Difference between revisions of "2008 Indonesia MO Problems/Problem 8"
Victorzwkao (talk | contribs) (→Solution 1) |
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since <math>f(2)>f(1)</math>, we have <math>2f(2)>2f(1)-1>0</math>. Or that <math>2>\frac{2f(1)-1}{f(2)}>0</math>. Thus, | since <math>f(2)>f(1)</math>, we have <math>2f(2)>2f(1)-1>0</math>. Or that <math>2>\frac{2f(1)-1}{f(2)}>0</math>. Thus, | ||
− | + | <cmath>\frac{2f(1)-1}{f(2)}=1</cmath> | |
<cmath>2f(1)-1=f(2)=f(1)f(1)-f(1)+1</cmath> | <cmath>2f(1)-1=f(2)=f(1)f(1)-f(1)+1</cmath> | ||
<cmath>f(1)^2-3f(1)+2=0</cmath> | <cmath>f(1)^2-3f(1)+2=0</cmath> |
Revision as of 15:11, 17 September 2024
Solution 1
Since , we know that
.
Let ,
be
,
, respectively. Then,
.
Let ,
be
,
, respectively. Then,
Let ,
be
,
, respectively. Then,
Let ,
be
,
, respectively. Then,
From the last 2 equations, we get that
Since , substituting, we get
Expanding the right side, we get
Simplifying and multiplying both sides by 2, we get
If we take modulo of f(2) on both sides, we get
Because , we also know that
. If
, then
.
Suppose :
since , we have
. Or that
. Thus,
Thus,
or
.
case 1:
Let , and
be an arbitrary integer
. Then,
Thus,
.
case 2:
Let , and
be an arbitrary integer
. Then,
This forms a linear line where
Thus,
Upon verification for , we get
Upon verification for , we get
Thus, both equations, and
are valid