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PEN I Problems
12
I 12
I 12
Source:
May 25, 2007
floor function
Problem Statement
Let
p
=
4
k
+
1
p=4k+1
p
=
4
k
+
1
be a prime. Show that
∑
i
=
1
k
⌊
i
p
⌋
=
p
2
−
1
12
.
\sum^{k}_{i=1}\left \lfloor \sqrt{ ip }\right \rfloor = \frac{p^{2}-1}{12}.
i
=
1
∑
k
⌊
i
p
⌋
=
12
p
2
−
1
.
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