Sum of one prime and two squares of primes in short intervals

partial
integration
we
obtain
A
(
α
) =
1
2
π
(4
−
w
)
u
4
−
w
u
2
−
N
−
2
+ 2
u
5
−
w
(
u
2
−
N
−
2
)
2
d
u
and
B
(
α
) =
1
4
−
w
u
4
−
w
(
u
2
−
N
−
2
)
3
/
2
+ 3
u
5
−
w
(
u
2
−
N
−
2
)
5
/
2
d
u
.
Hence
by
 
(12)
we
have
for
α
∈
[
η/
2
,
4
η
] that
A
(
α
)


u
2
−
1
/
1 +
|
γ
1
−
γ
2
|

α
2
−
1
/
1 +
|
γ
1
−
γ
2
|
and
B
(
α
)


α
1
−
1
/
1 +
|
γ
1
−
γ
2
|
,
(15)
where
A
(
α
) and
B
(
α
) satisfy
A
(
η/
4)
=
B
(
η/
4)
= 0,
and
from
 
(14)–(15)
we
obtain
G
1
(
α
)


α
2
−
1
/
1 +
|
γ
1
−
γ
2
|
2
(16)
for
α
∈
[
η/
2
,
4
η
].
From
 
(11)
and
 
(16)
we
get
J


η
1
−
1
/
1 + (
γ
1
+
γ
2
N η
)
2
1 +
|
γ
1
−
γ
2
|
2
exp
−
c
γ
1
+
γ
2
N η
,
hence
from
 
(9)
and
Stirling’s
formula
we
have
2
η
η
γ>
0
z
−
ρ/
Γ
ρ

2
d
α


η
1
−
1
/
γ
1
>
0
γ
2
>
0
|
γ
1
|
(1
−

)
/
(2

)
|
γ
2
|
(1
−

)
/
(2

)
1 + (
γ
1
+
γ
2
N η
)
2
1 +
|
γ
1
−
γ
2
|
2
exp
−
c
γ
1
+
γ
2
N η
.
(17)

---

52
A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
But
sorting
imaginary
parts
it
is
clear
that
|
γ
1
|
(1
−

)
/
(2

)
|
γ
2
|
(1
−

)
/
(2

)
1 +
γ
1
+
γ
2
N η
2
exp
−
c
γ
1
+
γ
2
N η


|
γ
1
|
(1
−

)
/
exp
−
c
2
γ
1
N η
,
hence
 
(17)
becomes


η
1
−
1
/
γ
1
>
0
|
γ
1
|
(1
−

)
/
exp
−
c
2
γ
1
N η
γ
2
>
0
1
1 +
|
γ
1
−
γ
2
|
2


N
1
/
ηL
2
,
(18)
since
the
number
of
zeros
ρ
2
= 1
/
2
+
iγ
2
with
n
≤ |
γ
1
−
γ
2
|
≤
n
+ 1 is
O
(log(
n
+
|
γ
1
|
)).
From
 
(6)–(8)
and
 
(18)
we
get
ξ
−
ξ
γ>
0
z
−
ρ/
Γ
ρ

2
d
α


N
1
/
ξL
2
,
(19)
and
 
Lemma 3
follows
from
 
(19)
.
2
We
will
also
need
the
following
result
based
on
the
Laplace
formula
for
the
Gamma
function,
see
 
[10]
.
In
fact
we
will
need
it
just
for
μ
= 2 but,
as
before,
we
write
the
more
general
case.
Lemma
4.
Let
N
be
a
positive
integer,
z
= 1
/N
−
2
πiα
,
and
μ
>
0
.
Then
1
/
2
−
1
/
2
z
−
μ
e
(
−
nα
) d
α
=
e
−
n/N
n
μ
−
1
Γ(
μ
)
+
O
μ
1
n
,
uniformly
for
n
≥
1
.
Proof.
We
start
with
the
identity
1
2
π
R
e
iDu
(
a
+
iu
)
s
d
u
=
D
s
−
1
e
−
aD
Γ(
s
)
,
which
is
valid
for
σ
=

(
s
)
>
0 and
a
∈
C
with

(
a
)
>
0 and
D >
0.
Letting
u
=
−
2
πα
and
taking
s
=
μ
,
D
=
n
and
a
=
N
−
1
we
find
R
e
(
−
nα
)
(
N
−
1
−
2
πiα
)
μ
d
α
=
R
z
−
μ
e
(
−
nα
) d
α
=
n
μ
−
1
e
−
n/N
Γ(
μ
)
.

---

A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
53
For
0
< X < Y
let
I
(
X, Y
) =
Y
X
e
iDu
(
a
+
iu
)
μ
d
u.
An
integration
by
parts
yields
I
(
X, Y
) =
1
iD
e
iDu
(
a
+
iu
)
μ
Y
X
+
μ
D
Y
X
e
iDu
(
a
+
iu
)
μ
+1
d
u.
Since
a
>
0,
the
first
summand
is

μ
D
−
1
X
−
μ
,
uniformly.
The
second
summand
is

μ
D
Y
X
d
u
u
μ
+1

μ
D
−
1
X
−
μ
.
The
result
follows.
2
We
remark
that
if
μ
∈
N
,
μ
≥
2,
 
Lemma 4
can
be
proved
in
an
easier
way
using
the
Residue
Theorem
(see,
e.g.
,
Languasco
 
[4]
or
Languasco
and
Zaccagnini
 
[6]
).
In
the
following
we
will
also
need
a
fourth-power
average
of
S
2
(
α
).
Lemma
5.
We
have
1
/
2
−
1
/
2
|
S
2
(
α
)
|
4
d
α

N L
2
.
Proof.
Let
P
2
=
{
p
j
:
j
≥
2
,
p
prime
}
and
r
0
(
m
) be
the
number
of
representations
of
m
as
a
sum
of
two
squares.
We
have
1
/
2
−
1
/
2
|
S
2
(
α
)
|
4
d
α
=
n
1
,n
2
,n
3
,n
4
≥
2
Λ(
n
1
)Λ(
n
2
)Λ(
n
3
)Λ(
n
4
)
e
−
(
n
2
1
+
n
2
2
+
n
2
3
+
n
2
4
)
/N
1
/
2
−
1
/
2
e
((
n
2
1
+
n
2
2
−
n
2
3
−
n
2
4
)
α
) d
α

p
1
,p
2
≥
2
log
p
1
log
p
2
e
−
2(
p
2
1
+
p
2
2
)
/N
p
3
,p
4
≥
2
p
2
1
+
p
2
2
=
p
2
3
+
p
2
4
log
p
3
log
p
4
+
n
1
,n
2
≥
2
n
1
∈P
2
Λ(
n
1
)Λ(
n
2
)
e
−
2(
n
2
1
+
n
2
2
)
/N
n
3
,n
4
≥
2
n
2
1
+
n
2
2
=
n
2
3
+
n
2
4
Λ(
n
3
)Λ(
n
4
)
= Σ
1
+ Σ
2
,
(20)

---

54
A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
say.
In
Σ
1
we
separately
consider
the
contribution
of
the
cases
p
1
p
2
=
p
3
p
4
and
p
1
p
2

=
p
3
p
4
;
hence
Σ
1

S
1
+
S
2
where,
by
partial
summation
and
the
Prime
Number
Theorem,
we
have
S
1
= 2
p
≥
2
(log
p
)
2
e
−
2
p
2
/N
2

1 +
+
∞
2
u
2
N
(log
u
)
e
−
2
u
2
/N
d
u
2

N L
2
,
and,
by
a
dissection
argument
and
Satz
3
on
p. 94
of
Rieger
 
[16]
,
we
also
obtain
S
2

y
≥
1
1
≤
x
≤
y
y
2
x
2
e
−
2
2
y
+1
/N
e
−
2
2
x
+1
/N
2
y
≤
p
1
<
2
y
+1
2
x
≤
p
2
<
2
x
+1
p
3
,p
4
≥
2
p
2
1
+
p
2
2
=
p
2
3
+
p
2
4
p
1
p
2
 
=
p
3
p
4
1

y
≥
1
y
4
e
−
2
2
y
+1
/N
p
1
,p
2
<
2
y
+1
p
3
,p
4
≥
2
p
2
1
+
p
2
2
=
p
2
3
+
p
2
4
p
1
p
2
 
=
p
3
p
4
1
1
≤
x
≤
y
e
−
2
2
x
+1
/N

y
≥
1
y
2
y
e
−
2
2
y
+1
/N
y
1
e
−
2
t
/N
d
t

y
≥
1
y
2
2
y
e
−
2
2
y
+1
/N

+
∞
2
(log
u
)
2
e
−
u/N
d
u

N L
2
.
Summing
up
Σ
1

N L
2
.
(21)
Recalling
that
r
0
(
m
)

m
ε
,
it
is
also
easy
to
see
that
Σ
2

n
1
,n
2
≥
2
n
1
∈P
2
Λ(
n
1
)Λ(
n
2
)(log(
n
2
1
+
n
2
2
))
2
r
0
(
n
2
1
+
n
2
2
)
e
−
2(
n
2
1
+
n
2
2
)
/N

n
1
,n
2
≥
2
n
1
∈P
2
n
ε
1
n
ε
2
e
−
2(
n
2
1
+
n
2
2
)
/N

j
≥
2
p
≥
2
p
jε
e
−
2
p
2
j
/N
n
≥
2
n
ε
e
−
2
n
2
/N

j
≥
2
e
−
2
2
j
/N
+
∞
2
t
jε
e
−
t
2
j
/N
d
t
N
1
/
2+
ε
+
∞
0
u
ε
−
1
/
2
e
−
u
d
u

N
1
/
2+2
ε
j
≥
2
N
1
/
(2
j
)
e
−
2
2
j
/N

N
1
/
2+2
ε
N
1
/
4
log
N
+
j>
(1
/
2) log
N
e
−
2
2
j
/N

N
3
/
4+3
ε
.
(22)
Combining
 
(20)–(22)
,
 
Lemma 5
follows.
2

---

A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
55
3.
Proof
of
 
Theorem 1
Let
H
≥
2,
H
=
o
(
N
) be
an
integer.
We
recall
that
we
set
L
= log
N
for
brevity.
Recalling
 
(1)
and
letting
R
(
n
) =
a
+
b
2
+
c
2
=
n
Λ(
a
)Λ(
b
)Λ(
c
)
,
we
have
(see,
e.g.
,
p. 14
of
 
[20]
)
that
r
(
n
) =
R
(
n
) +
O
n
3
/
4
(log
n
)
3
.
(23)
Then,
for
every
n
≤
2
N
,
we
can
write
r
(
n
) =
R
(
n
) +
O
n
3
/
4
(log
n
)
3
=
e
n/N
1
/
2
−
1
/
2
S
1
(
α
)
S
2
(
α
)
2
e
(
−
nα
) d
α
+
O
n
3
/
4
(log
n
)
3
.
From
this
equation,
the
Cauchy–Schwarz
inequality,
Lemma 5
and
the
Prime
Number
Theorem,
for
every
n
≤
2
N
we
also
have
r
(
n
)

1
/
2
−
1
/
2
|
S
1
(
α
)
||
S
2
(
α
)
|
2
d
α
+
N
3
/
4
L
3

1
/
2
−
1
/
2
|
S
1
(
α
)
|
2
d
α
1
/
2
1
/
2
−
1
/
2
|
S
2
(
α
)
|
4
d
α
1
/
2
+
N
3
/
4
L
3

N L
3
/
2
.
(24)
We
need
now
to
choose
a
suitable
weighted
average
of
r
(
n
).
We
further
set
U
(
α, H
) =
H
m
=1
e
(
mα
)
and,
moreover,
we
also
have
the
usual
numerically
explicit
inequality
|
U
(
α, H
)
| ≤
min
H
;
1
|
α
|
.
(25)
With
these
definitions
and
 
(23)
,
we
may
write
S
(
N, H
) :=
N
+
H
n
=
N
+1
e
−
n/N
r
(
n
) =
1
/
2
−
1
/
2
S
1
(
α
)
S
2
(
α
)
2
U
(
−
α, H
)
e
(
−
N α
) d
α
+
O
HN
3
/
4
L
3
.

---

56
A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
Using
 
Lemma 2
with

= 1
,
2 and
recalling
that
Γ(1)
= 1,
Γ(1
/
2)
=
π
1
/
2
,
we
can
write
S
(
N, H
) =
1
/
2
−
1
/
2
π
4
z
2
U
(
−
α, H
)
e
(
−
N α
) d
α
+
1
/
2
−
1
/
2
1
z
S
2
(
α
)
2
−
π
4
z
U
(
−
α, H
)
e
(
−
N α
) d
α
+
1
/
2
−
1
/
2
S
1
(
α
)
−
1
z
S
2
(
α
)
2
U
(
−
α, H
)
e
(
−
N α
) d
α
+
O
HN
3
/
4
L
3
=
I
1
+
I
2
+
I
3
+
O
HN
3
/
4
L
3
,
(26)
say.
From
now
on,
we
denote
E

(
α
) :=
S

(
α
)
−
Γ(1
/
)
z
1
/
.
Using
 
Lemma 4
we
immediately
get
I
1
=
π
4
N
+
H
n
=
N
+1
ne
−
n/N
+
O
H
N
=
πHN
4
e
+
O
H
2
.
(27)
Now
we
estimate
I
2
.
Using
the
identity
f
2
−
g
2
= 2
f
(
f
−
g
)
−
(
f
−
g
)
2
we
obtain
I
2

1
/
2
−
1
/
2
|
E
2
(
α
)
|
|
U
(
α, H
)
|
|
z
|
3
/
2
d
α
+
1
/
2
−
1
/
2
|
E
2
(
α
)
|
2
|
U
(
α, H
)
|
|
z
|
d
α
=
J
1
+
J
2
,
(28)
say.
Using
 
(3)
,
 
(25)
,
 
Lemma 3
and
a
partial
integration
argument
we
obtain
J
2

HN
1
/N
−
1
/N
|
E
2
(
α
)
|
2
d
α
+
H
1
/H
1
/N
|
E
2
(
α
)
|
2
d
α
α
+
1
/
2
1
/H
|
E
2
(
α
)
|
2
d
α
α
2

HN
1
/
2
L
2
+
HN
1
/
2
L
2
1 +
1
/H
1
/N
d
ξ
ξ
+
N
1
/
2
L
2
H
+
1
/
2
1
/H
d
ξ
ξ
2

HN
1
/
2
L
3
.
(29)
Using
the
Cauchy–Schwarz
inequality
and
arguing
as
for
J
2
we
get
J
1

HN
3
/
2
1
/N
−
1
/N
d
α
1
/
2
1
/N
−
1
/N
|
E
2
(
α
)
|
2
d
α
1
/
2
+
H
1
/H
1
/N
d
α
α
2
1
/
2
1
/H
1
/N
|
E
2
(
α
)
|
2
d
α
α
1
/
2
+
1
/
2
1
/H
d
α
α
4
1
/
2
1
/
2
1
/H
|
E
2
(
α
)
|
2
d
α
α
1
/
2

---

A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
57

HN
3
/
4
L
+
HN
3
/
4
L
1 +
1
/H
1
/N
d
ξ
ξ
1
/
2
+
H
3
/
2
N
1
/
4
L
1 +
1
/
2
1
/H
d
ξ
ξ
1
/
2

HN
3
/
4
L
3
/
2
+
H
3
/
2
N
1
/
4
L
3
/
2

HN
3
/
4
L
3
/
2
.
(30)
Combining
 
(28)–(30)
we
finally
obtain
I
2

HN
3
/
4
L
3
/
2
.
(31)
Now
we
estimate
I
3
.
By
the
Cauchy–Schwarz
inequality,
 
(25)
and
 
Lemma 5
we
obtain
I
3

1
/
2
−
1
/
2
|
S
2
(
α
)
|
4
d
α
1
/
2
1
/
2
−
1
/
2
|
E
1
(
α
)
|
2
|
U
(
α, H
)
|
2
d
α
1
/
2

N
1
/
2
L
H
2
1
/H
−
1
/H
|
E
1
(
α
)
|
2
d
α
+
1
/
2
1
/H
|
E
1
(
α
)
|
2
d
α
α
2
1
/
2

H
1
/
2
N L
2
,
(32)
where
in
the
last
step
we
used
 
Lemma 3
and
a
partial
integration
argument.
By
 
(26)–(27)
,
 
(31)
and
 
(32)
,
we
can
finally
write
S
(
N, H
) =
π
4
e
HN
+
O
H
1
/
2
N L
2
+
HN
3
/
4
L
3
+
H
2
.
Theorem 1
follows
since
the
exponential
weight
e
−
n/N
=
e
−
1
+
O
(
H/N
) for
n
∈
[
N
+ 1
,
N
+
H
] and
hence
by
(24)
it
can
be
removed
at
the
cost
of
inserting
an
extra
factor
O
H
2
L
3
/
2
in
the
error
term.
The
corollary
about
the
existence
in
short
intervals
follows
by
remarking
that
S
(
N,
H
)
>
0 if
L
4

H
=
o
N L
−
3
/
2
.
2
Acknowledgments
This
research
was
partially
supported
by
the
grant
PRIN2010-11
Arithmetic
Algebraic
Geometry
and
Number
Theory
.
We
wish
to
thank
the
referee
for
pointing
out
some
inaccuracies.
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---

Document Outline

• Sum of one prime and two squares of primes in short intervals
  • 1 Introduction
  • 2 Notation and lemmas
  • 3 Proof of Theorem 1
  • Acknowledgments
  • References