v_Enhance wrote: Find all triples $(f,g,h)$ of injective functions from the set of real numbers to itself satisfying \begin{align*} f(x+f(y)) &= g(x) + h(y) \\ g(x+g(y)) &= h(x) + f(y) \\ h(x+h(y)) &= f(x) + g(y) \end{align*} for all real numbers $x$ and $y$. (We say a function $F$ is injective if $F(a)\neq F(b)$ for any distinct real numbers $a$ and $b$.) Proposed by Evan Chen let $P(x,y): f(x+f(y))=g(x)+h(y)$ now put $Q(x,y,z)=P(x+g(z),y): f(x+g(z)+f(y))=h(x)+h(y)+f(z)$ now from $Q(x,y,z),Q(y,x,z)$ we have ( from injective) : $g(x)=x+k$ so ....

Nice -- this is the shortest solution I've yet seen. Do you mind if I add this to the official solutions packet? (You will be credited, of course.)

of course not my friend . you can add this solution if you want .

This is WAY too much information, yet one only has to pick the very little that's required. Take $f(0)=a, g(0) = b, h(0)=c$. Substituting $x \to x-a$ and $y \to 0$ and analogously in the others, we get \[ f(x) = g(x-a) + c \]\[ g(x) = h(x-b) +a \]\[h(x) = f(x-c) + b\]Now, we have \[f(x+f(y)) = g(x) + h(y) = h(x-b)+h(y)+a\]\[=h((y+b)-b)+h(x-b)+a = f(y+b+f(x-b))\]Thanks to injectivity, we get $x+f(y) = y+b+f(x-b) \implies f(y)-y = f(x-b) - (x-b) \forall x \in \mathbb{R}$. Therefore, $f(x)-x = a_0$ is constant. Similarly, $g(x)-x=b_0$, $h(x)-x = c_0$. Substituting all variables $=0$ in the given system, we have $2a_0 = b_0+c_0, 2b_0 = c_0 +a_0$ and $2c_0=a_0+b_0$. Therefore, $a_0 = b_0 = c_0$, implying $f(x), g(x), h(x) = x+k \forall x \in \mathbb{R}$.

dibyo_99 wrote: This is WAY too much information, yet one only has to pick the very little that's required. Indeed the hypotheses of the problem are fairly strong. I think in retrospect one of the hard parts of the problem was just that it looked very scary. But there are a lot of things that work (although apparently plenty of things also don't work).

alibez wrote: v_Enhance wrote: Find all triples $(f,g,h)$ of injective functions from the set of real numbers to itself satisfying \begin{align*} f(x+f(y)) &= g(x) + h(y) \\ g(x+g(y)) &= h(x) + f(y) \\ h(x+h(y)) &= f(x) + g(y) \end{align*} for all real numbers $x$ and $y$. (We say a function $F$ is injective if $F(a)\neq F(b)$ for any distinct real numbers $a$ and $b$.) Proposed by Evan Chen let $P(x,y): f(x+f(y))=g(x)+h(y)$ now put $Q(x,y,z)=P(x+g(z),y): f(x+g(z)+f(y))=h(x)+h(y)+f(z)$ now from $Q(x,y,z),Q(y,x,z)$ we have ( from injective) : $g(x)=x+k$ so .... mind blown... did you mean $f(x)=x+k$ because $f(x)+y=f(y)+x$ because of injectivity?

Konigsberg wrote: alibez wrote: v_Enhance wrote: Find all triples $(f,g,h)$ of injective functions from the set of real numbers to itself satisfying \begin{align*} f(x+f(y)) &= g(x) + h(y) \\ g(x+g(y)) &= h(x) + f(y) \\ h(x+h(y)) &= f(x) + g(y) \end{align*} for all real numbers $x$ and $y$. (We say a function $F$ is injective if $F(a)\neq F(b)$ for any distinct real numbers $a$ and $b$.) Proposed by Evan Chen let $P(x,y): f(x+f(y))=g(x)+h(y)$ now put $Q(x,y,z)=P(x+g(z),y): f(x+g(z)+f(y))=h(x)+h(y)+f(z)$ now from $Q(x,y,z),Q(y,x,z)$ we have ( from injective) : $g(x)=x+k$ so .... mind blown... did you mean $f(x)=x+k$ because $f(x)+y=f(y)+x$ because of injectivity? from $Q(x,y,z)-Q(y,x,z)$ we have : $f(x+g(z)++f(y))=f(z+g(x)+f(y))$ now from injective we have : $g(x)+z=g(z)+x$ so we have : $g(x)=x+k$

Label the equations (1), (2), (3). For all real numbers $x, y$ we have \[\begin{aligned} f(x+f(y)+g(y)+h(y)) &\stackrel{(1)}{=} g(x + g(y) + h(y)) + h(y) \\ &\stackrel{(2)}{=} h(x + h(y)) + f(y) + h(y) \\ &\stackrel{(3)}{=} f(x) + g(y) + f(y) + h(y) \\ &\stackrel{(3)}{=} f(x) + h(y) + h(y + h(y)) \\ &\stackrel{(2)}{=} h(y) + g(y + h(y) + g(x)) \\ &\stackrel{(1)}{=} f(y + h(y) + g(x) + f(y)) \end{aligned} \]Since $f$ is injective, we get \[x + f(y) + g(y) + h(y) = y + h(y) + g(x) + f(y).\]Fixing $y$, we see that $g(x)$ is of the form $g(x) = x + c$, for some constant $c$. By symmetry, the same is true for $f$ and $h$. Let $f(x) = x + c_1$, $g(x) = x + c_2$, and $h(x) = x + c_3$ for constants $c_1, c_2, c_3$. Substituting into the original equations, we see that these functions satisfy the equations if and only if $2c_1 = c_2+c_3$, $2c_2 = c_1+c_3$, and $2c_3=c_1+c_2$. These conditions imply that $c_1 = c_2 = c_3$. Therefore, the solution set is given by $f(x) = g(x) = h(x) = x + c$, for any constant $c$.

alibez wrote: v_Enhance wrote: Find all triples $(f,g,h)$ of injective functions from the set of real numbers to itself satisfying \begin{align*} f(x+f(y)) &= g(x) + h(y) \\ g(x+g(y)) &= h(x) + f(y) \\ h(x+h(y)) &= f(x) + g(y) \end{align*}for all real numbers $x$ and $y$. (We say a function $F$ is injective if $F(a)\neq F(b)$ for any distinct real numbers $a$ and $b$.) Proposed by Evan Chen let $P(x,y): f(x+f(y))=g(x)+h(y)$ now put $Q(x,y,z)=P(x+g(z),y): f(x+g(z)+f(y))=h(x)+h(y)+f(z)$ now from $Q(x,y,z),Q(y,x,z)$ we have ( from injective) : $g(x)=x+k$ so .... $Q(x,y,z)=Q(y,x,z)$ gives $f(x)=x+k$...Anyways, nice solution

Note that the equations give: $f(x + f(y)) + f(x) + f(y) = g(x + g(y)) + h(x + h(y))$. Call this $Q(f, x, y)$. Then, $Q(f, x, y), Q(g, x, y), Q(h, x, y)$ gives us a system of equations. Considering $f(x + f(y)), g(x + g(y)), h(x + h(y))$ to be the unknowns and solving gives $f(x + f(y)) = (g(x) + g(y) + h(x) + h(y))/2$. Switching $x$ and $y$ and using injectivity gives $x + f(y) = y + f(x)$ i.e. $f(x) - x = f(y) - y$. After, that it is easy to get $f(x) \equiv g(x) \equiv h(x) \equiv x + c$, where $c$ is a fixed real constant.

Not too different than above solutions, but posting for storage Consider adding $f(x) + f(y)$ to the first equation, $g(x) + g(y)$ to the second, and $h(x) + h(y)$ to the third. Their respective right hand sides will become $g(y + g(x)) + h(y + h(x))$, $h(y + h(x)) + f(y + f(x))$, and $f(y + f(x)) + g(y + f(x)$. \begin{align*} f(x + f(y)) + f(x) + f(y) &= g(y + g(x)) + h(y+h(x)). (1) \\ g(x+g(y)) + g(x) + g(y) &= h(y + h(x)) + f(y + f(x)). (2)\\ h(x+h(y))+h(x)+h(y) &= f(y+f(x))+g(y+g(x)). (3) \end{align*} If we swap $x$ and $y$ in $(1)$, $(2)$, and $(3)$, we obtain \begin{align*} f(y + f(x)) + f(x) + f(y) = g(x + g(y)) + h(x+ g(y)). (1') \\ g(y+g(x)) + g(x) + g(y) = h(x + h(y)) + g(x + g(y)). (2') \\ h(y + h(x)) + h(x) + h(y) = f(x + f(y)) + g(x + g(y)). (3') \end{align*}Now for brevity, let $f(y + f(x)) - f(x + f(y)) = A$, $g(y + g(x)) - g(x + g(y)) = B$, and $h(y + h(x)) - h(x + h(y)) = C$. Taking $(3') - (3)$, $(2') - (2)$, and $(1') - (1)$ gives \begin{align*} C &= A + B \\ B &= A + C \\ A &= B + C \end{align*}Solving gives $A = B = C = 0$, so $f(x + f(y)) = f(y + f(x))$ implying $x + f(y) = y + f(x)$ by the injectivity condition. It is not hard to find that $f(x) = x + c$ works, so our solution set is $f, g, h \equiv \boxed{x + c}$. $\square$

Let $x,y,z,t\in R$. Then using given equations with other variables we get $$f(x+f(y)+g(z)+h(t))=g(x+g(z)+h(t))+h(y)=h(x+h(t))+f(z)+h(y)=f(x)+g(t)+f(z)+h(y)$$Switching variables $x,z$ we have $$f(x+f(y)+g(z)+h(t))=f(z+f(y)+g(x)+h(t))$$Because $f$ is injective $$x+g(z)=z+g(x)$$Plugging $z=0$ we obtain $$\forall x\in R\ g(x)=x+g(0)$$Given system of equations is cyclic in triple $(f,g,h)$, so analogously $$\forall x\in R\ f(x)=x+f(0)$$$$\forall x\in R\ h(x)=x+h(0)$$Now we plug this back to the given system of equations and we can easily solve it obtaining $$f(0)=g(0)=h(0)$$Hence we got all solutions.

As I look at this now I think it's the almost the same as first solution, but with one additional step when getting this multifunctional equation. Problem gives indeed strong assumptions, but the problem here is to handle all conditions wisely. There are many dead ends. For me one of them was playing around with plugging zero or something, that would give the zero value of some function.

Plug \(x=0\) in the first equation and plug \(y=f(y)\) in the second equation. Summing and swapping \(x-y\) (with injectivity) gives \(g(f(x))\) is linear (with leading coefficient 1). Plugging \(x=f(x)\) in the first equation and swapping \(x-y\) (with injectivity) gives \(h(x)\) is linear(with leading coefficient 1) because \(g(f(x))\) is linear. We can do the same for other couples of equation which gives all functions are linear(with leading coefficient 1). The rest is putting \(f(x)=x+a\), \(g(x)=x+b\), \(h(x)=x+c\) and finding all functions are \(x+a\).

My solution is not shorter but it is different . Let $\omega$ denote the cyclic sum with $f,g,h$ Just by switching $x$ with $y$ and summing up. We have $\omega (f(x+f(y)))=\omega (f(y+f(x)))........(1)$ Now, $f(y+f(x))=g(y)+h(x)..........(2)$ $g(x+g(y))=h(x)+f(y)..........(3)$ $(2)-(3)$ gives, $f(y+f(x))-g(x+g(y))=g(y)-f(y).........(4)$ Now just using (4) after a rearrangement by sending $g,h$ from $L.H.S$ to the $R.H.S$ and sending $h$ from $R.H.S$ to $L.H.S$. we get, $h(x)-f(x)=h(y)-f(y)$, and we can conclude similar statements via symmetry. And also now we can rewrite $h,g$ as $f+constant$. Now in the first equation of the given equation ,interchanging x and y and subtracting turns out to be miraculous. $f(x+f(y))-f(y+f(x))=g(x)-g(y)+h(y)-h(x)=0$ Now using injectivity , we have , $x+f(y)=y+f(x)$, which gives $f(x)=x+a$ and thus we have $g(x)=x+b$ and $h(x)=x+c$ Now plugging these back into our system. We get $a=b=c$. Thus we have $f(x)=g(x)=h(x)=x+a$, where $a$ is a constant.

\begin{align*} f(a+f(b+h(c)))&=g(a)+h(b+h(c)) \\ &=g(a)+f(b)+g(c) \\ &=g(c)+h(b+h(a))\\ &=f(c+f(b+h(a))), \end{align*}and thus by injectivity $a+f(b+h(c))=c+f(b+h(a))$. Here, we set $b=0$ and obtain that $f(h(a))=a+k$, where $k$ is just a constant. Also note that, \begin{align*} f(h(a))+g(b)&=h(h(a)+h(b)) \\ &=h(h(b)+h(a)) \\ &=f(h(b))+g(a), \end{align*}hence $g(a)-a=g(b)-b$, which means that $g(a)=a+l$ for some constant $l$. We cyclically obtain that $h(a)=a+m$ and $f(a)=a+n$, where $m,n$ are some constants. Hence, we finally obtain \begin{align*} 2n&= m+l \\ 2l &= m+n \\ 2m &= n+l, \end{align*}thus $m=n=l$. Our only solutions are $f(a)=a+l,g(a)=a+l,h(a)=a+l$.

$P(x+g(0),y)\Rightarrow f(x+f(y)+g(0))=h(x)+h(y)+f(0)$, switching $x,y$ gives $f(x+f(y)+g(0))=f(y+f(x)+g(0))$ or $x+f(y)=y+f(x)$. Then setting $y=0$ gives that $f$ is linear. Cyclically, $g,h$ are also linear. Plugging back in, we find that $f(0)=g(0)=h(0)$, so the answer is $\boxed{(x+c,x+c,x+c)}$ for any $c\in\mathbb R$.

$P(x+g(0),y)\Rightarrow f(x+f(y)+g(0))=h(x)+h(y)+f(0)$, switching $x,y$ gives $f(x+f(y)+g(0))=f(y+f(x)+g(0))$ or $x+f(y)=y+f(x)$. Plugging in $y=0$ gives $\deg(f) \equiv 1$ with leading coefficient$1$ By symmetry $\deg(f,g,h) \equiv 1$ with leading coefficient $1$ hence $(f(x),g(x),h(x)) \equiv (x+c,x+c,x+c)$

Huh I never wrote up a solution. We claim that the only valid triples are $(f,g,h) = (x+c, x+c, x+c)$. These clearly work, giving $x+2c$ on both sides of each of the equality conditions. Let $a=f(0), b=g(0), c=f(0)$. Then, plugging in $y=0$ gives \begin{align} f(x+a) &= g(x) + c\\ g(x+b) &= h(x) + a \\ h(x+c) &= f(x)+b \end{align}Thus, the first equation becomes \[f(x+ f(y)) = f(x+a) - c + f(y-c) +b \]Selecting $x = y-c-f(y)$ gives \[c-b = f(y-c-f(y)+a)\]Since $f$ is injective, this implies that $f-f(y)$ is constant. Thus, $f(y)=y-a$. Similarly, $g(x) = x+b, h(x) = x+c$. Plugging this into the original gives $2a=b+c, 2b=a+c, 2c=a+b$, which gives $a=b=c$, so we have shown that our claimed solution set are the only valid triples. $\blacksquare$.

In post #12, if I'm not mistaken, it should be \begin{align*} -C &= A + B \\ -B &= A + C \\ -A &= B + C \end{align*}and not \begin{align*} C &= A + B \\ B &= A + C \\ A &= B + C \end{align*}Also in post #16 primesarespecial wrote: Now just using (4) after a rearrangement by sending $g,h$ from $L.H.S$ to the $R.H.S$ and sending $h$ from $R.H.S$ to $L.H.S$. we get, $h(x)-f(x)=h(y)-f(y)$, How exactly do we get $h(x)-f(x)=h(y)-f(y)$ from "Now just using (4) after a rearrangement by sending $g,h$ from $L.H.S$ to the $R.H.S$ and sending $h$ from $R.H.S$ to $L.H.S$". Can anyone please elaborate this part?

Solved from the walkthrough of OTIS; here are my notes that i typed down, put together should be full solution Let a=f(0), b=g(0), c=h(0) Our preliminary checks of comparing if f=ax+i, g=bx+j, h=cx+k compare coefficients of them we get a=b, a^2=c, and cyclically implying slopes are 1; By plugging in x=y=0 and solving, we get a=b=c. f(x+a)=f(x+f(0))=g(x)+h(0)=g(x)+c and similarly for h(x+c)=f(x)+b etc. then plug it in to get f in terms of itself (leave to reader) and then to get stuff to cancel set x=y-c-f(y), which implies from injectivity that constant=c-b=f(y-f(y)+a-c), implying that y-f(y) must stay constant. This is now reduced to linear solutions, and we're done! It's easily verified that this works

First we put $x=0$ Now the system of equation looks like \begin{align} &f(f(y)) = g(0) + h(y) \\ & g(g(y)) = h(0) + f(y) \\ & h(h(y)) = f(0) + g(y) \end{align} now we put $h(y)$ in $(1)$ .So the thing becomes \begin{align*} f(f(h(y))) &= g(0) +h(h(y)) \\ &=g(0) + f(0) +g(y)\\ &= g(y) +h(h(0)) \\&=f(y+f(h(0))) \end{align*}But $f$ is injective and so $f(h(y)) = y+ f(h(0))....(4)$ Now note that rhs of $(4)$ takes all real values so for all real $r$ there exist $s\in \mathbb{R}$ so that $f(s) = r$ and so $f$ is surjective and hence there exist $f_0$ so that $f(f_0)=0 $ .Similarly we can show that such reals exist for $g$ and $h$ too and so $\exists$ $f_0, g_0 , h_0 \in \mathbb{R}$ so that $f(f_0)=g(g_0)=h(h_0) = 0 $ Now from $4$ we will get that \begin{align*} f(h(y)) &= f(f(f_0)+h(y)) \\ &= g(h(y))+h(f_0) \\ & = g(h(y)+g(g_0)) +h(f_0) \\&=f(g_0) + h(h(y)) + h(f_0) \\ &= f(g_0) + f(0) + g(y) +h(f_0) \end{align*}So, $ f(g_0) + f(0) + g(y) +h(f_0) = y+f(h(0))$ thus we show that $\boxed{g(y) = y+c_2}$.Similarly we show for the rest two and putting back in the main equations we get that $c_1=c_2=c_3$ giving us $\boxed{f=g=h\equiv x+c} $$ \square$

Nice problem . The answer is $f=g=h\equiv x+a$ for some real constant $a$. Note that $h(h(z))=g(z)+f(0)$. The following simplification is very useful: \[f(f(x)+g(y)+h(z))=h(x)+g(h(z)+g(y))\]\[=h(x)+h(h(z))+f(y)=f(0)+f(y)+g(z)+h(x)\]\[=h(y+h(z))+g(x+g(0))=f(x+g(0)+f(y+h(z)))\]So, $f(x)+g(y)+h(z)=x+g(0)+f(y+h(z))$, since $f(x)$ is injective. Plugging $y=0$ here gives, \[f(x)+h(z)=x+f(h(z))\]and $f(z)+h(z)=z+f(h(z))$. Subtracting the two equations gives $f(x)-f(z)=x-z\Longrightarrow f(x)-x=f(z)-z=\text{constant}$. Similar arguments gives $g(x)-g(z)=x-z$ and $h(x)-h(z)=x-z$. So, there are constants $a,b,c$ such that $f(x)=x+a$, $g(x)=x+b$ and $h(x)=x+c$. Now, \[f(x+f(y))=f(x+y+a)=x+y+2a=g(x)+h(y)=x+b+y+c\]which means $2a=b+c$. Similarly, $2b=c+a$ and $2c=a+b$. Hence, $a=b=c$, and $f=g=h\equiv x+a$ for some real number $a$. To verify, we just need to check whether $f(x+f(y))=f(x)+f(y)$ works. Note, \[f(x+f(y))=f(x+y+a)=x+y+2a=x+a+y+a=f(x)+f(y),\]so our solution does work. Hence the only such functions are $\boxed{f=g=h\equiv x+a}$ where $a$ is constant real number.

Only one assertion Let $P(x, y)$ denote the assertion. Then $P(x+g(0), y) \implies f(x+g(0)+f(y))=g(x+g(0))+h(y)=h(x)+f(0)+h(y)$. Then by swapping variables, we obtain $f(x+g(0)+f(y))=f(y+g(0)+f(x))$, which by $f$'s injectivity implies $x+f(y)=y+f(x)$. Thus, $f(x)=x+c$ for some $c \in \mathbb{R}$, and by symmetry, $g$ and $h$ are also of the form $x+c$. By substituting back into the original equation, we can obtain $\boxed{f(x)=g(x)=h(x)=x+c}$.

Let call this P(x,y) f(x+f(y))=g(x)+h(y) g(x+g(y))=h(x)+f(y) h(x+h(y))=f(x)+g(y) P(-f(0), 0) gives us f(0)=g(-f(0))+h(0) P(-g(0), 0) gives us g(0)=h(-g(0))+f(0) P(-h(0), 0) gives us h(0)=f(-h(0))+g(0) If we sums all of this three, we will get g(-f(0))+h(-g(0))+f(-h(0))=0 Now, P(-f(0), -g(0)) gives us f(-f(0)+f(-g(0)))+f(-h(0))=0 let call this W Now, P(-h(0), y) gives us h(-h(0)+h(y))=f(-h(0))+g(y) Now, P(x, (-f(0)+f(-g(0)))) gives us g(x+g(-f(0)+f(-g(0))))=h(x)+f(-f(0)+f(-g(0))) If we sume last two, we will get h(-h(0)+h(y))+g(x+g(-f(0)+f(-g(0))))=h(x)+g(y)+f(-h(0)) +f(-f(0)+f(-g(0))) Now-from W we get h(-h(0)+h(y))+g(x+g(-f(0)+f(-g(0))))=h(x)+g(y) Let call this Q(x,y) Q(0,0) gives us h(0)+g(g(-f(0)+f(-g(0))))=h(0)+g(0) g(g(-f(0)+f(-g(0))))=g(0) Now from injectivity, we get g(-f(0)+f(-g(0)))=0 Come back to Q(x,y) gives h(-h(0)+h(y))+g(x)=h(x)+g(y)-call this T(x,y) T(x,x) gives us h(-h(0)+h(x))+g(x)=h(x)+g(x) By injectivity -h(0)+h(x)=x, i.e x+h(0)=h(x) Obviously h(0) is constant, which we call c Now x+c=h(x) Analogically for g and f Checking f(x+f(y))=? g(x)+h(y) f(x+f(y))=f(x+y+c)=x+y+2.c Now g(x)+h(y)=x+c+y+c=x+y+2.c I.E everything is Okey in my solution