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2023 OIM Problems/Problem 2: Difference between revisions

Tomasdiaz (talk | contribs)
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Created page with "== Problem == Let <math>\mathbb{Z}</math> be the set of integers. Find all functions <math>f: \mathbb{Z} \to \mathbb{Z}</math> such that: <cmath>2023f(f(x))+2022x^2=2022f(x)+..."
 
Eevee9406 (talk | contribs)
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solution
 
Line 7: Line 7:


== Solution ==
== Solution ==
{{solution}}
Consider the equation mod <math>2023</math>:
\begin{align*}
0f(f(x))+(-1)x^2&\equiv(-1)f(x)+0[f(x)]^2+1\pmod{2023}\\
-x^2 &\equiv -f(x)+1 \pmod{2023}\\
f(x) &\equiv x^2+1 \pmod{2023}\\
\end{align*}
Thus by definition, for every integer <math>x</math>, there exists some integer <math>k_x</math> such that
<cmath>f(x)=x^2+1+2023k_x</cmath>
If we substitute this into the initial functional equation, we get:
<cmath>2023((x^2+1+2023k_x)^2+1+2023k_x)+2022x^2=2022(x^2+1+2023k_x)+2023(x^2+1+2023k_x)^2+1</cmath>
<cmath>\Rightarrow 2023(1+2023k_x)+2022x^2=2022(x^2+1+2023k_x)+1</cmath>
<cmath>\Rightarrow2023(1+2023k_x)=2022(2023k_x)+2023</cmath>
<cmath>\Rightarrow2023(2023k_x)=2022(2023k_x)</cmath>
<cmath>\Rightarrow k_x=0</cmath>
Thus the only functional equation possible is <math>\boxed{f(x)\equiv x^2+1}</math>, which works upon substitution.
 
~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]


== See also ==
== See also ==
https://sites.google.com/associacaodaobm.org/oim-brasil-2023/pruebas
https://sites.google.com/associacaodaobm.org/oim-brasil-2023/pruebas

Latest revision as of 17:31, 8 April 2025

Problem

Let $\mathbb{Z}$ be the set of integers. Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that:

\[2023f(f(x))+2022x^2=2022f(x)+2023[f(x)]^2+1\]

for each integer $x$.

Solution

Consider the equation mod $2023$: \begin{align*} 0f(f(x))+(-1)x^2&\equiv(-1)f(x)+0[f(x)]^2+1\pmod{2023}\\ -x^2 &\equiv -f(x)+1 \pmod{2023}\\ f(x) &\equiv x^2+1 \pmod{2023}\\ \end{align*} Thus by definition, for every integer $x$, there exists some integer $k_x$ such that \[f(x)=x^2+1+2023k_x\] If we substitute this into the initial functional equation, we get: \[2023((x^2+1+2023k_x)^2+1+2023k_x)+2022x^2=2022(x^2+1+2023k_x)+2023(x^2+1+2023k_x)^2+1\] \[\Rightarrow 2023(1+2023k_x)+2022x^2=2022(x^2+1+2023k_x)+1\] \[\Rightarrow2023(1+2023k_x)=2022(2023k_x)+2023\] \[\Rightarrow2023(2023k_x)=2022(2023k_x)\] \[\Rightarrow k_x=0\] Thus the only functional equation possible is $\boxed{f(x)\equiv x^2+1}$, which works upon substitution.

~ eevee9406

See also

https://sites.google.com/associacaodaobm.org/oim-brasil-2023/pruebas

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