From Cauchy's functional equation, $f(q_0x) = q_0f(x)$ for any rational number $q_0$ and real $x$; we for any can then write \begin{align*} [f(x + q_0)]^{2013} &= [f(x) + f(q_0)]^{2013} \\ &= f(x)^{2013} + 2013f(x)^{2012}f(q_0) + \dots + f(q_0)^{2013} \\ &= f(x)^{2013} + 2013q_0f(1)f(x)^{2012} + \dots + [q_0f(1)]^{2013} \end{align*} and also \begin{align*} [f(x + q_0)]^{2013} &= f[(x + q_0)^{2013}] \\ &= f(x)^{2013} + 2013f(x^{2012}q_0) + \dots + f(q_0^{2013}) \\ &= f(x)^{2013} + 2013q_0f(x^{2012}) + \dots + [q_0f(1)]^{2013} \end{align*} For $x$ fixed, the difference between these two expressions is a polynomial in $q_0$; since it has infinitely many zeroes (these two expressions are equal for all rational $q_0$) it must be identically zero. Comparing linear coefficients we find $f(x^{2012}) = f(1)f(x)^{2012}$, so that $f$ has the same sign as $f(1)$ on the non-negative reals. This establishes boundedness, which together with Cauchy's FE means all solutions are of the form $f(x) = cx$ for a constant $c$. Substituting, we require $c^{2013} = c$ for $c$ real. Clearly $c \in \{-1, 0, 1\}$ all of which give valid solutions (i.e. $f(x) = -x$, $f(x) = 0$, and $f(x) = x$).

hyperbolictangent wrote: From Cauchy's functional equation, $f(q_0x) = q_0f(x)$ for any rational number $q_0$ and real $x$; we for any can then write \begin{align*} [f(x + q_0)]^{2013} &= [f(x) + f(q_0)]^{2013} \\ &= f(x)^{2013} + 2013f(x)^{2012}f(q_0) + \dots + f(q_0)^{2013} \\ &= f(x)^{2013} + 2013q_0f(1)f(x)^{2012} + \dots + [q_0f(1)]^{2013} \end{align*}and also \begin{align*} [f(x + q_0)]^{2013} &= f[(x + q_0)^{2013}] \\ &= f(x)^{2013} + 2013f(x^{2012}q_0) + \dots + f(q_0^{2013}) \\ &= f(x)^{2013} + 2013q_0f(x^{2012}) + \dots + [q_0f(1)]^{2013} \end{align*}For $x$ fixed, the difference between these two expressions is a polynomial in $q_0$; since it has infinitely many zeroes (these two expressions are equal for all rational $q_0$) it must be identically zero. Comparing linear coefficients we find $f(x^{2012}) = f(1)f(x)^{2012}$, so that $f$ has the same sign as $f(1)$ on the non-negative reals. This establishes boundedness, which together with Cauchy's FE means all solutions are of the form $f(x) = cx$ for a constant $c$. Substituting, we require $c^{2013} = c$ for $c$ real. Clearly $c \in \{-1, 0, 1\}$ all of which give valid solutions (i.e. $f(x) = -x$, $f(x) = 0$, and $f(x) = x$). I think you established increasing-ness that will show $f(x)=cx$ not boundedness?

TheDarkPrince wrote: I think you established increasing-ness that will show $f(x)=cx$ not boundedness? Increasing and additive implies linear, so yes.

hmm I have a weird solution (why do I have so many weird solutions nowadays). Also, 3-year bump . The answers are $f(x) \equiv 0$, $f(x) \equiv x$, and $f(x) \equiv -x$, which all work. First, we claim that $f$ is linear. We can easily check that if $f(x)$ satisfies the conditions, $-f(x)$ does too. So, we can assume WLOG that $f(1)$ is nonnegative. Let $x$ be a real number. We have \begin{align*} f\left((x+1)^{2013}+(-x+1)^{2013}\right) &= f\left((x+1)^{2013}\right)+f\left((-x+1)^{2013}\right) \\ &= f(x+1)^{2013}+f(-x+1)^{2013} \\ &= (f(x)+f(1))^{2013}+(-f(x)+f(1))^{2013} \\ &= \sum_{i=0}^{2013} \dbinom{2013}{i}f(x)^if(1)^{2013-i}+\sum_{i=0}^{2013} (-1)^{i}\dbinom{2013}{i}f(x)^if(1)^{2013-i} \\ &= 2\sum_{i=0}^{1006} \dbinom{2013}{2i}f(x)^{2i}f(1)^{2013-2i}, \end{align*}which is nonnegative. Since $(x+1)^{2013}+(-x+1)^{2013}$ can range from $2$ (when $x=0$) to arbitrarily large, we know that $f(x)$ is bounded on a nontrivial interval, so $f(x) \equiv f(1)x$ from Cauchy's functional equation. Since $f(1^{2013})=f(1)=f(1)^{2013}$, we know that $f(1) \in \{-1,0,1\}$. Thus, the possible answers are $f(x) \equiv 0$, $f(x) \equiv x$, and $f(x) \equiv -x$.

tigerzhang wrote: hmm I have a weird solution (why do I have so many weird solutions nowadays). Also, 3-year bump . The answers are $f(x) \equiv 0$, $f(x) \equiv x$, and $f(x) \equiv -x$, which all work. First, we claim that $f$ is linear. We can easily check that if $f(x)$ satisfies the conditions, $-f(x)$ does too. So, we can assume WLOG that $f(1)$ is nonnegative. Let $x$ be a real number. We have \begin{align*} f\left((x+1)^{2013}+(-x+1)^{2013}\right) &= f\left((x+1)^{2013}\right)+f\left((-x+1)^{2013}\right) \\ &= f(x+1)^{2013}+f(-x+1)^{2013} \\ &= (f(x)+f(1))^{2013}+(-f(x)+f(1))^{2013} \\ &= \sum_{i=0}^{2013} \dbinom{2013}{i}f(x)^if(1)^{2013-i}+\sum_{i=0}^{2013} (-1)^{i}\dbinom{2013}{i}f(x)^if(1)^{2013-i} \\ &= 2\sum_{i=0}^{1006} \dbinom{2013}{2i}f(x)^{2i}f(1)^{2013-2i}, \end{align*}which is nonnegative. Since $(x+1)^{2013}+(-x+1)^{2013}$ can range from $2$ (when $x=0$) to arbitrarily large, we know that $f(x)$ is bounded on a nontrivial interval, so $f(x) \equiv f(1)x$ from Cauchy's functional equation. Since $f(1^{2013})=f(1)=f(1)^{2013}$, we know that $f(1) \in \{-1,0,1\}$. Thus, the possible answers are $f(x) \equiv 0$, $f(x) \equiv x$, and $f(x) \equiv -x$. This is a wonderful answer.

The following solution is by Daniel Xia. It is much more slick than my own. By scaling, assume $f(1)=1$ (if $f(1)=0$, the proof is similar). Notice that $$f((1+x)^{2013} +(1-x)^{2013})=(f(1)+f(x))^{2013}+(f(1)-f(x))^{2013}=P(f(x))$$where $P(t)=(1+t)^{2013}+(1-t)^{2013}$ is a polynomial of degree $2012$ (the leading term is $4026t^{2012}$). Since $P$ has even degree, it is bounded below. So $f$ is bounded on a nontrivial interval (actually the image of $P$), hence linear.

The answer is $f(x)=\pm x$ and $f \equiv 0$, all of which clearly work. Let $x$ be rational and $y$ be real, so $f(x)=cx$ for some constant $c$. The second equation gives us $$f((x+y)^{2013})=f(x+y)^{2013}=(cx+f(y))^{2013}$$Since $f$ is additive, $f(qr)=qf(r)$ for all rational $q$ and real $r$. Using additivity and this property, we find that $$f((x+y)^{2013})=\sum_{i=0}^{2013} \binom{2013}{i}x^{2013-i}f(y^i)=\sum_{i=0}^{2013}\binom{2013}{i}(cx)^{2013-i}f(y)^i=(x+f(y))^{2013}.$$Fixing $y$ as an arbitrary real, this is again a polynomial equality in $x$, hence by comparing the $x^{2011}$ coefficients we find that $f(y^2)=c^{2011}f(y)^2$. This means $f$ is bounded on $\mathbb{R^+}$, since $f$ has the same sign as $c^{2011}$ (or is zero) over that interval. Hence $f$ is linear, so in fact $f(x)=cx$ over $\mathbb{R}$. The second equation then gives $c^{2013}=c$, or $c \in \{-1,0,1\}$, which is the advertised solution set.

An important detail that no one wrote correctly. Unless $f\equiv 0$, $f$ is not bounded over $\mathbb{R}^+$, it is either bounded below (if $c>0$) or bounded above (if $c<0$). Bounded means bounded below and above.

GianDR wrote: Bounded means bounded below and above. https://en.wikipedia.org/wiki/Abuse_of_notation

IAmTheHazard wrote: GianDR wrote: Bounded means bounded below and above. https://en.wikipedia.org/wiki/Abuse_of_notation This has absolutely no relation with what I said. Abuse of notation is (for example) when you don't use different symbols for the multiplication between two scalars and the multiplication of a scalar times a vector in a vector space, or when you omit symbols in order to get a cleaner notation, or any of the other examples given in the link. Saying a function is bounded when it's only bounded below or above is not abuse of notation, it's just wrong.

GianDR wrote: Saying a function is bounded when it's only bounded below or above is not abuse of notation, it's just wrong. I think it is convention, at least to the olympiad community.

ZETA_in_olympiad wrote: GianDR wrote: Saying a function is bounded when it's only bounded below or above is not abuse of notation, it's just wrong. I think it is convention, at least to the olympiad community. Hmm, I never saw this convention when I was a contestant. If you are a current contestant, then I'll accept this convention, but I must emphasize that writing "bounded" when you only have below or above boundedness is not mathematically correct outside the Olympiad (and it really shouldn't be considered correct inside the Olympiad either, but meh).

@GianDR The only place a contestant often uses boundedness in olympiad algebra is to justify linearity using Cauchy's FE, which, in my humble opinion, is not too bad. But I see your point that it is used incorrectly and I agree with it By the way, I am not a current contestant, but was just a while back. I only recently have some experience with functional equations, though.

From my experience, saying $f$ is bounded and saying $f$ is bounded from one direction are two different things and I have to learn it the hard way -- I wrote the former instead of the latter and result to a -4 for that problem that is otherwise correct.

See in Functional Equations by B J VENKATACHALA.

When I saw this on OTIS Excerpts, I thought it was an instasolve. First of all, from the additive identity, we get f(x)=kx. Then expanding and substituting, we have kx^2013=(kx)^2013 for all x, for nonzero x we have k=k^2013. Now, either k=0, so f(x)=0, or divide by k and we get k^2012-1=0. Since roots of unity are complex roots except for -1, 1, k=-1,1 (since the function only maps reals to reals). Our final solutions are f(x)=0,-x,x. @below sorry, I forgot that Cauchy’s equation was only over rationals unless over reals you need to prove boundedness. I’ll try to get it fixed

@above additive doesnt mean linear over reals, you have to establish boundedness over some interval first

Here is my revision: (Sorry I’m on and off of aops because busy so it’ll take a while to fix) From the additive function, we have f(qm)=qf(m) for rational q and real m. We have $\sum_{k=0}^{2013}\dbinom{2013}{k}(f(1)q)^{2013-k}f(x))^k=\sum_{k=0}^{2013}\dbinom{2013}{k}f(q)^{2013-k}f(x)^k=(f(x)+f(q))^{2013}=f(x+q)^{2013}=f((x+q)^{2013})=f(\sum_{k=0}^{2013}\dbinom{2013}{k}q^{2013-k}x^k=\sum_{k=0}^{2013}\dbinom{2013}{k}q^{2013-k}f(x^k)$. Then for arbitrary x, LHS-RHS=0, and because x is arbitrary, we must match up …f(x^i) with …f(x^i) since if otherwise f(x^i) with f(x^j) will not be equal, unless f is a constant, which gives us the solution f(x)=0x (must also satisfy Cauchy’s equation). Now when we match them up, we get f(x^2012)=f(1)f(x)^2012, so f(x^2012) has the same sign as f(1), and therefore f(x) has the same sign as f(1) on nonnegative numbres. Then there is a bound (either positive or negative, depending on f(1)) on f for the interval nonnegative reals, and we can now apply Bounded+Cauchy=Linear. Now, continue as my above post. This latex takes a VERY LONG TIME to write, so I can’t be bothered to latex any more of it.

The answer is $f(x) = 0, x, -x$. The key is to expand both sides of $f(x+q)^{2013}) = f(x+q)^{2013}$: let $f(1) = c$ and set $q \in Q$, so $$\sum q^{2013-k} {2013 \choose k} f(x^k) = (f(x+q)^{2013}) =f(x+q)^{2013} = \sum {2013 \choose k} (cq)^{2013-k} f(x)^k.$$By viewing this as a polynomial in $q$, it follows that $f(x^k) = c^{2013-k} f(x)^k$. By setting $k$ to be even, this implies $f(x)$ has the same sign as $c$ for $x$ nonnegative. Thus $f$ is bounded on a nontrivial interval and follows $f$ is linear, which yields the solution set.

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