Comment: A1?

BlazingMuddy wrote: Taking $a = 1$ for this one, we quickly obtain that $2(f(b+1) - f(b)) = f(2) - f(0)$, so $f$ is linear. Could you please explain this step? How does it prove linearity?

Obviously, we have $ f(f(x)) = f(2x)+2f(0) $. Also, $ f(2x) = 2f(x)-f(0) $ by symmetry and so the condition becomes $$ f(2a)+f(2b) = f(2a+2b)+f(0) $$It is equivalent to $$ (f(2a+2b)-f(0)) = (f(2a)-f(0))+(f(2b)-f(0)) $$This means $ f $ is linear on even numbers. Notice that $ f(x) = (f(2x)+f(0))/2 $, $ f $ is linear on integers. The rest is easy.

Math-wiz wrote: BlazingMuddy wrote: Taking $a = 1$ for this one, we quickly obtain that $2(f(b+1) - f(b)) = f(2) - f(0)$, so $f$ is linear. Could you please explain this step? How does it prove linearity? Umm... the function is $\mathbb{Z} \rightarrow \mathbb{Z}$, so letting $m = \frac{f(2) - f(0)}{2}$, we can see that $f(b + 1) - f(b) = m$ for all $b\in \mathbb{Z}$ and then you prove by induction in two directions that $f(x) = mx + f(0)$ for all $x\in \mathbb{Z}$ (divide the case when $x$ is negative and when $x$ is positive).

The answer is $f \equiv 0$ or $f(x)=2x+c$ for some constant $c$. Since these are the only linear solutions, proving that $f$ is linear kills the problem. Putting $a=0$ gives $$f(f(b))=2f(b)+f(0)$$, or for every $y$ in the image of $f$, $f(y)=2y+f(0)$. Hence, $$2f(a+b)+f(0)=f(f(a+b))=f(2a)+2f(b).$$A simple variable-switching and some translation gives the well-known Cauchy equation. Hence $f$ is linear.

Yes, my solution involves the function $g(x)=f(x)-f(0)$ after we reach the last step in @above's solution. Cauchy ends it. EDIT: I used that $f(2a)=2f(a)-f(0)$

A great olympiad level problem and absolutely appropriate for IMO! It does reward training, effort, hard work and good preparation and it does not rely on creativity and insight (which are overrated anyway). I think that I can train a neural network to solve functional equations of that type.

Pretty sure 95% of the contestants solved this one. Bronze = 19

Wictro wrote: Pretty sure 95% of the contestants solved this one. Bronze = 19 Bronze cutoff depends on how hard it is tu get partials on medium problems, it usually has nothing to do with easy probs unless they are hard

Notice f(f(x)) = f(2x) + 2f(0) and f(2x) = 2f(x) - f(0) so f(2a) + f(2b) = f(2a+2b) + f(0) so thus 2(f(a+1) - f(a))=f(2) - f(0) thus f(2) - f(0) = 2m so f(a+1) -f(a) = m so by induction it follow that f(a) = m(a-1) + c with c=f(0). Plug into problem condition: m(2a-1) + c + 2(c + m(b-1)) = f( m(a+b-1)+c ) = mc + m^2(a+b-1)+ c, which is equivalent to: m^2(s-1) - m(2s-2) + c(m-2) = 0 with s = a+b. If m=0 then c=0 so f(x)=0 or s-1 | c(m-2) but s is infinite so m=2 (c=0 above) so f(x)=2x+c with c in Z

by symetry we get f(2a)+2f(b)=f(2b)+2f(a)=f(f(a+b)) 2f(a)+f(0)=f(fa)) so f(fa+b))=2f(a+b)+f(0) we get then 2f(a)-f(0)+2f(b)=2f(a+b)+f(0) so f(a)+f(b)=f(a+b)+f(0) so h(a)+h(b)=h(a+b) with h(x)=f(x)-f(0) by cauchy h(x)=cx so f(x)=cx+f(0) by verification in the equation we get c=0 or c=2 and f(0)=0 so f(x)=0 or f(x)=2x

onechaline wrote: by symetry we get f(2a)+2f(b)=f(2b)+2f(a)=f(f(a+b)) 2f(a)+f(0)=f(fa)) so f(fa+b))=2f(a+b)+f(0) we get then 2f(a)-f(0)+2f(b)=2f(a+b)+f(0) so f(a)+f(b)=f(a+b)+f(0) so h(a)+h(b)=h(a+b) with h(x)=f(x)-f(0) by cauchy h(x)=cx so f(x)=cx+f(0) by verification in the equation we get c=0 or c=2 and f(0)=0 so f(x)=0 or f(x)=2x F(0) need not to be 0

Swap $a$ and $b$, we have: $$f(2a)+2f(b)=f(2b)+2f(a)$$=>$f(2a)=2f(a)-f(0)$ =>$2f(a)+2f(b)-f(0)=f(f(a+b)$(1) =>$f(f(a))=2f(a)+f(0)$(2) Replace $a$ by $f(a)$ and $b=0$ in (1) and use (2), we have $f(0)=0$ =>$2f(a)+2(b)=f(f(a+b)=2f(a+b)$ =>$f(x)=cx+d$, c,d is constant. Test, we have $d=0$ and$c=0$ or $c=2$ So, we have two function:$f(x)=2x$, $f(x)=0$. Comment: It is too easy for IMO and follow my solution, the problem can replace to real number with continuos.

Very very extremely ridiculously standard FE. Set $a = 0$ to get $f(f(b)) = 2f(b) + f(0)$. Thus the given rewrites as $f(2a) + 2f(b) = 2f(a + b) + f(0)$. Now let $a = b$ to get $f(2a) = 2f(a) - f(0)$, so the given becomes $f(a) + f(b) = f(a + b) + f(0)$. Letting $g(x) = f(x) - f(0)$ this is just $g(a + b) = g(a) + g(b) $ over integers implying $g(x) = cx$ and $f(x) = cx + f(0)$. Now substitute and do boring calculations to get $c = 2$ or $c = 0$ and $f(x) \equiv 0$ or $f(x) \equiv 2x + d$ for an arbitrary integer $d$.

Let $P(x,y)$ be the given assertion of the function. $P(0,y)$ and $P(a,0)\implies f(2x)+z=2f(z) \forall a\in \mathbb{Z}$ where $f(0)=c$. So the equation is equivalent to $$2(f(a)+f(b))=c+f(f(a+b))\cdots(1)$$Setting $b=0$ we get $f(f(a))=2f(a)+c$. Thus, $(1)$ is actually $f(a)+f(b)=f(a+b)+c$. Now defining $g:\mathbb Z \mapsto \mathbb Z$ with $f(a)=g(a)+z$ we easily see that $g$ is additive and so $g(a)=ka+d$. Substituting and using that $g(0)=0$ we get that either $f(a)=2a+z$ or $k=0\implies f$ is constant and $f\equiv 0$.

15min to solve it( and I am really bad with FE)...is this really P1?

bel.jad5 wrote: 15min to solve it( and I am really bad with FE)...is this really P1? Exactly. Seems like APMO and IMO jury was the same . Even APMO P1 was a stupid one. Atleast they could keep it APMO P5 level difficulty

I feel that I am kinda biased towards saying that this is A0, but I think 2017A1 is as easy as this(?)

Are they going to have 2 functional equations in this year's IMO? We had 2 in APMO (though I think P1 is more of divisibility rather than functional).

We know Cauchy+Continuity $\implies $ Linearity. Place $a=n, b=0$ $$f(2n) + 2f(0)=f(f(n)) \dots (i)$$Place $a=2n, b= -n$ $$f(4n)+2f(-n)=f(f(n)) \dots (ii)$$So we have $f(2n)+2f(0)=f(4n)+2f(-n) \implies f(4n)-f(2n)=2(f(0)-f(-n))$ So $f$is linear. Now it is trivial to note that $f(x)=2x+c ~ \text{or} ~ f(x)=0$

Note that this FE still holds after a linear shift, so shift f(x) such that f(0)=0. Then by subbing in a=0, b=0 we get f(0)=0, and by subbing in alternatively a=0 b=b etc we get f(2x)=f(f(x))=2f(x), so our original FE becomes f(2a)+f(2b)=f(2a+2b), which is Cauchy on evens. By f(0)=0 we have that f(x) is in the form f(x)=cx and testing all c, only c=0 and c=2 work (no piecewise since Cauchy). Assume f(x) isn't just the zero function, then pick a even number not divisible by 4 call it N, let N=2n. We try to find f(x) for odd x, plug in an odd number call it a. f(N)+2f(a)=f(f(n+a)), n+a is even since both terms are odd, so the LHS is just 4n+2f(a) and the RHS is just 4n+4a, so 2f(a)=4a so f(a)=2a for odd a. Therefore we win. Taking the linear shift back we get that f(x)=2x+k for any k, or f(x)=0 constantly.

A fanumtastic and interesting IMO problem! Here is my solution. ||Letting $a=0$ gives $f(f(b))=2f(b)$. Also, $f(2a)+2f(0)=f(f(a))=2f(a)$. Now, let $b=a+b$ and $a=0$ so $f(0)+2f(a+b)=f(f(a+b))$, but also $f(0)+2f(a+b)=f(2a)+2f(b)$. So $2f(a+b)+3f(0)=2f(a)+2f(b)$. This is Cauchy's equation with a shift vertically. So I am done and the solutions are $f(x)=2x+c$ where $c$ is any constant.

Let $P(x,y)$ denote the assertion: $$f(2a)+2f(b)=f(f(a+b))$$$$P(x,0) \Rightarrow f(2x)+2f(0)=f(f(x))$$$$P(0,x) \Rightarrow f(0)+2f(x)=f(f(x))$$Combining the two equations, we get: $$\boxed{f(2x)+f(0)=2f(x)} \ldots (i)$$From the original FE, using $(i)$, we have: $$f(2a)+2f(b)=f(f(a+b)) \Rightarrow 2f(a)-f(0)+2f(b)=f(0)+2f(a+b)$$$$\Rightarrow 2f(a)-2f(0)+2f(b)=2f(a+b) \Rightarrow f(a)-f(0)+f(b)-f(0)=f(a+b)-f(0)$$Let $g(x)=f(x)-f(0)$ $$\Rightarrow g(a)+g(b)=g(a+b)$$(Cauchy's functional equation over integers) $$\Rightarrow g(x)=kx\Rightarrow f(x)=kx+f(0)$$Let $f(0)=c \Rightarrow f(x)=kx+c$ Substituting the value of $f(x)$ in the original equation, we get: $2ka+c+2kb+2c=k^2(a+b)+(k+1)c$ Solving this equation for $k$, we get $\boxed{k=2 \text{ or } k=c=0}$ Therefore, the solutions are: $\boxed{f(x)=2x+c}$ or $\boxed{f(x)\equiv 0}$, and we're done!

The answer is $f \equiv 0 $ or $f(x) = 2x + c$ for some integer $c$, both of which work. Let $P(a, b)$ denote the original assertion. From $P(a, b)$ and $P(0, a + b)$ we have $$f(2a) + 2f(b) = f(f(a + b)) = f(0) + 2f(a + b).$$Thus $f(2a) + 2f(b) = f(0) + 2f(a + b)$. If we let $g \equiv f - f(0)$, then subtracting $3f(0)$ from both sides of this equation gives $g(2a) + 2g(b) = 2g(a + b)$. Since $g(0) = f(0) - f(0) = 0$, by setting $b = 0$ we have $g(2a) = 2g(a)$. Therefore $2g(a) + 2g(b) = 2g(a + b)$, or $g(a) + g(b) = g(a + b)$. Since $g$ is from $\mathbb{Z}$ onto $\mathbb{Z}$, this implies that $g(x) = mx$ for some integer $m$ by Cauchy's FE. Letting $n = f(0)$, we have $f(x) = mx + n$; plugging this back into the original equation yields $$2am + 2bm + 2n = m^2(a + b) + mn.$$Since this must hold for all $a$ and $b$, we have $2m = m^2$ and $2n = mn$. Taking cases on $m$ yields the solutions described at the start.

Easiest IMO prob in recent years (maybe tied with IMO 2017/1)

I claim that the answer is that the only solutions are $f(x)=0\forall x\in \mathbb{Z}$ and $f(x)=2x+c\forall x\in \mathbb{Z}$, where $c$ is any integer. First, we plug in $(0,x)$ to get that \[f(0)+2f(x)=f(f(x)),\]\[\implies f(f(a+b))=2f(a+b)+f(0).\] Then, we can plug in $(a,a)$ to get that \[f(2a)+2f(a)=f(f(2a)),\]and plug in $(0,2a)$ to get that \[f(0)+2f(2a)=f(f(2a)),\]and setting these two to be equal, we can get that \[f(2a)+2f(a)=f(0)+2f(2a),\]\[\iff f(2a)=2f(a)-f(0).\] Substituting our above equalities into the equation gives that \[f(2a)+2f(b)=f(f(a+b)),\]\[\iff [2f(a)-f(0)]+2f(b)=[2f(a+b)+f(0)],\]\[\iff f(a)+f(b)=f(a+b)+f(0),\]and if we let $g(x)=f(x)+f(0)$, we get that \[g(a)+g(b)=g(a+b).\] By Cauchy's FE over the integers, we know that all solutions to $g(x)$ are in the form of $cx$ for some constant integer $c$. This gives us that \[g(x)=cx,\]\[\iff f(x)=cx+d,\]where $d$ is an integer. Plugging this back into the original equation, we get that \[f(2a)+2f(b)=f(f(a+b)),\]\[\iff (2ac+d)+2(bc+d)=c(c(a+b)+d)+d,\]\[\iff 2c(a+b)+3d=c^2(a+b)+(c+1)d.\]Clearly, if the two sides are equal, since they are both linear, they must have the same slope. This occurs when \[2c=c^2,\]which only occurs at $c=0$ and $c=2$. Plugging in the former gives that $d=0$ only and plugging in the latter gives that $d$ can be any integer. Finalizing this, this means that the only solutions $f$ are $f(x)=0\forall x\in \mathbb{Z}$ and $f(x)=2x+c\forall x\in \mathbb{Z}$, where $c$ is any integer, as desired. This finishes the problem.

The answer is $f(x)=2x+c$ for $c\in\mathbb{Z}$ and $f(x)=0$, and it is easy to check that these satisfy the equation. Let $P(x,y)$ denote the assertion. $P(a,2a)$ gives us $3f(2a)=f(f(3a))$. Then, $P(a,0)$ gives $f(2a)+2f(0)=f(f(a))$ and $P(0,a)$ gives $f(0)+2f(a)=f(f(a))$, so we get $f(2a)+f(0)=2f(a)$. Now, adding $f(0)$ to both sides of the original equation, we get \[f(2a)+2f(b)+f(0)=2f(a)+2f(b)=f(f(a+b))+f(0)=3f\left(\frac{2}{3}(a+b)\right) + f(0).\]Now, dividing both sides by $2$ and subtracting $2f(0)$ from both sides gives us \[g(a)+g(b)=\frac{3}{2}g\left(\frac{2}{3}(a+b)\right),\]where $g(x)=f(x)-f(0)$. Plugging in $b=2a$, we get $g(a)+g(2a)=\frac{3}{2}g(2a)$, giving $2g(a)=g(2a)$, giving us $2^ng(a)=g(2^na)$ by induction. From this, we get \[g(a)+g(b)=3g\left(\frac{a+b}{3}\right),\]and by plugging in $b=8a$, we get $g(a)+g(8a)=9g(a)=3g(3a)$, so $g(3a)=3g(a)$, which gives us \[g(a)+g(b)=g(a+b).\]Now, since $g$ satisfies Cauchy and is over $\mathbb{Z}$, we have that $g(x)=kx$, giving $f(x)=kx+c$. Plugging this into the original equation, we get \[2k(a+b)+3c=k^2(a+b)+c(k+1),\]so either $k=0$ or $k=2$ and either $c=0$ or $k=2$ in order for both $k$ and $c$ to work. We see that the only cases which work is when $(k,c)=(2,c), (0,0)$, so $f(x)=2x+c$ and $f(x)=0$ are our only solutions.

We claim that the only answers are $\boxed{f(x) = 0}$ and $\boxed{f(x) = 2x + c}$ for any constant $c$. $\underline{P}(a, a) \implies \boxed{f(2a) + 2f(a) = f(f(2a))} \ldots \ldots \textcircled{1}$ $\underline{P}(a, 0) \implies \boxed{f(2a) + 2f(0) = f(f(a))} \ldots \ldots \textcircled{2}$ $\underline{P}(0, a) \implies \boxed{f(0) + 2f(a) = f(f(a))} \ldots \ldots \textcircled{3}$ $\textcircled{2} - \textcircled{3} \implies \boxed{f(2a) + f(0) = 2f(a)}$ $\textcircled{1} \implies f(f(2a)) = f(2a) + f(2a) + f(0) \implies f(f(a)) = 2f(a) + f(0)$ $\implies 2 f(a+b) + f(0) = f(2a) + 2f(b) = 2f(a) + 2f(b) - f(0)$ $\implies f(a+b) + f(0) = f(a) + f(b)$ $\implies f(a+b) - f(0) = (f(a) - f(0)) + (f(b) - f(0))$ Let $\boxed{g(x) := f(x) - f(0)}$. Then the above equation implies $g(a+b) = g(a) + g(b)$. Since $g$ satisfies Cauchy's additive functional equation over $\mathbb{Z}$, it must be of the form $g(x) = kx$. Hence $f(x) = g(x) + f(0) = kx + c$. Substituting this back into the original equation, we get the desired solutions.

Let a = 0, b = x. We get that $f(0) + 2f(x) = f(f(x))$. Let a = x, b = 0. We get $f(2x) + 2f(0) = f(f(x))$. By these two we get that $f(0) + 2f(x) = f(2x) + 2f(0)$ $\Rightarrow$ $2f(x) = f(2x) + f(0)$. Now we want to plug that in the original equation $f(2a) + 2f(b) = f(f(a+b))$ and we get $2f(a) - f(0) + 2f(b) = f(0) + 2f(a+b)$. From that we have $f(a) + f(b) = f(0) + f(a+b)$ and we can transform it to $f(a) - f(0) + f(b) - f(0) = f(a+b) - f(0)$. Let $g(x) = f(x) - f(0)$ $\Rightarrow$ we get that $g(a) + g(b) = g(a+b)$, which is Cauchy equation over integers $\Rightarrow$ $g(x) = kx$ $\Rightarrow$ $f(x) = kx + f(0)$ $\Rightarrow$ $f(x) = kx + c$. Now plugging in the original equation we get that $k^2(a+b) + c(k+1) = 2k(a+b) + 3c$. If $k > 2$, we get that $LHS > RHS$ - impossible. Now checking the values for k we get that k = 2 works and if k = 0, c = 3c $\Rightarrow$ c = 0 $\Rightarrow$ $f(x) = 2x + c$ or $f(x) = c = 0$ $\Rightarrow$ $f(x) = 2x + c$ and $f(x) = 0$ are the only solutions. Checking these work $\Rightarrow$ we are ready.

The solutions are $f(x) = 2x+C$ for some integer constant $C$ and $f(x) = 0$. Taking $P(0,a)$ gives us that $f(f(a)) = f(0) + 2f(a)$. Substituting this for the RHS of the given equation, we have that \[f(2a)+2f(b) = f(0) + 2f(a+b).\]Setting $a=b$ in the equation above gives us that $f(2a) = 2f(a) - f(0)$ – substituting that into the above equation gives us \[f(a)+f(b) = f(0) + f(a+b),\]which means that $f(x)-f(0)$ satisfies Cauchy's equation – therefore, $f(x)$ is linear. Plugging this in gives us the claimed solutions.

I claim the only solutions are $f(x) = 2x + c$ for all positive integers $c$. It is trivial to verify these all work. Let $P(a,b)$ be the assertion, then we take $P(a,b), P(b,a)$ and subtract to get $f(2a) - f(2b) = 2f(a) - 2f(b) $, setting $b = 0$ gives $f(2a) = 2f(a) - f(0)$. We can then write $2f(a) + 2f(b) = f(f(a + b)) + f(0)$. So we have $f(a) + f(b) = f(a + 1) + f(b -1)$ for $a + b =a + 1 + b - 1$, so we can write $f(a + 1) - f(a ) = f(b) - f(b - 1)$ . Thus $f$ is linear, since all consecutive differences are equal. Plugging in $f(x) = kx + c$ gives $2k(a + b) + 3c = k^2(a + b) + kc + c$, forcing $k = 2$, so we are done.

lminsl wrote: The answer is $f \equiv 0$ or $f(x)=2x+c$ for some constant $c$. Since these are the only linear solutions, proving that $f$ is linear kills the problem. Putting $a=0$ gives $$f(f(b))=2f(b)+f(0)$$, or for every $y$ in the image of $f$, $f(y)=2y+f(0)$. Hence, $$2f(a+b)+f(0)=f(f(a+b))=f(2a)+2f(b).$$A simple variable-switching and some translation gives the well-known Cauchy equation. Hence $f$ is linear. Hmm same solution , I messed up the translation part oof .

Let $P(a,b)$ be the given relation. Comparing $P(n,n+m)$ and $P(m,2n)$, we obtain $2f(m+n)=f(2m)+f(2n)$ for all $m,n\in\mathbb{Z}$. Call this relation $Q(m,n)$. Plugging in $Q(k-1,k+1)$ yields $f(2k+2)=2f(2k)-f(2k-2)$, so $f$ is linear over the even integers. Plugging in $Q(k,0)$ yields $2f(k)=f(2k)+f(0)$, so $f$ is linear over $\mathbb{Z}$. By inspection it follows that $f$ is either the zero function or $f=2x+c$ for all $x$, where $c$ is a fixed constant independent of $x$. $\blacksquare$