Supercongruences involving Euler polynomials
Authors: Sun Zhi-Hong 
Published: November 2015

Let

p
>
3

p>3

be a prime, and let

a
a

be a rational

p
p

-adic integer. Let

{

E
n

(
x
)
}

\{E_n(x)\}

denote the Euler polynomials given by
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="StartFraction 2 e Superscript x t Baseline Over e Superscript t Baseline plus 1 EndFraction equals sigma-summation Underscript n equals 0 Overscript normal infinity Endscripts upper E Subscript n Baseline left-parenthesis x right-parenthesis StartFraction t Superscript n Baseline Over n factorial EndFraction">

2

e

x
t

e
t

+
1

=

∑

n
=
0

∞

E
n

(
x
)

t
n

n
!

\frac {2\text {e}^{xt}}{\text {e}^t+1}=\sum _{n=0}^{\infty }E_n(x)\frac {t^n}{n!}

. In this paper we show that
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="StartLayout 1st Row 1st Column Blank 2nd Column a m p semicolon sigma-summation Underscript k equals 0 Overscript p minus 1 Endscripts StartBinomialOrMatrix a Choose k EndBinomialOrMatrix StartBinomialOrMatrix negative 1 minus a Choose k EndBinomialOrMatrix identical-to left-parenthesis negative 1 right-parenthesis Superscript mathematical left-angle a mathematical right-angle Super Subscript p Superscript Baseline plus left-parenthesis a minus mathematical left-angle a mathematical right-angle Subscript p Baseline right-parenthesis left-parenthesis p plus a minus mathematical left-angle a mathematical right-angle Subscript p Baseline right-parenthesis upper E Subscript p minus 3 Baseline left-parenthesis negative a right-parenthesis left-parenthesis mod p cubed right-parenthesis comma 2nd Row 1st Column Blank 2nd Column a m p semicolon sigma-summation Underscript k equals 0 Overscript p minus 1 Endscripts StartBinomialOrMatrix a Choose k EndBinomialOrMatrix left-parenthesis negative 2 right-parenthesis Superscript k Baseline identical-to left-parenthesis negative 1 right-parenthesis Superscript mathematical left-angle a mathematical right-angle Super Subscript p Superscript Baseline minus left-parenthesis a minus mathematical left-angle a mathematical right-angle Subscript p Baseline right-parenthesis upper E Subscript p minus 2 Baseline left-parenthesis negative a right-parenthesis left-parenthesis mod p squared right-parenthesis for a not-identical-to 0 left-parenthesis mod p right-parenthesis comma EndLayout">

a
m
p
;

∑

k
=
0

p
−
1

(

a
k

)

(

−
1
−
a

k

)

≡
(
−
1

)

⟨
a

⟩
p

+
(
a
−
⟨
a

⟩
p

)
(
p
+
a
−
⟨
a

⟩
p

)

E

p
−
3

(
−
a
)

(
mod

p
3

)
,

a
m
p
;

∑

k
=
0

p
−
1

(

a
k

)

(
−
2

)
k

≡
(
−
1

)

⟨
a

⟩
p

−
(
a
−
⟨
a

⟩
p

)

E

p
−
2

(
−
a
)

(
mod

p
2

)

for

a
≢
0

(
mod

p
)
,

\begin{align*} &\sum _{k=0}^{p-1}\binom ak\binom {-1-a}k\equiv (-1)^{\langle a\rangle _p}+ (a-\langle a\rangle _p)(p+a-\langle a\rangle _p)E_{p-3}(-a)\pmod {p^3}, \\&\sum _{k=0}^{p-1}\binom ak(-2)^k\equiv (-1)^{\langle a\rangle _p}-(a-\langle a\rangle _p)E_{p-2}(-a) \pmod {p^2}\quad \text {for}\quad a\not \equiv 0\pmod p, \end{align*}

where

⟨
a

⟩
p

∈
{
0
,
1
,
…
,
p
−
1
}

\langle a\rangle _p\in \{0,1,\ldots ,p-1\}

satisfying

a
≡
⟨
a

⟩
p

(
mod

p
)

a\equiv \langle a\rangle _p\pmod p

. Taking

a
=
−

1
3

,
−

1
4

,
−

1
6

a=-\frac 13,-\frac 14,-\frac 16

in the first congruence, we solve some conjectures of Z. W. Sun. We also establish a congruence for

∑

k
=
0

p
−
1

k

(

a
k

)

(

−
1
−
a

k

)

\sum _{k=0}^{p-1}k\binom ak\binom {-1-a}k

modulo

p
3

p^3

.