Problem 11333
(American Mathematical Monthly, Vol.114, December 2007) Proposed by R. Tauraso (Italy).
Letα and β be positive irrational numbers. Show that for any positive integer n,
⌊n/α⌋−1
X
k=0
⌈(k + {n/α})α⌉
β
=
⌊n/β⌋−1
X
k=0
⌈(k + {n/β})β⌉
α
where{x} denotes the fractional part of x.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
We first note that
⌈(k + {n/α})α⌉ = ⌈(k + n/α − ⌊n/α⌋)α⌉ = ⌈n − jα⌉ = n + ⌈−jα⌉ = n − ⌊jα⌋
where j = ⌊n/α⌋ − k and therefore
⌊n/α⌋−1
X
k=0
⌈(k + {n/α})α⌉
β
=
⌊n/α⌋
X
j=1
n − ⌊jα⌋
β
.
Since (n − ⌊jα⌋)/β is never an integer then for any positive integer k
[⌊jα⌋ + ⌊kβ⌋ < n] = [⌊kβ⌋ < n − ⌊jα⌋] = [kβ < n − ⌊jα⌋] = [k < (n − ⌊jα⌋)/β] = [k ≤ ⌊(n − ⌊jα⌋)/β⌋]
where [·] is the indicator function. Hence
⌊n/β⌋
X
k=1
[⌊jα⌋ + ⌊kβ⌋ < n] =
⌊n/β⌋
X
k=1
[k ≤ ⌊(n − ⌊jα⌋)/β⌋] = n − ⌊jα⌋
β
.
Finally
⌊n/α⌋−1
X
k=0
⌈(k + {n/α})α⌉
β
=
⌊n/α⌋
X
j=1
n − ⌊jα⌋
β
=
⌊n/α⌋
X
j=1
⌊n/β⌋
X
k=1
[⌊jα⌋ + ⌊kβ⌋ < n]
which is symmetric with respect to α and β and the statement has been proved.
