a2 +r 2 ≡0 (mod d)
(a,r)=1
e
ha
r
+
ka ,
rd
where
Z
ky Φ(r 2 + y2 , h)e –
dy.
rd
R
From Chinese Remainder Theorem, the exponential sum mod rd can be expressed as
b r, h, d) =
Φ(k;
S(hd, k; r)%(k, d),
where S(h, k; r) is the usual Kloosterman sum. Thus,
X X λd X 1 X
b r, h, d)S(hd, k; r)%(k, d) + O(LX –1+2η+ε ),
Ψ2 (W ) = X
Φ(k;
d
r
d6D h6H
r>1
k∈Z
ON THE GREATEST PRIME FACTORS OF n2 + 1
7
From integration by parts, we have
√ –A
b r, h, d) √1 1 + |k| L
Φ(k;
rd
L
for any A > 0. The contribution from k = 0 is thus
X
1+ε X X
X X λd X 1
(h, r)
b r, h)S(hd, 0; r)%(d) X√
Φ(0;
d
r
L h6H √ r
d6D h6H
r>1
r L
√ η+ε
LX .
Hence
Ψ2 (W ) = Ψ∗2 (W ) + O(LX –1+3η +
√
LX 3η ).
where
Ψ∗2 (W ) = X
X X X X λd %(k, d)Φ(k;
b r, h, d)S(hd, k; r)
rd
d6D h6H r>1 k6=0
.
We would like to prove that
1
3
1
1
1
1
1
Ψ∗2 (W ) X ε (DX 2 + 2 η + D 2 X 2 +2η + D 2 L 4 X 2 +3η ).
(9)
By dyadic device and smooth partition of unity, it suffices to consider
X
X X X X r λd %(k, d)Φ(k;
b r, h, d)S(hd, k; r)
ξ
,
R
rd
d∼D h∼H r>1 k6=0
√
where R L and ξ is a smooth function with compact support in [1, 2]. Put K = DRL–1/2 X η ,
then for h ∼ H, d ∼ D, r ∼ R and |k| > K , we find
√ 1
K
L –10/η
b r, h, d) √ 1 +
Φ(k;
X –10 ,
RD
L
from which we can truncate the sum over k to 0 < |k| 6 K , and thus consider
Ψ(D, H, K , R) = X
X X X X r λd %(k, d)Φ(k;
b r, h, d)S(hd, k; r)
ξ
.
R
rd
d∼D h∼H k∼K r>1
If we follow the approach of Hooley, it suffices to apply Weil’s bound to S(hd, k; r) for each fixed
h, d, k, r, and then sum over each variable trivially. Thanks to the work of Deshouillers and
Iwaniec, we may control the bias among Kloosterman sums, for which we appeal to the following
estimate of quartilinear sums (see [DI1, Theorem 10]).
Lemma 4. Let C, M, N , S, Q be positive numbers and let g(c, m, n, l) be a function of C ∞ class with compact
support in [C, 2C] × (0, +∞)3 such that
∂ ν1 +ν2 +ν3 +ν4
g(c, m, n, s) c–ν1 m–ν2 n–ν3 s–ν4 Q
∂cν1 ∂mν2 ∂nν3 ∂sν4
for any νi > 0, 1 6 i 6 4, the implied constant in depends at most on ν1 , ν2 , ν3 , ν4 . For any coefficient
(αn,l ), we have
8
PING XI
X XX
αn,s
m∼M n∼N s∼S
X
1
g(c, m, n, s)S(±ms, n; c) (CMNS)ε kαkM 2 Q · ∆(C, M, N , S),
(c,s)=1
where
kαk =
XX
n
|αn,s |2
1/2
s
and
∆(C, M, N , S) =
√
r
MNS + C
p
S(M + N + S) + C
3
1
MNS
+ C 2 (S(N + S)) 4 .
+ MN
2
C 2S
Note that
r Φ(k;
b r, h, d)
∂ ν1 +ν2 +ν3 +ν4
1
√
ξ
r –ν1 h–ν2 k–ν3 d –ν4
ν
ν
ν
ν
1
2
3
4
∂r ∂h ∂k ∂d
R
rd
DR L
for any νi > 0, 1 6 i 6 4. We now apply Lemma 4 with
(c, m, n, s) → (r, h, k, d),
(C, M, N , S) → (R, H, K , D),
r Φ(k;
b r, h, d)
g(c, m, n, s) → ξ
,
R
rd
αn,s → λd %(k, d)
1
√ ,
Q =
DR L
getting
√
p
X 1+ε KD n√
√
Ψ(D, H, K , R) HKD + R D(H + K + D)
DR L
r
o
3
1
HKD
2
+R
+ R 2 (D(K + D)) 4
2
R D + HK
1
3
1
1
1
1
1
X ε (DX 2 + 2 η + D 2 X 2 +2η + D 2 L 4 X 2 +3η ),
from which we may conclude (9).
Hence we obtain
1
3
1
1
1
1
1
Ψ2 (W ) X ε (DX 2 + 2 η + D 2 X 2 +2η + D 2 L 4 X 2 +3η ) + LX –1+3η +
from which we find
2
Ψ2 (W ) X 1–η ,
provided that
(10)
1
1
D 6 min{X 2 –2η , X 1–7η L– 2 },
in which case we have
2
Ψ(W ) 6 Ψ1 (W ) + O(X 1–η ).
√
LX 3η ,
ON THE GREATEST PRIME FACTORS OF n2 + 1
9
4.3. Evaluation of Ψ1 (W ). The evaluation of Ψ1 (W ) is based on the asymptotic analysis of
the congruence sum
X W (l) log l
Ψ1 (W , d) =
%(l).
l
l≡0 (mod d)
By Mellin inversion, we may write
Z
W (l) log l
1
=
l
2πi
T (s)l –s ds
(σ)
for σ > 1, where
Z
T (s) =
W (ξ)
R
log ξ s–1
ξ dξ (|s| + 1)–2 Lσ–1 log X .
ξ
It then follows that
Ψ1 (W , d) =
(11)
1
2πi
Z
T (s)Kd (s)ds,
(σ)
where
Kd (s) =
X
n≡0 (mod d)
%(n)
, <s > 1.
ns
For each fixed squarefree number d, we have
1 X %(nd) %(d) ζ(s)L(s, χ4 ) Y 1 –1
Kd (s) = s
=
1
+
,
d
ns
ds
ζ(2s)
ps
p|d
n>1
where χ4 is the primitive character mod 4. Note that Kd (s) admits a meromorphic continuation
to the right half plane <s > 1/2 with a simple pole at s = 1. Moving the integral line in (11) to
<s = 1/2, we find
Z
1
Ψ1 (W , d) = Res T (s)Kd (s) +
T (s)Kd (s)ds
s=1
2πi ( 12 )
Z
1
%(d) L(1, χ4 ) Y 1 –1
log ξ
=
1+
W (ξ)
dξ + O(τ (d)2 (dL)– 2 log X ),
d
ζ(2)
p
ξ
R
p|d
from which we derive that
X
Ψ1 (W ) = bg (0)X
λd Ψ1 (W , d)
d6D
= bg (0)X
L(1, χ4 )
ζ(2)
Z
W (ξ)
R
1
log ξ X %(d) Y 1 –1
dξ
λd
1+
+ O((D/L) 2 X 1+ε )
ξ
d
p
d6D
p|d
subject to the constraints in (10). From the construction of λd in Rosser-Iwaniec sieve, we have
1 Y X %(d) Y 1 –1
%(p) 1 –1 λd
1+
= 2eγ 1 + O
1–
1+
d
p
log D
p
p
p|d
d6D
p6D
1 Y ζ(2) 1
= 2eγ
1+O
1–
,
L(1, χ4 )
log D
p
p6D
10
PING XI
from which and Mertens’ theorem, we obtain
Z
1
X
log ξ
Ψ1 (W ) = 2bg (0)
W (ξ)
dξ(1 + o(1)) + O((D/L) 2 X 1+ε ).
log D R
ξ
4.4. Conclusion: Upper bound for H2 (X ). Collecting the above evaluations, we have
Z
1
X
log ξ
Ψ(W ) 6 2bg (0)
W (ξ)
dξ(1 + o(1)) + O((D/L) 2 X 1+ε )
log D R
ξ
subject to the constraints in (10). For L X 1–η , we may take
1
D = X 1–7η L– 2 .
Summing over W , it follows that
Xϑ
Z
log t
1 dt(1 + o(1))
t log(X 1–7η t – 2 )
Z ϑ
θdθ
= 2bg (0)X log X
(1 + o(1)).
1
1–η 1 – 2 θ – 7η
H2 (X ) 6 2bg (0)X
X 1–η
5. Related works
Let ω(n) be a positive function that tends to infinity arbitrarily slowly. Let f + (n) > 0 be an
admissible function such that
P + (f (n))
lim inf
> 0.
n→+∞ n · f + (n)
The following table presents the historical choices of f + for given irreducible polynomials f with
deg(f ) > 2.
f + (n)
Authors
Years
2
ω(n)
Chebyshev/Markov
1895
2
0.1
Hooley
1967
Deshouillers-Iwaniec
1982
Hooley
1978
Heath-Brown
2001
Irving
2015
Dartyge
2015
Irreducible f
n +1
n +1
n
n2 + 1
3
n +2
3
n +2
n3 + 2
n4 – n2 + 1
n0.202
n
1
30
(conditionally)
n
10–303
n10
n10
–52
–26531
deg(f ) = 4, Galois group
' Z/2Z × Z/2Z
nδ , ∃δ > 0
de la Bretèche
2015
deg(f ) > 2
(log n)θ , ∀θ < 1
Nagell
1921
deg(f ) > 2
exp(c log2 n log3 n), ∃c > 0
Erdős
1952
Erdős & Schinzel
1952
Tenenbaum
1990
deg(f ) > 2
deg(f ) > 2
exp exp(c(log2 )
2/3
), ∃c > 0
0.6137
exp((log n)
)
ON THE GREATEST PRIME FACTORS OF n2 + 1
11
In the case of n2 + 1, Dartyge [Da1] proved that there exist infinitely many n, whose prime
factors are at least n1/12.2 , such that P + (n2 + 1) > n1+δ with certain δ > 0. In a recent joint work
with J. Wu, we are able to reduce 12.2 to 11.2, which yields there exist infinitely many P11 such
2
1+δ
that P + (P11
+ 1) > P11
. On the other hand, we also proved that P + (p2 + 1) > p0.847 infinitely
often.
Appendix A. Construction of smooth functions
We now construct explicitly a smooth function φ ∈ C ∞ (R), which is identically 1 in [1, 2] and
vanishes in (–∞, 12 ] ∪ [ 52 , +∞). To do so, we first introduce an auxiliary function
( 1
e– x ,
x > 0,
ξ(x) =
0,
x 6 0.
One can check ξ ∈ C ∞ (R).
Put
φ(x) =
ξ((x – 12 )( 52 – x))
ξ((x –
1 5
2 )( 2
We may find φ ∈ C ∞ (R) and


φ(x) = 1,
φ(x) = 0,


φ(x) ∈ [0, 1],
– x)) + ξ((x – 1)(x – 2))
.
x ∈ [1, 2],
x ∈ (–∞, 12 ] ∪ [ 52 , +∞),
otherwise.
On the other hand, we can also construct another smooth function ψ defined in [0, 1], such
that
ψ(0) = 1,
ψ(1) = 0,
ψ (ν) (0+ ) = ψ (ν) (1– ) = 0 (ν = 1, 2, · · · ).
This can be realized by taking
ψ(x) = e · ξ(1 – x), x ∈ [0, 1].
The function ψ is useful in proving Lemma 2, for which we can take (d’après Fouvry [Fo])

t – ∆`–1 

, ∆`–1 < t 6 ∆` ,
1 – ψ
` – ∆`–1

∆


t – ∆` W`,∆ (t) =
ψ
,
∆` 6 t < ∆`+1 ,


`+1 – ∆`
∆




0,
otherwise.
References
[dlB]
[Da1]
[Da2]
[DI1]
[DI2]
[Er]
[ES]
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Q
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12
PING XI
[Fo]
[HB]
[Ho1]
[Ho2]
[Iw]
[Ma]
[Na]
[Sm]
[Te]
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A. A. Markov, Uber die Primteiler des Zahlen von der Form 1 + 4x2 , Bull. Acad. Sci. St. Petersburg 3, (1895) 55–59.
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Department of Mathematics, Xi’an Jiaotong University, Xi’an 710049, P.R. China
E-mail address: [email protected]