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2
D 2
D 2
Source:
May 25, 2007
modular arithmetic
Putnam
group theory
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Problem Statement
Suppose that
p
p
p
is an odd prime. Prove that
∑
j
=
0
p
(
p
j
)
(
p
+
j
j
)
≡
2
p
+
1
(
m
o
d
p
2
)
.
\sum_{j=0}^{p}\binom{p}{j}\binom{p+j}{j}\equiv 2^{p}+1\pmod{p^{2}}.
j
=
0
∑
p
(
j
p
)
(
j
p
+
j
)
≡
2
p
+
1
(
mod
p
2
)
.
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