Difference between revisions of "2019 IMO Problems/Problem 1"

(Added Solution 2.)
(Fixed up latex. Fixed error in solution 2.)
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'''''Problem:'''''
 
'''''Problem:'''''
  
''Let Z be the set of integers. Determine all functions f : Z Z such that, for all
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''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))''
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''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:
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Let us substitute <math>0</math> in for <math>a</math> to get
f(0) + 2f(b) = f(f(b))  
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<cmath>f(0) + 2f(b) = f(f(b)).</cmath>
  
Now, let x = f(b) to get and f(0) equal some constant c:
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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).
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<cmath>c + 2x = f(x).</cmath>
Therefore, we have found that '''all''' solutions must be of the form f(x) = 2x + c.
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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.
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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 both of these both work, so the answer is <math>f(x)=0</math> or <math>f(x)=2x+c</math> for some integer <math>c.</math>
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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 07:05, 20 July 2019

Problem:

Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that, for all integers $a$ and $b$, \[f(2a) + 2f(b) = f(f(a + b)).\]

Solution 1:

Let us substitute $0$ in for $a$ to get \[f(0) + 2f(b) = f(f(b)).\]

Now, since the domain and range of $f$ are the same, we can let $x = f(b)$ and $f(0)$ equal some constant $c$ to get \[c + 2x = f(x).\] Therefore, we have found that all solutions must be of the form $f(x) = 2x + c.$

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.

Solution 2: We plug in $a=-b=x$ and $a=-b=x+k$ to get \[f(2x)+2f(-x)=f(f(0)),\] \[f(2(x+k))+2f(-(x+k))=f(f(0)),\] respectively.

Setting them equal to each other, we have the equation \[f(2x)+2f(-x)=f(2(x+k))+2f(-(x+k)),\] and moving "like terms" to one side of the equation yields \[f(2(x+k))-f(2x)=2f(-x)-2f(-(x+k)).\] Seeing that this is a difference of outputs of $f,$ we can relate this to slope by dividing by $2k$ on both sides. This gives us \[\frac{f(2x+2k)-f(2x)}{2k}=\frac{f(-x)-f(-x-k)}{k},\] which means that $f$ is linear.

Let $f(x)=mx+n.$ Plugging our expression into our original equation yields $2ma+2mb+3n=m^2a+m^2b+mn+n,$ and letting $b$ be constant, this can only be true if $2m=m^2 \implies m=0,2.$ If $m=0,$ then $n=0,$ which implies $f(x)=0.$ However, the output is then not all integers, so this doesn't work. If $m=2,$ we have $f(x)=2x+n.$ Plugging this in works, so the answer is $f(x)=2x+c$ for some integer $c.$