1961 IMO Problems/Problem 3
Problem
Solve the equation
where is a given positive integer.
Solution 1
Since , we cannot have solutions with
and
. Nor can we have solutions with
, because the sign is wrong. So the only solutions have
or
, and these are:
multiple of
, and
even;
even multiple of
and
odd;
= even multiple of
and
odd.
Solution 2 (Calculus)
First consider . Squaring both sides yields
, and utilizing identities yields
. This results in solutions
for integers
. After manually checking the solutions in the original equation, we conclude that only
and
are solutions to the equation.
Next, we consider . This is equivalent to
, so the solutions are
for integers
. Once again, substituting into the equation results in a valid identity, so these are the only solutions.
Let . Taking the derivative yields
for
. We wish to consider extrema, so we find all
-values such that
by checking the zeros of each factor. In
,
for even
when
, and for odd
when
. The differences between the two come from the
factor, reaching zeros when
. This is impossible when
is even, and if
is odd, it implies that
, hence the solutions.
We then calculate the value of at every such
. For even
, the
-values
result in
-values of
. Since these are the only possible extrema, we conclude that
is always a maximum and
is always a minimum; thus the only solutions are the maxima, which occur at
when
; more generally, we have
for integers
.
Next, for odd , the
-values
yield
-values of
. Obviously
are all solutions; we show that on our closed domain these are the only solutions. On
to
, the function is always decreasing; if it were increasing at any point, an extremum would be required to change the sign of the derivative; however, there does not exist an extremum between these two points (since we have already found all the extrema!); thus only
satisfies the equation. We can repeat this argument for all values between two extrema; eventually, we can conclude that the only times that
are the above (since no segment contains
at any point other than an endpoint; as a result, the function cannot equal
at any point other than an endpoint of some segment, which are the extrema). As a result, extending this to all
, the solutions are
and
for integers
.
Since these results match our and
cases, we conclude that when
is odd, the solutions are
and
for integers
, and when
is even, the solutions are
for integers
.
~eevee9406
1961 IMO (Problems) • Resources | ||
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