1 AMM problem 11651
Solvers: TCDmath problem group, Mathematics, Trinity College, Dublin 2, Ireland.
Show that
⌊n/φ⌋
⌊ φ ⌋
n+1
n
⌊n/φ⌋
(1.1)
⌊
⌋=n−⌊ ⌋+⌊
⌋−⌊
⌋...
φ
φ
φ
φ
√
holds for every nonnegative integer n if and only if φ = (1 + 5)/2.
Answer. The right-hand side may be written as E(n), and clearly E(n) = n − E(⌊n/φ⌋). This
will not converge unless φ > 1. We assume from now on that φ > 1.
If (1.1) holds for all n, then
(1.2)
⌊
⌊n/φ⌋ + 1
n+1
⌋+⌊
⌋=n
φ
φ
On the other hand, if (1.2) holds for all n then
⌊
n+1
⌊n/φ⌋ + 1
⌋=n−⌊
⌋
φ
φ
⌊
n
⌊n/φ⌋ + 1
⌋=⌊ ⌋−⌊
⌊
φ
φ
⌋
⌊n
φ
φ
⌋+1
φ
⌋...
whence the identity (1.1) can be ‘unrolled.’ We discard the original identity in favour of the equivalent
(1.2).
The latter identity implies
n
n
+ 2 = n + O(1)
φ φ
for all n. Dividing by n,
1
1
+ 2 = 1 + O(1/n)
φ φ
√
for arbitrarily large n, which is only possible, since φ > 1, if φ is the golden section (1 + 5)/2.
To deal with the converse, we assume that φ √
is indeed the golden section. We reserve ψ to denote
the other root of x2 − x − 1 = 0, i.e., ψ = (1 − 5)/2 = −1/φ.
Let m = ⌊n/φ⌋. We need to prove
⌊
n+1
m+1
⌋+⌊
⌋=n
φ
φ
for every nonnegative integer n. Write
n
= m + α,
φ
so
n+1
1
=m+α+ .
φ
φ
1
Since n/φ + n/φ2 = n,
m+α+
n
=n
φ2
Case (i): α + 1/φ < 1, in which case m = ⌊(n + 1)/φ⌋, and it is enough to show that
n−m=α+
or
α+
m+1
n
=⌊
⌋.
2
φ
φ
m+1
n
n
<
< α + 2 + 1.
2
φ
φ
φ
Since m/φ < n/φ2 , the second inequality is obvious. The first is equivalent to
n
n/φ + 1 − α
+α<
2
φ
φ
or
α<
1−α
φ
But α < 1 − 1/φ = 1/φ2 and 1 − α > 1 − 1/φ2 = 1/φ, so this is correct.
Case (ii): α + 1/φ > 1. Then ⌊(n + 1)/φ⌋ = m + 1, and we need to show
m+1+⌊
m+1
n
n
⌋ − − 2 = 0.
φ
φ φ
So we need to show that
n/φ − α + 1 n
n
n
−α+1+
− − 2
φ
φ
φ φ
1−α
= (1 − α)(1 + 1/φ) = (1 − α)φ
=1−α+
φ
is between 0 and 1. It is positive, and α > 1−1/φ = 1/φ2 , so 1−α < 1−φ2 = 1/φ, as required.
2