FE solution too simple?

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Source: 101 Algebra Problems from the AMSP

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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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#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

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

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This is in fact from Iran TST.

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