To solve this beautifull problem you have to solve the pseudo causchy's equation : $g(x+y+s)=g(x)+g(y)$ where $g(x)=f(x)-r$

the only solutions are f(x)=x+r+s and f(x)=-x+r-s.f(x+f(y))=f(x+r)+y+s(*) int (*) let x=0.then f(f(x))=f(r)+x+s.now in (*)y=f(y)=>f(x+f(f(y)))=f(x+y+f(r)+s)=f(x+r)+f(y)+s.(1)in (1) let x=x-r.<=>f(x+y+f(r)-r+s)=f(x)+f(y)+s.denote w=f(r)-r+s.so we have f(x+y+w)=f(x)+f(y)+s.let g(x)=f(x)+s.then (1) is equivalent with g(x+y+w)=g(x)+g(y).(2).in (2)let x=y=0.then g(w)=2g(0).let g(0)=a,where a is a rational number(we will find a later in the solution).then g(w)=2a.in(2) let y=0.then g(x+w)=g(x)+a(3).but using (3) (2)becomes g(x+y)+a=g(x)+g(y).putting h(x)=g(x)-a (2)becomes h(x+y)=h(x)+h(y) with x,y rational numbers.but this is just cauchy so h(x)=ex,where e is a rational number(we will find e later).so g(x)=h(x)+a=ea+a.if we plug this in relation (2)then we find that e=a/w.so g(x)=(a/w)x+a.so f(x)=g(x)-s=(a/w)x+a-s.we plug this into (*) and we get after some easy computations that(a^2/w^2)y+a^2/w-s(a/w)-s=(a/w)r+y.so a^2/w^2 must be 1.so w=a or w=-a. case 1:w=a.we get a=2s+r.we easily get f(x)=x+s+r for every rational x. case 2:w=-a.=>f(x)=-x+r-s.that's all ps:could someone put this into latex pls

Yes that is right Who is the designer of this problem?

bodom wrote: the only solutions are $f(x)=x+r+s$ and $f(x)=-x+r-s$ $f(x+f(y))=f(x+r)+y+s$ (*) int (*) let $x=0$ Then $f(f(x))=f(r)+x+s$ now in (*) $y=f(y)$ and we have $f(x+f(f(y)))=f(x+y+f(r)+s)=f(x+r)+f(y)+s$ (1) in (1) let $x=x-r$ <=> $f(x+y+f(r)-r+s)=f(x)+f(y)+s$ Now denote $w=f(r)-r+s$ So we have $f(x+y+w)=f(x)+f(y)+s$ Let $g(x)=f(x)+s$ .Then (1) is equivalent with $g(x+y+w)=g(x)+g(y)$ (2) in (2)let $x=y=0$ and then $g(w)=2g(0)$ Let $g(0)=a$,where a is a rational number(we will find a later in the solution).then $g(w)=2a$ In(2) let $y=0$ Then $g(x+w)=g(x)+a$ (3).but using (3) (2)becomes $g(x+y)+a=g(x)+g(y)$ .Putting $h(x)=g(x)-a$ (2)becomes $h(x+y)=h(x)+h(y)$ with x,y rational numbers.But this is just cauchy so $h(x)=ex$,where e is a rational number(we will find e later).so $g(x)=h(x)+a=ea+a$ If we plug this in relation (2) then we find that $e=\frac{a}{w}$ so $g(x)=(\frac{a}{w})x+a$. so $f(x)=g(x)-s=(\frac{a}{w})x+a-s$. we plug this into (*) and we get after some easy computations that $\frac{a^2}{w^2}y+\frac{a^2}{w}-s(\frac{a}{w})-s=\frac{a}{w}r+y$ .so $\frac{a^2}{w^2}$ must be 1 .so $w=a$ or $w=-a$. case 1:w=a.we get a=2s+r.we easily get $f(x)=x+s+r$ for every rational x. case 2:w=-a.=> $f(x)=-x+r-s$ .that's all ps:could someone put this into latex pls

Adding z-r to both sides of the definning equation and taking f, we get f(z-r+f(x+f(y)))=f(z+y+s-r+f(x+r)). Applying the definning equation to each side, we have f(z)+x+f(y)+s=f(z+y+s)+x+r+s, which simplifies to f(y)+f(z)=f(y+z+s)+r. Let g(x)=f(x-s)-r, then g(u)+g(v)=g(u+v), where u=y+s and v=z+s. Hence g(x)=cx where c is a constant. Substituting f(x)=c(x+s)+r into the defining equation, we find c=1 or -1. This solution may reveal how the problem was created.

Let $g(x)=f(x)-r-s$ So now we have $g(x+g(y))=g(x-s)+y+s$ and $g^2(x+g(y))=x+g(y)$ Now as we can express all rational in terms of $x+g(a)$ with $a$ fixed. So $g$ is bijective. So now replacing $y$ by $g(y)$ we have by previous result $g(x+y)=g(x-s)+g(y)+s=g(y-s)+g(x)+s$ Hence $g(x)-g(x-s)=c$ , where $c$ is constant. Hence finally we get $g(x+y)=g(x)+g(y)-c$ so $g(x)=ax+b$ for some constant $a,b$. Now putting this into original equation we get $(a,b)=(1,0),(-1,-2s)$ So clearly all solutions are given by $f(x)=x+r+s$ or $f(x)=-x+r-s$.