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QUestion:

Find all the function $f:N\to N$, such for any $m,n\in N$, have $$2f(mn)\ge f(m^2+n^2)-(f(m))^2-(f(n))^2\ge 2f(m)f(n)$$

This problem is from Mathematical olympiad 2014(chongqing provinces)

My try: let $m=n=0$,then we have $$2f(0)\ge f(0)-2f(0)^2\ge 2(f(0))^2$$ so $$f(0)\ge 0$$

I guess $f(n)=n^2$.But I can't prove it so seem this follow can't usefull,so How find this function?

math110
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  • Actually you can get f(0)=0 and f(1)<=1. My guess is that f=0 or f(n)=n^2. – Chen Jiang Aug 04 '14 at 15:23
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    You have more than this. Let's assume 2 cases. First is $f(0)=0$. Second is $f(0)>0$, then you can divide by $2f(0)$ and get $1\ge f(0)$, and since it is not 0, it is $f(0)=1$, impossible due to the middle term! So you know that $f(0)=0$ Now that you know that, you can test with $m=0$ and $n>0$ and deduce that $f(n²)=(f(n))²$ for n in N – Martigan Aug 04 '14 at 16:20
  • @Martigan how do you get $f(0)=1$ from $f(0) \le 1$ and $f(0) \ne 0$? – BeaumontTaz Aug 04 '14 at 17:07
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    @BeaumontTaz, $f(0)\neq 0$, $1\ge f(0)$ and $f:N\to N$ – Martigan Aug 04 '14 at 17:13
  • @Martigan, I forgot that we were in $\mathbb{N}$. Thank you for the clarification. – BeaumontTaz Aug 04 '14 at 17:14

1 Answers1

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Let's start with $m=n=0$.

We have then $f(0)\ge 0$.

If $f(0) \neq 0$, then $2f(0)\ge f(0)-2f(0)^2\ge 2(f(0))^2$ becomes $2\ge 1-2f(0)\ge 2(f(0))$, that is $1\ge4f(0)>0$, which is impossible since $f:N\to N$.

Hence $f(0)=0$

Then take $m=0$ and $n>0$

You have $2f(0)\ge f(n^2)-(f(n))^2\ge 2(f(0))^2$, which is $f(n^2)=(f(n))^2$

You take $m=n\neq0$ and immediatly also have $f(2n^2)=4f(n^2)$

For $n=1$ and $m=0$ you deduce than either $f(1)=0$ or $f(1)=1$

You can deduce easily that if $f(1)=0$, then $f(2)=0$ as well, and with the initial inequality and $f(n^2)=(f(n))^2$ you conclude quickly that $f(n)=0$ for all n.

If $f(1)=1$, use the first equation with $m=n=1$ to find out that $f(2)=4$, then $f(4)=16$ and $f(5)=25$ very quickly.

You don't have yet $f(n)=n^2$ for every $n$. Assume that $f(n)=n^2 +g(n)$, with $g(0)=g(1)=g(2)=0$. By developing the first inequality with the chosen $m$ and $n$ you find out that $g(n)=0$ for all $n$.

Martigan
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