FE solution too simple?

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Yiyj1

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Yiyj1

#1 h Apr 9, 2025, 3:26 AM

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Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $f(f(x)+y) = f(x^2-y)+4f(x)y$ holds for all pairs of real numbers $(x,y)$ .

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?

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InterLoop

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InterLoop

#2 h Apr 9, 2025, 3:37 AM • 1 Y

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You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$ .
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

Another example is simply the fact that you have not "excluded" the solution $f(x) \equiv 0$ from the equation $f(f(x)) =f(x^2)$ in any way - so $f(x) = x^2$ is wrong for that function as well. (thus $f(x) \equiv 0$ is not injective)

This post has been edited 2 times. Last edited by InterLoop, Apr 9, 2025, 3:39 AM

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Yiyj1

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Yiyj1

#3 h Apr 9, 2025, 3:39 AM

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InterLoop wrote:

You cannot immediately "cancel" the $f$ without further conclusions.

For example $f(3) = f(2) = 1$ is possible for a function - this does not mean that $3 = 2$ .
The property that leads you to $f(a) = f(b) \implies a = b$ or the "cancellation" of $f$ is called injectivity. You have to prove the function is injective first before cancellation.

ahh ic. I'll try to prove the injectivity. ty!

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AshAuktober

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AshAuktober

#4 h Apr 9, 2025, 4:40 AM

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This is in fact from Iran TST.

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davichu

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davichu

#5 h Apr 22, 2025, 10:39 AM

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Clearly, $f(x)\equiv0$ is a trivial solution, from now on, we assume it is not the case
Let $P(x,y)$ denote the assertion $f(f(x)+y) = f(x^2-y)+4f(x)y$
$P(x,-f(x))\rightarrow f(0)=f(x^2+f(x))-4f(x)^2$ $P(x,x^2)\rightarrow f(x^2+f(x))=f(0)+4f(x)x^2$ Adding these two together we get:
$4f(x)^2=4f(x)x^2\rightarrow f(x)^2=f(x)x^2$
Since $f(x)\neq0$ ,we can divide by $f(x)$ on both sides to get $f(x)=x^2$
so the only solutions are $f(x)\equiv0$ and $f(x)=x^2\forall x \in \mathbb{R}$

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Primeniyazidayi

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Primeniyazidayi

#6 h Apr 22, 2025, 11:08 AM

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davichu wrote:

Since $f(x)\neq0$ ,we can divide by $f(x)$ on both sides to get $f(x)=x^2$

You must at first prove that $f(x) =0 \text{ iff } x=0$ (or simply avoid pointwise trap).

This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 22, 2025, 11:12 AM

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Primeniyazidayi

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Primeniyazidayi

#7 h Apr 22, 2025, 11:25 AM

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The finish for @2above(hopefully correct):We will avoid pointwise trap.We of course have $f(0) =0$ .Let $f(t) =0$ for $t \neq 0$ . $P(t,y)$ gives $f(y) =f(t^2-y)$ .Take some $u$ such that $f(u) =u^2 \neq 0$ .Then we have $u^2=t^2(t^2-2u) +u^2$ or $u=\frac{t^2}{2}$ .But $P(0, x)$ gives that $f$ is even which means $\frac{t^2}{2}=-\frac{t^2}{2}$ or $t=0$ , contradiction. Thus we are done.

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ariopro1387

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ariopro1387

#8 h Apr 22, 2025, 4:04 PM

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Let $P(x,y)$ be the assertion of the problem.
$P(x,\frac{x^2-f(x)}{2});$ $\frac{x^2-f(x)}{2}.f(x) = 0$
$\forall x \in \mathbb{R}$
1. $f(x)\equiv0$
2. $f(x)=x^2$
we have to just check that both won't happen:
if $f(x_{1}) = 0:$
$P(x_{1},y);$ $f(y) = f(x_{1}^2-y)$
then by changing $y$ value we get that $x_{1} = 0$ or $f(x)\equiv C$ (Just $C=0$ works).