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4
I 4
I 4
Source:
May 25, 2007
floor function
Problem Statement
Show that for all positive integers
n
n
n
,
⌊
n
+
n
+
1
⌋
=
⌊
4
n
+
1
⌋
=
⌊
4
n
+
2
⌋
=
⌊
4
n
+
3
⌋
.
\lfloor \sqrt{n}+\sqrt{n+1}\rfloor =\lfloor \sqrt{4n+1}\rfloor =\lfloor \sqrt{4n+2}\rfloor =\lfloor \sqrt{4n+3}\rfloor.
⌊
n
+
n
+
1
⌋
=
⌊
4
n
+
1
⌋
=
⌊
4
n
+
2
⌋
=
⌊
4
n
+
3
⌋
.
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