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
,
Solution 1:
Let us substitute in for
to get
Now, since the domain and range of are the same, we can let
and
equal some constant
to get
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
respectively.
Setting them equal to each other, we have the equation and moving "like terms" to one side of the equation yields
Seeing that this is a difference of outputs of
we can relate this to slope by dividing by
on both sides. This gives us
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