Let \(\mathbb{R}\) be the set of real numbers. Determine all functions \(f:\mathbb{R}\to\mathbb{R}\) that satisfy the equation

\[f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\]

for all real numbers \(x\) and \(y\).

Let \(\mathbb{R}\) be the set of real numbers. Determine all functions \(f:\mathbb{R}\to\mathbb{R}\) that satisfy the equation

\[f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\]

for all real numbers \(x\) and \(y\).

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TopNewestLet \(P(x,y)\) be the above FE.

\[P(0,0)\implies f(f(0))+f(0)=f(0)\implies f(f(0))=0\] \[P(0, f(0))\implies 2f(0)=f(f(0))+f(0)^2\implies f(0)^2-2f(0)=0\implies f(0)=0, 2\] Case 1: \(f(0)=0\). Then \[P(0, y)\implies f(f(y)) + f(0)=f(y)+yf(0)\implies f(f(y))=f(y)\] Since the range of \(f:\mathbb{R}\to\mathbb{R}\) is \(\mathbb{R}\), then \(\boxed{f(x)=x}\).

Case 2: \(f(0)=2\). Then let \(g(x)=f(x)-2\) which changes the FE to \[P(x):= g(g(x+y)+x+2)+f(xy)+2=x+g(x+y)+yg(x)+2y\] [I'll finish later] – Daniel Liu · 2 years, 2 months ago

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– Patrick Corn · 2 years, 2 months ago

It does look like \( f(x) = 2-x \) satisfies the functional equation. My guess would be \( f(x) = x \) or \( f(x) = 2-x \), but I do agree with the comment that case 1 still has some work to be done to get to \( f(x) = x \).Log in to reply

– Daniel Liu · 2 years, 2 months ago

This problem is quite a bit harder than I perceived it to be. I don't think I can find a solution.Log in to reply

Let \(P(x,y)\) be the assertion \(f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)\)

\(P(0,0)\Rightarrow f(f(0))=0 \\P(0,f(0))\Rightarrow 2f(0)=f(0)^2\Rightarrow f(0)=0 \text{or} 2\)

case 1.

\(f(0)=2\Rightarrow f(2)=0 \\ P(x,1)\Rightarrow f(x+f(x+1))=x+f(x+1) \\ P(0,f(x+1)+x)\Rightarrow f(x+1)+x+2=f(x+1)+x+2(f(x+1)+x) \\ \Rightarrow f(x)=2-x\forall \in\mathbb{R}\)

Case 2.\(f(0)=0\)

\(P(x,0)\Rightarrow f(x+f(x))=x+f(x) \\P(x,1)\Rightarrow f(x+f(x+1))=x+1+f(x) \\ P(1,f(x+1)+x)\Rightarrow f(1+f(1+x+f(x+1)))+f(x+f(x+1))=1+f(x+1+f(x+1))+f(x+1)+x \\ \Rightarrow f(f(x)+x+1)=f(x)+x+1 \\ P(x,-1)\Rightarrow f(x+f(x-1))+f(-x)=x+f(x-1)-f(x) \\ \Rightarrow -f(x)=f(-x)\\ P(x,-x)\Rightarrow f(x)+f(-x^2)=x-xf(x)\\ P(-x,x)\Rightarrow f(-x)+f(-x^2)=-x+x(-x) \\ \Rightarrow f(x)-f(-x)=2x-x(f(x)+f(-x)) \\ \Rightarrow f(x)=x \forall x\in\mathbb{R}\)

hence, \(f(x)=2-x\) and \(f(x)=x\) are solutions. – Sualeh Asif · 2 years, 2 months ago

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– Daniel Liu · 2 years, 2 months ago

Oh dang, I never thought of putting something as unusual as \(f(x+1)+x\) in the FE.Log in to reply

@Daniel Liu and that is why this is problem 5 of the IMO! – Sualeh Asif · 2 years, 2 months ago

That is what most people never thought!Log in to reply

– Daniel Liu · 2 years, 2 months ago

Okay, so Both problems #2 and #5 were bashy. :(Log in to reply

However problems #1, #4 and #6 have been generally appreciated – Sualeh Asif · 2 years, 2 months ago

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@Patrick Corn !

Yes you are right\(f(x)= x\) or \(2-x\quad \forall x\) – Sualeh Asif · 2 years, 2 months ago

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– Jubayer Nirjhor · 2 years, 2 months ago

In case 1, you must prove that \(f\) is surjective to make that conclusion because \(\mathbb R\) is the codomain of \(f\), which is not necessarily the range.Log in to reply

– Daniel Liu · 2 years, 2 months ago

Oh right. I'll think about this problem some more.Log in to reply

It is not very hard to see that f(x) = x and f(x) = 2 - x are solutions. The difficult part is to show that they are the only solutions. In my eyes, Sualeh Asif demonstrates that f(0) = 2 implies that f(x) = x - 2. My compliments! But I am not convinced by his reasoning that f(0) = 0 implies that f(x) = x for all real x. In particular, I do not see the consequence of P(x,1) on line 2 of case 2. Neither the consequence of P(1,f(x+1)+x) on line 4 of case 2. – Gerard Renardel · 2 years ago

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