ISL 2002 A5
by Wolstenholme, Aug 2, 2014, 9:46 PM
Let 
be a positive integer that is not a perfect cube. Define real numbers 
by
![\[a=\root3\of n\kern1.5pt,\qquad b={1\over a-[a]}\kern1pt,\qquad c={1\over b-[b]}\kern1.5pt,\]](//latex.artofproblemsolving.com/d/6/7/d67e71071e91739bbc317c28727cecea1ccddb71.png)
where![$[x]$](//latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
denotes the integer part of 
. Prove that there are infinitely many such integers with the property that there exist integers 
, not all zero, such that 
.
Proof:
Since
is the "nastiest" of the variables we want to artificially make it nice. A good method to do so would be to force ![$ [b] = 1 $](//latex.artofproblemsolving.com/2/e/3/2e39977e77ef42e6b338ed21e1c0a85b8bc65fd3.png)
. Let 
denote the fractional part of a real number. Since 
we want 
so for now assume this to be the case.
Now, let
and ![$ m = [a] $](//latex.artofproblemsolving.com/8/0/4/80414d254cfdaab1ddf17f147eee4a66f6c72eb4.png)
. We have that 
and 
and ![$ c = \frac{\alpha}{1 - [b]\alpha} = \frac{\alpha}{1 - \alpha} $](//latex.artofproblemsolving.com/b/c/d/bcd64960c76eeb2c3572ef76e32bcb439846a3ca.png)
. So we want 
. Multiplying both sides by 
we get that 
.
Since we want to figure out what
should be in terms of 
we should choose 
so that we can find a 
in our equation. So we want to solve the simultaneous set of equations in terms of :
![\[ -r = -1 \]](//latex.artofproblemsolving.com/8/a/3/8a35c878f899d4e1bdf5c24b86d5b10194c23aee.png)
![\[ r - rm + t = -3m \]](//latex.artofproblemsolving.com/3/3/4/3343f56638b34ad65c6ea2997032e79c454e66cc.png)
![\[ mr - s = -3m^2 \]](//latex.artofproblemsolving.com/c/1/e/c1ec8109ab4bcfcd1c5569c21e17ba069bd63f49.png)
Therefore we find that
Plugging these in our equation becomes 
. Now we need to check that 
like we assumed but this is clear since 
for all positive integers .
Therefore for every of the form 
for some positive integer , letting 
yields a desired triple 
. Therefore we have found an infinite number of solutions and so we are done.
be a positive integer that is not a perfect cube. Define real numbers 
by![\[a=\root3\of n\kern1.5pt,\qquad b={1\over a-[a]}\kern1pt,\qquad c={1\over b-[b]}\kern1.5pt,\]](http://latex.artofproblemsolving.com/d/6/7/d67e71071e91739bbc317c28727cecea1ccddb71.png)
where
![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
denotes the integer part of 
. Prove that there are infinitely many such integers with the property that there exist integers 
, not all zero, such that 
.Proof:
Since
is the "nastiest" of the variables we want to artificially make it nice. A good method to do so would be to force ![$ [b] = 1 $](http://latex.artofproblemsolving.com/2/e/3/2e39977e77ef42e6b338ed21e1c0a85b8bc65fd3.png)
. Let 
denote the fractional part of a real number. Since 
we want 
so for now assume this to be the case.Now, let
and ![$ m = [a] $](http://latex.artofproblemsolving.com/8/0/4/80414d254cfdaab1ddf17f147eee4a66f6c72eb4.png)
. We have that 
and 
and ![$ c = \frac{\alpha}{1 - [b]\alpha} = \frac{\alpha}{1 - \alpha} $](http://latex.artofproblemsolving.com/b/c/d/bcd64960c76eeb2c3572ef76e32bcb439846a3ca.png)
. So we want 
. Multiplying both sides by 
we get that 
.Since we want to figure out what
should be in terms of 
we should choose 
so that we can find a 
in our equation. So we want to solve the simultaneous set of equations in terms of :![\[ -r = -1 \]](http://latex.artofproblemsolving.com/8/a/3/8a35c878f899d4e1bdf5c24b86d5b10194c23aee.png)
![\[ r - rm + t = -3m \]](http://latex.artofproblemsolving.com/3/3/4/3343f56638b34ad65c6ea2997032e79c454e66cc.png)
![\[ mr - s = -3m^2 \]](http://latex.artofproblemsolving.com/c/1/e/c1ec8109ab4bcfcd1c5569c21e17ba069bd63f49.png)
Therefore we find that
Plugging these in our equation becomes 
. Now we need to check that 
like we assumed but this is clear since 
for all positive integers .Therefore for every
of the form 
for some positive integer , letting 
yields a desired triple 
. Therefore we have found an infinite number of solutions and so we are done.
August 2016
June 2016