Difference between revisions of "2019 IMO Problems/Problem 1"
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'''''Problem:''''' | '''''Problem:''''' | ||
| − | ''Let Z be the set of integers. Determine all functions f : Z | + | ''Let <math>\mathbb{Z}</math> be the set of integers. Determine all functions <math>f : \mathbb{Z} \to \mathbb{Z}</math> such that, for all |
| − | ''integers a and b, f(2a) + 2f(b) = f(f(a + b))'' | + | ''integers <math>a</math> and <math>b</math>, <cmath>f(2a) + 2f(b) = f(f(a + b)).</cmath>'' |
'''Solution 1:''' | '''Solution 1:''' | ||
| − | Let us substitute 0 in for a to get | + | Let us substitute <math>0</math> in for <math>a</math> to get |
| − | f(0) + 2f(b) = f(f(b)) | + | <cmath>f(0) + 2f(b) = f(f(b)).</cmath> |
| − | Now, let x = f(b) | + | Now, since the domain and range of <math>f</math> are the same, we can let <math>x = f(b)</math> and <math>f(0)</math> equal some constant <math>c</math> to get |
| − | c + 2x = f(x). | + | <cmath>c + 2x = f(x).</cmath> |
| − | Therefore, we have found that '''all''' solutions must be of the form f(x) = 2x + c. | + | Therefore, we have found that '''all''' solutions must be of the form <math>f(x) = 2x + c.</math> |
| − | Plugging back into the original equation, we have: 4a + c + 4b + 2c = 4a + 4b + 2c + c which is true. Therefore, we know that f(x) = 2x + c satisfies the above for any '''integral''' constant c, and that this family of equations is unique. | + | Plugging back into the original equation, we have: <math>4a + c + 4b + 2c = 4a + 4b + 2c + c</math> which is true. Therefore, we know that <math>f(x) = 2x + c</math> satisfies the above for any '''integral''' constant c, and that this family of equations is unique. |
'''Solution 2:''' | '''Solution 2:''' | ||
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Setting them equal to each other, we have the equation <cmath>f(2x)+2f(-x)=f(2(x+k))+2f(-(x+k)),</cmath> and moving "like terms" to one side of the equation yields <cmath>f(2(x+k))-f(2x)=2f(-x)-2f(-(x+k)).</cmath> Seeing that this is a difference of outputs of <math>f,</math> we can relate this to slope by dividing by <math>2k</math> on both sides. This gives us <cmath>\frac{f(2x+2k)-f(2x)}{2k}=\frac{f(-x)-f(-x-k)}{k},</cmath> which means that <math>f</math> is linear. | Setting them equal to each other, we have the equation <cmath>f(2x)+2f(-x)=f(2(x+k))+2f(-(x+k)),</cmath> and moving "like terms" to one side of the equation yields <cmath>f(2(x+k))-f(2x)=2f(-x)-2f(-(x+k)).</cmath> Seeing that this is a difference of outputs of <math>f,</math> we can relate this to slope by dividing by <math>2k</math> on both sides. This gives us <cmath>\frac{f(2x+2k)-f(2x)}{2k}=\frac{f(-x)-f(-x-k)}{k},</cmath> which means that <math>f</math> is linear. | ||
| − | Let <math>f(x)=mx+n.</math> Plugging our expression into our original equation yields <math>2ma+2mb+3n=m^2a+m^2b+mn+n,</math> and letting <math>b</math> be constant, this can only be true if <math>2m=m^2 \implies m=0,2.</math> If <math>m=0,</math> then <math>n=0,</math> which implies <math>f(x)=0.</math> If <math>m=2,</math> we have <math>f(x)=2x+n.</math> Plugging | + | Let <math>f(x)=mx+n.</math> Plugging our expression into our original equation yields <math>2ma+2mb+3n=m^2a+m^2b+mn+n,</math> and letting <math>b</math> be constant, this can only be true if <math>2m=m^2 \implies m=0,2.</math> If <math>m=0,</math> then <math>n=0,</math> which implies <math>f(x)=0.</math> However, the output is then not all integers, so this doesn't work. If <math>m=2,</math> we have <math>f(x)=2x+n.</math> Plugging this in works, so the answer is <math>f(x)=2x+c</math> for some integer <math>c.</math> |
Revision as of 08:05, 20 July 2019
Problem:
Let 
be the set of integers. Determine all functions 
such that, for all
integers 
and 
, ![\[f(2a) + 2f(b) = f(f(a + b)).\]](http://latex.artofproblemsolving.com/0/8/e/08ea0a361613dcad402c0319f965d7f58f9f4238.png)
Solution 1:
Let us substitute 
in for to get
![\[f(0) + 2f(b) = f(f(b)).\]](http://latex.artofproblemsolving.com/4/f/a/4face31600c59bd7c8a21452c3764a31e499db30.png)
Now, since the domain and range of 
are the same, we can let 
and 
equal some constant 
to get
![\[c + 2x = f(x).\]](http://latex.artofproblemsolving.com/b/9/7/b971efe0ec65376d1a868fb942beb8a0726923f6.png)
Therefore, we have found that all solutions must be of the form 
Plugging back into the original equation, we have: 
which is true. Therefore, we know that 
satisfies the above for any integral constant c, and that this family of equations is unique.
Solution 2:
We plug in 
and 
to get
![\[f(2x)+2f(-x)=f(f(0)),\]](http://latex.artofproblemsolving.com/b/9/4/b94dc917d0959f19f194f74a1fd4e29760a627ff.png)
![\[f(2(x+k))+2f(-(x+k))=f(f(0)),\]](http://latex.artofproblemsolving.com/d/b/0/db0bb9b81d64e59a0aeefcabb03a033fc30ca6fc.png)
respectively.
Setting them equal to each other, we have the equation ![\[f(2x)+2f(-x)=f(2(x+k))+2f(-(x+k)),\]](http://latex.artofproblemsolving.com/4/5/e/45edc14cede8c8feff9f67a2cfc70f0f2c621a38.png)
and moving "like terms" to one side of the equation yields ![\[f(2(x+k))-f(2x)=2f(-x)-2f(-(x+k)).\]](http://latex.artofproblemsolving.com/b/4/b/b4b9de51e64636e17fd2aa78d71fa5b180f4f051.png)
Seeing that this is a difference of outputs of 
we can relate this to slope by dividing by 
on both sides. This gives us ![\[\frac{f(2x+2k)-f(2x)}{2k}=\frac{f(-x)-f(-x-k)}{k},\]](http://latex.artofproblemsolving.com/0/5/8/058e44fd4e422d210ff2b0a28e89ab0790754a87.png)
which means that is linear.
Let 
Plugging our expression into our original equation yields 
and letting be constant, this can only be true if 
If 
then 
which implies 
However, the output is then not all integers, so this doesn't work. If 
we have 
Plugging this in works, so the answer is 
for some integer 
