Difference between revisions of "2008 Indonesia MO Problems/Problem 8"
Victorzwkao (talk | contribs) (→Solution 1) |
Victorzwkao (talk | contribs) (→Solution 1) |
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Since <math>f: \mathbb{N}\rightarrow\mathbb{N}</math>, we know that <math>f(n)\ge 1</math>. | Since <math>f: \mathbb{N}\rightarrow\mathbb{N}</math>, we know that <math>f(n)\ge 1</math>. | ||
− | Let <math>m</math>, <math>n</math> be <math>1</math>, <math> | + | Let <math>m</math>, <math>n</math> be <math>1</math>, <math>1</math>, respectively. Then, <math>f(1) + f(2) = f(1)f(1) + 1</math>. |
Let <math>m</math>, <math>n</math> be <math>1</math>, <math>2</math>, respectively. Then, <math>f(2) + f(3) = f(1)f(2)+1</math> | Let <math>m</math>, <math>n</math> be <math>1</math>, <math>2</math>, respectively. Then, <math>f(2) + f(3) = f(1)f(2)+1</math> | ||
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From the last 2 equations, we get that <math>\frac{1}{2}(f(2)^2+1)=f(4)=f(3)(f(1)-1)+1</math> | From the last 2 equations, we get that <math>\frac{1}{2}(f(2)^2+1)=f(4)=f(3)(f(1)-1)+1</math> | ||
− | Since <math>f(3) = f(2)(f(1)-1)+1</math>, substituting, we get | + | Since <math>f(3) = f(2)(f(1)-1)+1</math>, substituting, we get |
− | + | \begin{align*} | |
− | + | \frac{1}{2}(f(2)^2+1)&=(f(2)(f(1)-1)+1)(f(1)-1)+1\ | |
− | + | \frac{1}{2}(f(2)^2+1)&=f(2)f(1)^2-f(2)f(1)+f(1)-f(2)f(1)-f(2)-1+1\ | |
− | + | f(2)^2+1&=2f(2)f(1)^2-2f(2)f(1)+2f(1)-2f(2)f(1)-2f(2) | |
− | + | \end{align*} | |
− | |||
− | |||
If we take modulo of f(2) on both sides, we get | If we take modulo of f(2) on both sides, we get |
Latest revision as of 15:19, 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
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