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

(This is Problem 1 of the 2019 IMO along with one solution. Please feel free to add more solutions which may be more elegant, and to correct this one if it is deemed incorrect.)
 
(Added Solution 2.)
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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: 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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'''Solution 2:'''
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We plug in <math>a=-b=x</math> and <math>a=-b=x+k</math> to get
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<cmath>f(2x)+2f(-x)=f(f(0)),</cmath>
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<cmath>f(2(x+k))+2f(-(x+k))=f(f(0)),</cmath>
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respectively.
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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.
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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> 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>

Revision as of 01:06, 20 July 2019

Problem:

Let Z be the set of integers. Determine all functions f : Z → 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, let x = f(b) to get and f(0) equal some constant c: 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 $f(x)=0.$ If $m=2,$ we have $f(x)=2x+n.$ Plugging both of these both work, so the answer is $f(x)=0$ or $f(x)=2x+c$ for some integer $c.$

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