Yay! My first (first) IMOSL solution! First we let $f(0) = \alpha$. $f(f(x)+y) = 2x + f(f(y)-x)$ $(1)$ Plugging in $x=0$ into (1): $f(y+\alpha) = f(f(y))$ $(2)$ Plugging in $x=f(y)$ into (1): $f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(2y+\alpha) = 2f(y) + \alpha$ $(3)$ Now plugging in $y=-\alpha$ into (3): $f(-\alpha) = 2f(-\alpha) + \alpha \Rightarrow f(-\alpha) = -\alpha$ Plugging in $y=-\alpha$ into (2): $\alpha = f(f(-\alpha)) \Rightarrow \alpha = -\alpha \Rightarrow \alpha = 0$ So (2) can be rewritten as: $f(f(x)) = x$ $(4)$ Plugging in $y=0$ into (1): $f(f(x)) = 2x + f(-x) \Rightarrow f(-x) = -x$ Plugging in $x=-x$ for the last equation gives $f(x)=x$ for all $x$. It is easy to see that this function indeed satisfies the initial conditions given in the problem.

Philip_Leszczynski wrote: $f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(2y+\alpha) = 2f(y) + \alpha$ $(3)$ You are wrong here sorry you miscalculate I think. The correct is $f(f(f(y))+y) = 2f(y) + f(f(y)-f(y)) \Rightarrow f(f(y+\alpha)+y)=2f(y)+\alpha$ from where we can find (plugging where $y=-a$) that $f(-\alpha)=0$ and finally the solutions in not $f(x)=x$ but $f(x)=x+c$

Setting $ y=-f(x)$, we have $ f(0)=2x+f(f(-x)-x))$, so $ f(0)-2x=f(f(-x)-x)$, and $ f(0)-2x$ is a surjective function of $ x$, so $ f$ is surjective as well. Then, by surjectivity, we can let $ c$ be a number for which $ f(c)=0$. Setting $ x=c$, we have $ f(y)=2c+f(f(y)-c)$. Since $ f(y)$ can take on all real values $ r$, $ r=2c+f(r-c)$ for all real numbers $ r$, implying $ f(r-c)=r-2c$. Letting $ s=r-c$, we find $ \boxed{f(s)=s-c}$ for some constant $ c$ for all real numbers $ s$. Also, it can be easily verified that all functions of this form satisfy this equation.

Let $P(x,y)$ denote the assertion $f(f(x)+y) = 2x + f(f(y)-x)$. Now, \[P(0,x) \implies f(x+f(0)) = f(f(x))\] \[P(f(y),y) \implies f(f(f(y)) + y ) = 2f(y)+f(0) \implies f(f(y+f(0))+y) = 2f(y) + f(0)\] Taking $y=-f(0)$ here, implies $f(0) = 2f(-f(0)) +f(0) \implies f(-f(0)) = 0$. Next, \[P(-\frac x2, -f(-\frac x2) - f(0)) \implies x = f(f(-f(-\frac x2) - f(0)) + \frac x2)\] Therefore, $f$ is onto i.e. covers all values of $x$. Now,let $f(0) = c$, and define a function $g : \mathbb{R} \rightarrow \mathbb{R}$ as $g(x) = f(x - c)$. Therefore, our original assertion $P(x,y)$ for $f$ changes into the equivalent assertion $Q(x,y)$ for $g$, as being \[g(g(x+c) + y + c) = 2x + g(g(y+c) - x +c)\] As $f$ is onto, therefore $g$ is also onto. Also, $g(0) = 0$. And then finally, \[Q(-c,x-c) \implies g(x) = -2c + g(g(x) + 2c) \implies g(g(x) + 2c) = g(x) + 2c\] As $g$ is onto, therefore, $g(x)$ covers all values of $x$, and therefore, $g(x) + 2c$ also covers all values of $x$, Thus, $g(x) = x$ for every $x \in \mathbb{R}$. And therefore, $f(x) = x+f(0)$ for every $x \in \mathbb{R}$ and this is indeed a solution.

Set $y=-f(x)$ then $-2x=f(f(-f(x))-x)$ so $f$ is $surjective$. So there exist $a,b$ such that $f(a)=0$ and $f(b)=a$. Then set $x=a$ and $y=b$ $\Rightarrow$ $a=2a+f(0)$ so $f(-f(0))=0$. Now let $x=-f(0)$ $\Rightarrow$ $f(y)=-2f(0)+f(f(y)+f(0))$ let $f(0)=c$ $\Rightarrow$ $(f(y)+c)+c=f(f(y)+c)$. Since $f$ is $surjective$ we get that $f(y)+c$ is also $surjective$ which implies $f(x)=x+c$ $\forall$ $x$ $\in$ $\mathbb{R}$ where $c=f(0)$.

orl wrote: Find all functions $f$ from the reals to the reals such that \[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\] for all real $x,y$. This is my solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3376958#p3376958

Notice that we can take $x$ such that $x=\frac{1}{2}\cdot f(y)$.Also it is impossible to be $f(x)>0$ because we can always take $x$ so small to make that absurd.Putting $y=y_1$,($y_1$ is such that $f(y_1)=0$),and $x->x_1=\frac{1}{2}\cdot f(y)$ we easily get $f(f(x_1)+y)=2x_1+f(\frac{1}{2}\cdot f(y))$.Lets assume that there is $x_2$ such that $x_1\neq x_2$ and $f(x_1)=f(x_2)$.From that it follows: $2x_1+f(\frac{1}{2}\cdot f(y))=f(f(x_1)+y)=f(f(x_2)+y)=2x_2+f(\frac{1}{2}\cdot f(y))$,then we get $2x_1=2x_2$.Clearly the function is an injection. Putting $x=0$ we get $f(f(0)+y)=f(f(y))$ which is equivalent to $f(x)=x+f(0)$,(lets take $f(0)=c$). Now it remains to check if that is a solution:$f(f(x)+y)=f(x+c+y)=x+y+2c$ and $RHS=f(f(y)-x)+2x=f(y-x+c)+2x=y-x+2c+2x=y+x+2c$.Obviously $LHS=RHS$.Hence the only solution is $f(x)=x+c$

please correct me if I am wrong

is this solution above ok?

Let $f(0) = a$. We will first prove that $f$ is surjective i.e. the range of $f$ is $\mathbb{R}$. Let $y = -f(x)$ and we get \[a - 2x = f(f(-f(x)) - x)\] and because $a-2x$ can range over all the reals, there must exist some real $z$ such that $f(z) = a-2x$. Now let $r$ be some real constant such that $f(r) = 0$. Let $x = r$ and we have $f(y) = 2r + f(f(y) - r)$. If we let $y$ be some number such that $f(y) = r$, then we have $r = 2r + a$ so $r = -a$, so $f(-a) = 0$. Hence, we can replace $r$ with $-a$ and get $f(y) = -2a + f(f(y)+a)$, and since $f(y)$ can take on any real value, let's just replace $f(y)$ with $z$ so we have $f(z+a) = z+2a$. Letting $x = 0$ and $y = z$ in the original FE gives us \[f(z+a) = f(f(z))\] so we have $f(f(z)) = z+2a$. Now we show that $f$ is injective. Suppose we have $f(x) = f(y)$ for some $x, y$, so then $f(f(x)) = x+2a = f(f(y)) = y+2a$ which implies that $x = y$. So now, since $f(z+a) = f(f(z))$, we have $f(z) = z+a$. Therefore, our solutions are $f(x) = x+a$ for some $a \in \mathbb{R}$. Checking that these work is trivial.

fclvbfm934 wrote: Hence, we can replace $r$ with $-a$ and get $f(y) = -2a + f(f(y)+a)$, and since $f(y)$ can take on any real value, let's just replace $f(y)$ with $z$ so we have $f(z+a) = z+2a$. . I think you could have stopped here, as $a$ is constant and $z$ can be chosen arbitrarily, so for any $x$ , we may let $z=x-a$, and we get the solution .. Am I messing something ?

I think you are right @Nanas.

By putting $P(x;0)$ we have $f(f(x))=2x+f(f(0)-x)$ which begets $f$ is injective and by $P(0,y)$ we have $f(f(0)+y)=f(f(y))$. Obviously $f(y)=y+f(0)$. It is easy to verify that it is indeed a solution.

Mukhammadiev wrote: By putting $P(x;0)$ we have $f(f(x))=2x+f(f(0)-x)$ which begets $f$ is injective and by $P(0,y)$ we have $f(f(0)+y)=f(f(y))$. Obviously $f(y)=y+f(0)$. It is easy to verify that it is indeed a solution. How do we get the function as injective ?

Am i badly mistaken or this works: Put $f(x)$ =$g(x) +x$. Simplifying the equation and then putting $x=0$ gives that $g$ is a constant function. Done.

Here's a solution that I believe avoids injectivity and surjectivity altogether. Let $x=0$ to get $f(f(0)+y)=f(f(y))$ and let $y=0$ to get $f(f(x))=2x+f(f(0)-x)$. Substituting our expression for $f(f(x))$ into the first equation yields \[f(f(0)+y)=f(f(y))=2y+f(f(0)-y).\] Now make the change of variables $m=f(0)+y$, $n=f(0)-y$. Then $2y=m-n$, so \[f(m)=m-n+f(n)\implies f(m)-m=f(n)-n.\] Let $g(x)=f(x)-x$. This implies $g(m)=g(n)$ for all pairs of real numbers $(m,n)$, so $g$ is constant. Thus $f(x)-x=c\implies f(x)=x+c$, where $c$ is a constant. Checking yields that all such functions work.

djmathman wrote: Now make the change of variables $m=f(0)+y$, $n=f(0)-y$. [...] This implies $g(m)=g(n)$ for all pairs of real numbers $(m,n)$ Note that $m$ and $n$ are not independent of each other since your choice implies $m+n=2f(0)$. So you can't conclude $g(m)=g(n)$ for all pairs $(m,n)$. (That's as always the risk when substituting new variables and forgetting about conditions )

Let $f(0) = c$. Plug in $y = -f(x)$, so $c = 2x + f(f(-f(x)) - x)$ and $f$ is surjective. Thus there exist $a, b$ such that $f(a) = 0$ and $f(b) = a$. Setting $x = a$ and $y = b$ gives $a = 2a + c$ so $a = -c$. Set $x=0$ and $y=0$ to get $f(c + x) = f(f(x))$ and $f(f(x)) = 2x + f(c - x)$ respectively, so $f(c + x) = 2x + f(c - x)$. Setting $y = k - f(x)$ in the original equation gives us $$f(k) = 2x + f(f(k - f(x)) - x)$$Now let $x = -c$, then $f(k) = -2c + f(f(k) + c) = -2c + 2f(k) + f(c - f(k))$ by the previously obtained identity and $f(-c) = 0$. Let $t = f(k)$, then we have $f(c - t) + t = 2c$, and since $f$ is surjective, this holds for all real $t$. Setting $t = c - z$ for any real $z$ gives $f(z) = z + c$ which clearly satisfies.

We claim $f(x) = x + c$ for some constant $c$, which clearly works. Substitute $y = -f(x)$. This yields \[f(0) - 2x = f(f(-f(x)) - x)\text{.}\]This implies $f$ is surjective. Let $x$ satisfy $f(x) = 0$ and $y$ satisfy $f(y) = a + x$ for a fixed $a$. Directly substituting yields \begin{align*} f(y) &= 2x + f(a + x - x) \\ a + x - 2x &= f(a) \\ f(a) &= a - x\text{.} \end{align*}Since $a$ can be anything, $f$ must be linear with slope $1$. (Note that this also implies $x$ is unique).

Let $P(x,y)$ denote the given assertion. Claim: $f$-is surjective

Since $f$ is surjective we get that $\exists \alpha : f(\alpha)=0$ $P(\alpha,\alpha) \implies f(-\alpha)=-2\alpha$ $...(1)$ $P(\alpha,f(\alpha)) \implies f(-\alpha)=f(0)-2\alpha$ Combining with $(1)$ we get: $f(0)=0$ $P(0,x) \implies f(x)=f(f(x))$. Since $f$-is surjective we get that there exists a $z$ s.t: $\exists z:z=f(x)$ So $f(z)=z \implies f(x)=x \forall x \in \mathbb{R} $

The answer is $f(x) = x + c$ for any real $c$, which works. Denote the original assertion as $P(x, y)$. Claim: $f$ is surjective Proof: From $P(x, -f(x))$ we have $f(0) = 2x + f(f(-f(x)) - x)$ or $f(f(-f(x)) - x) = f(0) - 2x$. Since $f(0)$ is constant, from varying $x$ we see that $f$ can take on any real output. Now, from surjectivity, we can choose $t$ such that $f(t) = 0$. Then $P(t, x)$ gives $f(x) = 2t + f(f(x) - t)$. Letting $x_1 = f(x) - t$, this simplifies to $x_1 + t = 2t + f(x_1)$, or $f(x_1) = x_1 - t$. Since $x_1$ can take on any real value by surjectivity, this implies the solution set.

missed the slick solution We have $f(f(x))=f(x+f(0))$. Putting $(x,y)=(x-f(0),0)$ we get $f(2f(0)-x)=f(x)-2x+2f(0)$, since $f(f(x-f(0)))=f(x)$. Now we put $(x,y)=(-x,f(0))$ we find \begin{align*} f(f(f(0))+x) &=2x+f(f(-x)+f(0))=2x+f(f(f(-x)))=2x+f(f(-x+f(0))) \\ &=2x+f(-x+2f(0))=2x+f(x)-2x+2f(0)=f(x)+2f(0)\end{align*}Also putting $x=f(0)$ gives $f(f(f(0))+x)=2f(0)+f(f(x)-f(0))$, so $f(f(x)-f(0))=f(x)$. If we put $y=-f(x)$, we find $f(\text{blah})=-2x+f(0)$ thus $f$ is surjective. Now by surjectivity $f(x-f(0))=x$ for any real $x$. It follows that $f(x)=x+a$ for all real $x$ and some constant $a$, this is a valid solution as easily checked.

The answers are $f(x) = x+c$ for all $x \in \mathbb{R}$ and any fixed constant $c$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We denote by $P(x,y)$ the assertion that $f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)$. Then, we can start off by proving the following properties of $f$. Claim : The function $f$ is surjective. Proof : Note that, $P\left(f\left(\frac{t-f(0)}{2}\right),\frac{t-f(0)}{2}\right)$ gives, \[f\left(f\left(f\left(\frac{t-f(0)}{2}\right)\right)+\frac{t-f(0)}{2}\right)=2\left(\frac{t-f(0)}{2}\right)+f\left(f\left(\frac{t-f(0)}{2}\right)-f\left(\frac{t-f(0)}{2}\right)\right)=t-f(0)+f(0)=t\]which clearly implies the claim. Claim : The function $f$ is injective. Proof : Say there exists real numbers $\alpha<\beta$ such that $f(\alpha)=f(\beta)$. Then, $P(x,\alpha)$ and $P(x,\beta)$ will yield \[f(f(x)+\alpha)=2x+f(f(\alpha)-x)=2x=f(f(\beta)-x)=f(f(x)+\beta)\]Now, since we have already shown that $f$ is surjective, there must exist some real number $x$ such that $f(x)=t-\alpha$ for every $t\in \mathbb{R}$. Considering such $x$ will give \[f(t)=f(f(x)-\alpha)=f(f(x)-\beta)=f(t+(\beta-\alpha))\]from which it is clear that $f$ is periodic with period $T=|\beta-\alpha|$. Now, comparing $P(x+T,y)$ gives us, \[2(x+T)=f(f(x+T)+y)-f(f(y)-(x+T))=f(f(x)+y)-f(f(y)-x)=2x\]which is a clear contradiction is $T\neq 0$. Thus, there cannot exist such $\alpha$ and $\beta$ proving the claim. Now, simply look at $P(0,x)$ to see that, \[f(f(0)+x)=f(f(x)) \implies f(x)=x+f(0)\]since $f$ is injective. Thus, all solutions are indeed of the claimed forms and we are done.

orl wrote: Find all functions $f$ from the reals to the reals such that \[f\left(f(x)+y\right)=2x+f\left(f(y)-x\right)\] for all real $x,y$. $P(x,-f(x))$ gives $f(0)=2x+f(f(-f(x))-x)$ so $f$ is surjective. Let $f(a)=f(b)$ then: $P(x,a),P(x,b)$ gives:$f(f(x)+a)=f(f(x)+b)\Rightarrow f(x+c)=f(x)$ with $c=a-b$ then: $P(x+c,y)$ gives $c=0$ so $f$ is one to one. $P(0,y)$ so $f(f(0)+y)=f(f(y))\Rightarrow f(y)=y+f(0)$ wich is the sollution

We claim that the function must be of the form $f(x)=x+c$. This clearly works since both sides evaluate to $x+y+2c$. We start by showing $f$ must be surjective. Let $y=-f(x)$, then the equation becomes \[f(0)=2x+f(f(-f(x))-x).\]After moving the $2x$ to the left side, by varying $x$ the left side can achieve all real values, while the right side contains only a function term. Thus $f$ takes all real values and is surjective. Because $f$ is surjective, we can consider a value $u$ such that $f(u)=0$, and let $x=u$. Then the equation simplifies to \[f(y)=2u+f(f(y)-u).\]Again using the fact that $f$ is surjective, we can make the substitution $t=f(y)-u$. Because $u$ is a constant, $f(y)-u$ must take on all real values. This substitution gives us \begin{align*} t+u&=2u+f(t) \\ f(t)&=t-u \end{align*}Thus the function $f$ must be of the desired form, and we are done.

Setting $y=-f(x)$, we get \[f(f(x)-f(f(x))=f(0)=2x+f(f(-f(x))-x),\]so by varying $x$ across all reals, we get that $f$ is surjective. Now, using that $f$ is surjective, we must have some $t\in\mathbb{R}$ such that $f(t)=0$. Then, plugging in $x=t$ gives us \[f(f(t)+y)=f(y)=2t+f(f(y)-t),\]and letting $x=f(y)$, we have $x-2t=f(x-t)$, which can be written as $f(x)=x-t$. Replacing the constant $-t$ with $c$ and plugging it back into the original equation, we see that all such equations work, so our answer is $f(x)=x+c$ for all $x$.

Let y = -f(x). We get $f(0) = 2x + f(f(-f(x) - x)$ $\Rightarrow$ $f(f(-f(x) - x) = f(0) - 2x$ and now f(0) is fixed and we can vary x however we want $\Rightarrow$ f goes trough all values $\Rightarrow$ f is surjective. Now since f is surjective there $\exists$ t, for which $f(t) = 0$. Let x = t. We get $f(y) = 2t + f(f(y) - t)$. Let f(y) - t = u which goes trough all values $\Rightarrow$ $u - t = f(u)$ $\Rightarrow$ we get that $f(x) = x - t$, for every t. Now we have to check it so plugging in we get x - t + y - t = 2x + y - t - x - t, which is correct $\Rightarrow$ we are ready.

We claim the answer is all functions of the form $f(x) = x + c$, it is easy to verify all of these functions work. Set $y =-f(x)$, then we get $f(0) - 2x = f(f(y) - x)$, varying $x$ shows that $f$ is surjective. Now we show $f$ is injective, assume otherwise and we have $f(a) = f(b)$ for unequal $a,b$, then set $y = c$ such that $f(c) = a + b$, then we get $ 2a + f(b)= f(f(b) + c)= f(f(a) + c)=2b + f(a) = 2b + f(b)$, giving $a = b$, contradiction. Now set $x = 0$, and undo a layer of $f$ via bijectivity, we get $f(0) + y = f(y)$, so $f(y) = y + c$ for some $c$.

This is nice The answer is $f(x)=x-c$, which trivially works. If $y=-f(x)$, we get that $f(0)=2x+f(f(-f(x)-x))$, so $f$ is surjective. Now, we take a constant $c$ such that $f(c)=0$. $P(c,y)$ gives $f(y)=2c+f(f(y)-c)$ which rearranges to $f(y)-c=f(f(y)-c)+c$. As $f$ is surjective, for all $x$ we can find $y$ such that $x=f(y)-c$. Thus, for all $x$, $f(x)=x-c$. It is easily verified that $c$ can be any real number, so we are done.

$P(x, -f(x))$ gives $f(0) = 2x + f(f(-f(x)) - x)$ which means $f$ is surjective. Take $y$ such that $f(y) = x$, and we get $f(f(x) + y) = 2x + f(0)$. Say $f(a) = f(b)$, plugging this in into the new equation, we get $a = b$, so $f$ is injective. $P(0, y)$ gives $f(y + f(0)) = f(f(y)) \implies f(y) = y + f(0)$, so $f(x) = x + c$, which clearly works.

alexanderhamilton124 wrote: . Take $y$ such that $f(y) = x$, and we get $f(f(x) + y) = 2x + f(0)$. Say $f(a) = f(b)$, plugging this in into the new equation, we get $a = b$, You cant say a=b becoz y is not fixed you taken y such that f(y) = x ,

Note that $f$ is surjective. Let $f(a)=0$. $P(a, y)$ gives: $f(y) = 2a+f(f(y)-a)$. Since $f$ is surjective, let $t=f(y)-a \in \mathbb R$. Thus: $f(t)=t-a$ for all $t \in \mathbb R$ for some constant $a$.

Anshu_Singh_Anahu wrote: alexanderhamilton124 wrote: . Take $y$ such that $f(y) = x$, and we get $f(f(x) + y) = 2x + f(0)$. Say $f(a) = f(b)$, plugging this in into the new equation, we get $a = b$, You cant say a=b becoz y is not fixed you taken y such that f(y) = x , Thank you for spotting the error! Will try to fix today.