2003 Pan African MO Problems/Problem 1
Problem
Let . Find all functions:
such that:
(1) , all
;
(2) ;
(3) , all
.
Solution
Because , we must have
, so
. Thus,
.
By trying smaller values, we suspect that for all values in the set
. To prove this, we would use induction to show that
for
for all integer values of
.
The base case works because . For the inductive step, assume that
for
for an integer value of
. To prove this case, we need to prove that all values from
to
also work.
Note that . Therefore, we must have
. Additionally, given that
is an integer that satisfies
, we must have
. Since
is within the range of
, we must have
. Furthermore, we must also have
. Thus, we must have
. Since
can be any integer that satisfies
, the value
can be any even integer. Thus all numbers from
to
are represented in the proof, so the inductive step holds.
Therefore, for
for any integer value of
, so the only function
that satisfies the three conditions is
.
See Also
2003 Pan African MO (Problems) | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All Pan African MO Problems and Solutions |