And one more thing- I wanted to know, when do we need to check for point-wise traps? Only when there are multiple solutions, right?

Math-wiz wrote: $P(x,0)\implies f(f(x))=x+f(0)$ Not necessarily true. Rather, $P(x,0)\implies f(f(x))=x+f(xf(0)).$

BobThePotato wrote: Math-wiz wrote: $P(x,0)\implies f(f(x))=x+f(0)$ Not necessarily true. Rather, $P(x,0)\implies f(f(x))=x+f(xf(0)).$ Oopsie! Biiig blunder. So sorry for disturbing you

Math-wiz wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, $$f(xy+f(x))=f(xf(y))+x$$ Let $P(x,y)$ be the assertion $f(xy+f(x))=f(xf(y))+x$ Let $a=f(0)$ $P(0,0)$ $\implies$ $f(a)=a$ $P(a,0)$ $\implies$ $f(a^2)=0$ $P(a,a)$ $\implies$ $f(a^2+a)=a$ If $f(x)=0$ for some $x$, subtracting $P(a^2,x)$ from $P(x,a^2)$ implies $x=a^2$ $P(x-a,a^2)$ $\implies$ $f(a^2(x-a)+f(x))=x$ and so $f(x)$ is surjective Since surjective, let $x$ such that $f(x)=a^2$ Subtracting $P(x,0)$ from $P(x,y)$ $\implies$ $f(x(a^2+a)+a^2)=0$ And so $x(a^2+a)=0$ and so Either $x=0$, and so $a=f(0)=a^2$ and so $a\in\{0,1\}$ Either $a^2+a=0$ and so $a\in\{-1,0\}$ And so $a\in\{-1,0,1\}$ If $a=-1$ : $P(0,0)$ $\implies$ $f(-1)=-1$ $P(-1,0)$ $\implies$ $f(1)=0$ $P(1,0)$ $\implies$ $f(-1)=-2$ Contradiction If $a=1$ : $P(0,0)$ $\implies$ $f(1)=1$ $P(1,0)$ $\implies$ $f(1)=0$ Contradiction So $a=0$ and then : $P(x,0)$ $\implies$ $f(f(x))=x$ $P(1,x-f(1))$ $\implies$ $f(x)=x+1-f(1)$ and since $f(0)=0$, this implies $\boxed{f(x)=x\quad\forall x}$ which indeed is a solution

Thanks, @pco. Btw, can someone help with this? Math-wiz wrote: And one more thing- I wanted to know, when do we need to check for point-wise traps? Only when there are multiple solutions, right?

There is nothing to check. Just always remember that $f(x)g(x)=0$ $\forall x$ implies "$\forall x$, either $f(x)=0$, either $g(x)=0$" And not "either $f(x)=0$ $\forall x$, either $g(x)=0$ $\forall x$" If you always carefully avoid this error, you'll clearly get that if you want the second conclusion (which is not always true, btw), you must prove that it does not exists $x,y$ such that $f(x)=0$ and $g(x)\ne 0$ and $f(y)\ne 0$ and $g(y)=0$

BobThePotato wrote: Math-wiz wrote: $P(x,0)\implies f(f(x))=x+f(0)$ Not necessarily true. Rather, $P(x,0)\implies f(f(x))=x+f(xf(0)).$ implies f(f(x))=x+f(xf(0)).$,does it means that f is injective,if not why then?

Maxito12345 wrote: ... $f(f(x))=x+f(xf(0))$,does it means that f is injective,if not why then? If your question is "Why functional equation $f(0)=a$ and $f(f(x))=f(x)+f(ax)$ does not imply in all cases injectivity of $f(x)$ ?", this is a really strange question. In my opinion, you certainly think you have a trivial proof of injectivity and instead giving us this proof for checking, you ask this question ... . The only answer to this question would be, according to me, to provide a non injective such funtion. Here it is : Let $a,b$ any two algebraically independent transcendental numbers (whose existence is proved). Let $A$ the $\mathbb Q$ vectorspace whose basis is $B=\bigcup_{i\in\mathbb Z}\{a^i,ba^i\}$ We can define $f(x)$ as : $\forall x\notin A$ : $f(x)=\frac{a+\sqrt{a^2+4}}2 x$ $f(0)=a$ $\forall x\in A\setminus\{0\}$, $f(x)$ is a linear application whose value is determined by values in the basis $B$. And define $f(x)$ over the basis $B$ in the following manner : $\forall n\in\mathbb Z_{<0}$ : $f(a^n)=ba^n$ and $f(ba^n)=a^n+ba^{n+1}$ $f(1)=b$ and $f(b)=a+1$ $f(a)=a$ and $f(ab)=a$ $\forall n\in\mathbb \mathbb Z_{>0}$ : $f(a^{n+1})=f(f(a^n))-a^n$ and $f(a^{n+1}b)=f(f(a^nb))-a^nb$ (easy to show that these indeed both are valid inductions, remembering that $f(x)$ is linear over $A$). It is easy to check that this is a consistant definition and that $f(f(x))=x+f(xf(0))$ $\forall x$ And obviously this function is not injective since for example $f(0)=f(a)=f(ab)=a$

pco wrote: Maxito12345 wrote: ... $f(f(x))=x+f(xf(0))$,does it means that f is injective,if not why then? If your question is "Why functional equation $f(0)=a$ and $f(f(x))=f(x)+f(ax)$ does not imply in all cases injectivity of $f(x)$ ?", this is a really strange question. In my opinion, you certainly think you have a trivial proof of injectivity and instead giving us this proof for checking, you ask this question ... . The only answer to this question would be, according to me, to provide a non injective such funtion. Here it is : Let $a,b$ any two algebraically independent transcendental numbers (whose existence is proved). Let $A$ the $\mathbb Q$ vectorspace whose basis is $B=\bigcup_{i\in\mathbb Z}\{a^i,ba^i\}$ We can define $f(x)$ as : $\forall x\notin A$ : $f(x)=\frac{a+\sqrt{a^2+4}}2 x$ $f(0)=a$ $\forall x\in A\setminus\{0\}$, $f(x)$ is a linear application whose value is determined by values in the basis $B$. And define $f(x)$ over the basis $B$ in the following manner : $\forall n\in\mathbb Z_{<0}$ : $f(a^n)=ba^n$ and $f(ba^n)=a^n+ba^{n+1}$ $f(1)=b$ and $f(b)=a+1$ $f(a)=a$ and $f(ab)=a$ $\forall n\in\mathbb \mathbb Z_{>0}$ : $f(a^{n+1})=f(f(a^n))-a^n$ and $f(a^{n+1}b)=f(f(a^nb))-a^nb$ (easy to show that these indeed both are valid inductions, remembering that $f(x)$ is linear over $A$). It is easy to check that this is a consistant definition and that $f(f(x))=x+f(xf(0))$ $\forall x$ And obviously this function is not injective since for example $f(0)=f(a)=f(ab)=a$ Your opinion about me is based in... Dont judge not only you can solve FE and your not a god.

You're welcome. Glad to have helped you.

Math-wiz wrote: IMOC A1 wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, $$f(xy+f(x))=f(xf(y))+x$$ f(xy+f(x))=f(xf(y))+x Solution: P(0,0) f(f(0))=f(0). Let f(0)=a Therefore:f(a)=a P(a,0) a=f(a^2)+a f(a^2)=0 P(a^2,a) f(a^3)=f(a^3)+a^2 a^2=0 ==> a=0 Therefore: f(0)=0 P(x,0) ff(x)=x P(1,x-f(1)) f(x-f(1)+f(1))=f(f(x-f(1))+1 f(x)=x-f(1)+1. Let f(1)=c So, f(x)=x-c+1 Now substitute it in the equation we will get that c=1 Therefore: f(x)=x-1+1 f(x)=x and we have done .

nice

Let the world see the nice solution in $\LaTeX$. Bst wrote: $$f(xy+f(x))=f(xf(y))+x$$Solution: $P(0,0)$ $$f(f(0))=f(0).$$Let $f(0)=a$. Therefore: $$f(a)=a$$$P(a,0)$ $$a=f(a^2)+a$$$$f(a^2)=0$$$P(a^2,a)$ $$f(a^3)=f(a^3)+a^2$$$$a^2=0 \implies a=0$$Therefore: $$f(0)=0$$$P(x,0)$ $$f(f(x))=x$$$P(1,x-f(1))$ $$f(x-f(1)+f(1))=f(f(x-f(1)))+1$$$$f(x)=x-f(1)+1.$$Let $f(1)=c$ So, $$f(x)=x-c+1$$Now substitute it in the equation we will get that $c=1$ Therefore: $$f(x)=x-1+1$$$$f(x)=x$$and we have done . $1748

Let $a,b$ any two algebraically independent transcendental numbers (whose existence is proved). What are those numbers? Never heard of them.

Lukas8r20 wrote: Let $a,b$ any two algebraically independent transcendental numbers (whose existence is proved). What are those numbers? Never heard of them. See for example Lindemann Weierstrass theorem

Is this enough to prove injectivity ? $P(0,y) \rightarrow f(f(0))=f(0) $ $P(f(0),-1) \rightarrow f(-f(0) +f(f(0))) = f(f(0)f(-1))+f(0) \Leftrightarrow f(f(0)f(-1))=0$ Let $u=f(0)f(-1)$, $P(x,u) \rightarrow f(ux+f(x)) = x+f(0)$, Assume $\exists a,b \in \mathbb{R} $, such that $ua+f(a) = ub+f(b)$, from $P(x,u)$, this implies $a=b$, thus $$ua+f(a)=ub+f(b) \Leftrightarrow a=b \Leftrightarrow f(a)=f(b) \quad ( f \quad \text{is injective})$$

SomeUser221104 wrote: Is this enough to prove injectivity ? $P(0,y) \rightarrow f(f(0))=f(0) $ $P(f(0),-1) \rightarrow f(-f(0) +f(f(0))) = f(f(0)f(-1))+f(0) \Leftrightarrow f(f(0)f(-1))=0$ Let $u=f(0)f(-1)$, $P(x,u) \rightarrow f(ux+f(x)) = x+f(0)$, Assume $\exists a,b \in \mathbb{R} $, such that $ua+f(a) = ub+f(b)$, from $P(x,u)$, this implies $a=b$, thus $$ua+f(a)=ub+f(b) \Leftrightarrow a=b \Leftrightarrow f(a)=f(b) \quad ( f \quad \text{is injective})$$ No I think it is wrong because if you have $f(a)=f(b)$. How do you prove "a=b"

$P\left(x,-\frac{f(x)}{x}\right)\implies f\left(xf\left(-\frac{f(x)}{x}\right)\right) = f(0) - x$ for all $x\neq 0$ so $f$ is surjective $P(0,y)\implies f(f(0)) = f(0)$ For $f(u) = 0$, $P(u, f(0))\implies u = 0\implies f(0) = 0$ $P(x,0)\implies f(f(x)) = x$ $P(1,y)\implies f(y+f(1)) = y + 1$ so $f$ is linear with slope $1$ Setting $f(x) = x+c$ we get that the only solution is $\boxed{f(x) = x}$

Let $P(x,y)$ the assertion of the given F.E. $P(0,0)$ $$f(f(0))=f(0)$$$P(f(0),-1)$ $$f(f(-1)f(0))=0$$If we let $f(-1)f(0)=c$ then by $P(c,f(0))$ $$c=0 \implies f(0)=0$$$P(x,0)$ $$f(f(x))=x \implies f \; \text{involution}$$$P(1,x-f(1))$ $$f(x)=x-f(1)+1 \implies f(x)=x+d$$Now since $f(0)=0$ we have $d=0$ thus $f(x)=x$ is the only solution. Thus we are done

$(0,0)$. $$f(f(0))=f(0)$$$(f(0),0)$. Exist $a$ $f(a)=0$. $(a,f(0))$. $$a=0$$$$f(0)=0$$$$f(f(x))=x$$$x,1$. $$f(x+f(1))=x+1=f(f(x+1))$$$$f(x)=x$$

Let $P(x,y)$ be the assertion $f(xy+f(x))=f(xf(y))+x$ $P(0,0) \implies f(f(0))=f(0)$ $P(x,0) \implies f(f(x))=f(0)+x$ Let $f(a)=f(b)$ for any constants $a,b$. $f(f(a))=f(f(b))=f(0)+b=f(0)+a \implies a=b$ Thus, $f$ is injective. $f(f(0))=f(0) \implies f(0)=0$ So we get $f(f(x))=x$. So $f$ is an involution. Plugging in $f(x)=kx+c$ in the original equation we get $k=1$ and $c=0$ So the only function satisfying the equation is $\boxed{f(x)=x}$.

HasnatFarooq wrote: Let $P(x,y)$ be the assertion $f(xy+f(x))=f(xf(y))+x$ $P(0,0) \implies f(f(0))=f(0)$ $P(x,0) \implies f(f(x))=f(0)+x$ No, exactly the same error than indicated in post #3 $P(x,0)$ implies $f(f(x))=f(xf(0))+x$