Outline: Easily get f(f(x))=x, f(x^2015)=f(x)^2015+f(0)^2015, f(x+y)=f(x)+f(y)-f(0). f is additive, So we need another condition. Idea: Try to calculate f((x+1)^2015-x^2015). Then f( {2014 deg polynomial for x} ) = {2014 deg polynomial for f(x)}. Since 2014 deg polynomial is bounded, we get bounding condition. Therefore f has form ax+b, and substituting we get f(x)=x or -x.

Isogonics wrote: Outline: Easily get f(f(x))=x, f(x^2015)=f(x)^2015+f(0)^2015, f(x+y)=f(x)+f(y)-f(0). f is additive, So we need another condition. Idea: Try to calculate f((x+1)^2015-x^2015). Then f( {2014 deg polynomial for x} ) = {2014 deg polynomial for f(x)}. Since 2014 deg polynomial is bounded, we get bounding condition. Therefore f has form ax+b, and substituting we get f(x)=x or -x. Are you sure that your solution is already true ? I can find $f(0)$ $f(0)=t$ so $f(t)=0$ . $f(u+v)+t=f(u)+f(v)$ so $f(2t)=-t$ in the question put $y=2t,x=t$ We have $f(0)=(2t)^{2015}$ now , put $x=t , y=0$ so $f(2t^{2015})=0=f(t)$ so $t=2t^{2015}$ so $2t^{2015}=2^{2015}t^{2015}$ so $t=0$

At the actual contest, I proved that $f$ is bijective, and got stuck there. I don't even get why $f(f(x)) = x$ is true.

rkm0959 wrote: At the actual contest, I proved that $f$ is bijective, and got stuck there. I don't even get why $f(f(x)) = x$ is true. $f(x^{2015} + (f(y))^{2015}) = (f(x))^{2015} + y^{2015}$ $f(y^{2015} + (f(x))^{2015}) = (f(y))^{2015} + x^{2015}$ so $ f(f(f(x^{2015} + (f(y))^{2015}))=x^{2015} + (f(y))^{2015} $ because $x^{2015} + (f(y))^{2015}$ can make all of real number so for every real number like $a$ we have $f(f(a))=a$

Wow, I shouldn't have walked out of the testing room so quickly. T^T Can you show me how to get the other equations?

Start with $$f(f(x^{2015}+f(y)^{2015}))=f(f(x)^{2015}+y^{2015})=x^{2015}+f(y)^{2015}$$and the fact that $x^{2015}$ is surjective over the reals to get $f(f(x))=x$ and that $f$ is bijective for all $x \in \mathbb{R}$. Now let $f(0)=a$, $f(a)=0$. Taking $P(x,a)$ gives $f(x^{2015})=f(x)^{2015}+a^{2015}$. Now taking $P(x,f(y))$, we get $$f(x^{2015}+y^{2015})=f(x)^{2015}+f(y)^{2015}=f(x^{2015})+f(y^{2015})-2a^{2015}$$or using that $x^{2015}$ is surjective, we have $$f(x+y)=f(x)+f(y)-2a^{2015}$$Plugging in $x=y=0$ gives $a=2a^{2015}$, so using that, we have $f(x+y)=f(x)+f(y)-a$ for all $x,y$. Claim: $a=0$. Proof of Claim: Using $x=y=a$, we have $f(2a)=-a$. Plugging in $x=a$, $y=2a$ gives $2a^{2015}=a=f(0)=(2a)^{2015}$, so $a=0$ as desired. Now we have $f(x+y)=f(x)+f(y)$ and $f(x)^{2015}=f(x^{2015})$. Surprise! By checking, we get $f(x) \equiv x$ and $f(x) \equiv -x$ as answers. $\blacksquare$

I think that in the proof of the claim, we must take $x=-a,y=2a$ And how was the equality $f(0)=(2a)^{2015}$ established?

We claim the solutions are $f(x)=\pm x$ which can easily be checked to work. Now, let $P(x,y)$ be the FE and suppose $f$ is a solution. Also, let $N=2015$ for convenience. Note that \[f(f(x^N+f(y)^N))=f(f(x)^N+y^N)=x^N+f(y)^N.\]By tuning $x$, we can make $x^N+f(y)^N$ take any value, so in fact $f\circ f=\mathrm{id}$. This means that the FE can be re-written as \[f(x^N+y^N)=f(x)^N+f(y)^N.\]Plugging $y=0$ into this, we get that $f(x^N)=f(x)^N+f(0)^N\equiv f(x)^N+k$. Thus, the FE becomes \[f(x^N+y^N)=f(x^N)+f(y^N)-2k,\]or simply $f(x+y)=f(x)+f(y)-2k$. Thus, we have $f(x)=cx+2k$ for all $x\in\mathbb{Q}$. Since $f$ is an involution, we have that $f(x)\equiv x$ or $f(x)\equiv 2k-x$ over $\mathbb{Q}$. Suppose it was $2k-x$. Then, since $f(x)^N+k=f(x^N)$, we have \[(2k-x)^N+k=2k-x^N\]for all $x\in\mathbb{Q}$. This implies that it is a polynomial identity, and by looking at the $x$ term, we see that $k=0$, so in all cases, $f(0)=0$. Thus, we have $f(x+y)=f(x)+f(y)$ and $f(x^N)=f(x)^N$ for $N\ge 2$. This is actually ELMO SL 2013 A1, but we'll do the proof here for completeness. Note that $f(qx)=qf(x)$ for any rational $q$. But we also have \[f((x+q)^N)=f(x+q)^N,\]which by the binomial theorem is equivalent to \[\sum_{i=0}^N\binom{N}{i}q^{N-i}(f(x^i)-f(x)^i)=0.\]This is true for all $q$ rational, so we in fact have $f(x^i)=f(x)^i$ for all $0\le i \le N$. Thus, $f(x^2)=f(x)^2\ge 0$, so $f$ is bounded on an interval, so $f(x)=cx$ for some $c$. We must have $f\circ f=\mathrm{id}$, so our only solutions are $f(x)\equiv x,-x$, as desired.

AHZOLFAGHARI wrote: rkm0959 wrote: At the actual contest, I proved that $f$ is bijective, and got stuck there. I don't even get why $f(f(x)) = x$ is true. $f(x^{2015} + (f(y))^{2015}) = (f(x))^{2015} + y^{2015}$ $f(y^{2015} + (f(x))^{2015}) = (f(y))^{2015} + x^{2015}$ so $ f(f(f(x^{2015} + (f(y))^{2015}))=x^{2015} + (f(y))^{2015} $ because $x^{2015} + (f(y))^{2015}$ can make all of real number so for every real number like $a$ we have $f(f(a))=a$ Quote: $ f(f(f(x^{2015} + (f(y))^{2015}))=x^{2015} + (f(y))^{2015} $ It's not correct. How do you show $f(f(x)) = x$? @Below lol, ok, thx. I was literally dizzy then. I guess sleeping is the solution.

@above That's just a typo. There should be two 'f'. Then, his/her argument works right.

Fun problem. We claim that the only solution is $f(x) = x$ and $-x$ which easily satisfies the equation. Let $P(x,y)$ be the assertion of $x$ and $y$ to the given problem. Now, $f$ is clearly injective : just check $P(k,a)$ and $P(k,b)$. $P(x,f(y))$ and swapping $x$ and $y$ gives us $f(f(n)) = n$ for any $n$ real number. Therefore, $P(x,f(y))$ and injectivity gives us $f$ is additive. Now, we then have $f(x^{2015}) + f(y^{2015}) = f(x)^{2015} + f(y)^{2015}$. This forces $f(x^{2015}) = f(x)^{2015}$. Now, we want to classify all functions which are additive and $f(x^{2015}) = f(x)^{2015}$. Since $f$ is additive, then $f(qx) = qf(x)$ for every rational number $q$. Furthermore, let's WLOG $f(1) = 1$. Now, \[ f((x+q)^{2015}) = f(x+q)^{2015} = (f(x) + q)^{2015}\]\[ f \left( \sum_{i = 0}^{2015} \binom{2015}{i} x^i q^{2015 - i} \right) = \sum_{i = 0}^{2015} \binom{2015}{i} f(x)^i 1^{2015 - i} \]\[ \sum_{i = 0}^{2015} f \left( \binom{2015}{i} x^i q^{2015 - i} \right) = \sum_{i = 0}^{2015} \binom{2015}{i} f(x)^i q^{2015 - i} \]\[ \sum_{i = 0}^{2014} f \left( \binom{2015}{i} x^i q^{2015 - i} \right) = \sum_{i = 0}^{2014} \binom{2015}{i} f(x)^i q^{2015 - i} \]\[ \sum_{i = 0}^{2014} \binom{2015}{i} q^{2015 - i} f(x^i) = \sum_{i = 0}^{2014} \binom{2015}{i} f(x)^i q^{2015 - i} \]\[ \sum_{i = 0}^{2014} \binom{2015}{i} q^{2015 - i} (f(x^i) - f(x)^i) = 0 \]for any rational number $q$. Therefore fixing a number $x$ and consider \[ P(n) = \sum_{i = 0}^{2014} \binom{2015}{i} n^{2015 - i} (f(x^i) - f(x)^i) \]is identical to the zero polynomial. This is enough to force $f(x^i) = f(x)^i$ for every $0 \le i \le 2014$. Hence, $f(x^2) = f(x)^2 \ge 0$ for any $x$. Therefore, we get that $f$ is additive and increasing, forcing $f(x) = kx$. Plugging back to our equation, we have $f(x) = x$ or $f(x) = -x$.

The answer is $f(x)=\pm x$, both of which clearly work. To start, note that by exploiting the symmetry in the condition, we have $$f(f(x^{2015}+f(y)^{2015}))=f(y^{2015}+f(x)^{2015})=x^{2015}+f(y)^{2015}.$$If we fix $y$ and vary $x$, $x^{2015}+f(y)^{2015}$ attains all real values, hence it follows that $f$ is an involution. Hence by substituting $y \to f(y)$, we obtain $$f(x^{2015}+y^{2015})=f(x)^{2015}+f(y)^{2015}.$$By letting $y=0$, we find that $f(x^{2015})=f(x)^{2015}+c^{2015}$, where $c:=f(0)$. Substituting, $$f(x^{2015}+y^{2015})=f(x^{2015})+f(y^{2015})-2c^{2015} \implies f(x+y)=f(x)+f(y)-2c^{2015}.$$Let this assertion be $Q(x,y)$. Setting $x=y=0$ yields $c=2c^{2015}$, so we can rewrite $Q(x,y)$ as $$f(x+y)=f(x)+f(y)-c.$$This implies that $g(x):=f(x)-c$ is additive, so $g(x)=kx$ for all $x \in \mathbb{Q}$. Further from $f(x^{2015})=f(x)^{2015}+c^{2015}=f(x)^{2015}+c/2$, by substitution we obtain $$g(x^{2015})+c=(g(x)+c)^{2015}+c/2 \implies kx^{2015}+c=(kx+c)^{2015}+c/2~\forall x \in \mathbb{Q}$$This is a polynomial equality in $x$, since it holds for infinitely many $x$. The coefficient of $x$ on the LHS is zero, and the coefficient of $x$ on the RHS is $2015kc^{2014}$, hence either $k=0$ or $c=0$. If $k=0$, then $f$ is constant over $\mathbb{Q}$, but this contradicts the fact that it is involutive (and thus injective). Thus $c=0$, so $k=\pm 1$ by substituting. Hence $f(x)=\pm x$ over $\mathbb{Q}$, and using $c=0$ we have the system of functional equations $$f(x+y)=f(x)+f(y) \text{ and } f(x^{2015})=f(x)^{2015}.$$This is ELMO Shortlist 2013/A3, but for completeness we provide a proof (also because I've never written a proof to the original). Since we previously proved $f(1)=\pm 1$, and $-f$ works iff $f$ works, WLOG let $f(1)=1$, so $f(x)=x~\forall x \in \mathbb{Q}$. Let $x$ be rational and $y$ be an arbitrary real. Then (from the second equation), $$f((x+y)^{2015})=f(x+y)^{2015}=(f(x)+f(y))^{2015}=(x+f(y))^{2015}.$$Since $f$ is additive, $f(qr)=qf(r)$ for all rational $q$ and real $r$. Using additivity and this property, we find that $$f((x+y)^{2015})=\sum_{i=0}^{2015} \binom{2015}{i}x^{2015-i}f(y^i)=\sum_{i=0}^{2015}\binom{2015}{i}x^{2015-i}f(y)^i=(x+f(y))^{2015}.$$Fixing $y$ as an arbitrary real, this is again a polynomial equality in $x$, hence by comparing the $x^{2013}$ coefficients we find that $f(y^2)=f(y)^2$. This means $f$ is bounded as $z \geq 0 \implies f(z)=f(\sqrt{z})^2 \geq 0$, hence $f$ is linear and thus $f(x)=x$. Since $-f$ works as well we also obtain $f(x)=-x$, as desired. $\blacksquare$

I claim that the only solutions are $\boxed{f(x) = \pm x}$. Denote by $P(x, y)$ the assertion of the equation. Applying $f$ to the initial equation, we obtain that \[f(f(x^{2015} + f(y)^{2015})) = f(f(x)^{2015} + y^{2015}) = x^{2015} + f(y)^{2015}. \]Now observe that if we fix $y$, then as $x^{2015}$ is surjective, we have $f$ must be surjective also. Further, we obtain that $f$ is an involution ($f(f(x)) = x$), and so $f$ is really a bijection. Now consider $P(x, f(y))$: \[P(x, f(y)) \rightarrow f(x^{2015} + y^{2015}) = f(x)^{2015} + f(y)^{2015}.\]As $x^{2015}$ is a surjection, we have \[f(x + y) = f(x) + f(y).\]Also consider $P(x, f(0))$: \[f(x^{2015}) = f(x)^{2015} + f(0)^{2015}.\]We easily obtain that as $f$ is additive, we must have $f(0) = 0$. Therefore we have that \[f(x + y) = f(x) + f(y) \text{ and } f(x^{2015}) = f(x)^{2015}.\] Let $q \in \mathbb{Q}$. Then we must have that \begin{align*} f(x + q) &= (f(x) + f(y))^{2015} \\ &= f(x)^{2015} + 2015f(q)f(x)^{2014} + \cdots + f(q)^{2015}, \\ \end{align*}and also that \begin{align*} f(x + q)^{2015} &= f((x + q)^{2015}) \\ &= f(x^{2015} + 2015x^{2014}q + \cdots + q^{2015}) \\ &= f(x)^{2015} + 2015qf(x^{2014}) + \cdots + f(1)q^{2015}. \\ \end{align*}Thus in comparing coefficients, we procure that $f(x^{2014}) = cf(x)^{2014}$, where $c = f(1)$. Now $cf(x)^{2014} \geq 0 \implies f(x^{2014}) \geq 0$, or if $c<0$ then the reverse inequality. Now as $f$ is additive and bounded, we have that $f$ is linear, and in particular, given by the equation $f(x) = cx$, where $c = f(1)$. We determine via substitution that $f(1) = \pm 1$, so that we obtain the desired/claimed solutions. $\blacksquare$

Solved by a walkthrough from someone anonymous (not a class teacher). Here is the walkthrough rephrased.

$\blacksquare$ Also there were so many brackets i had to put cuz 2015 in exponent... at this point i think it wouldve been smarter to substitute like for convenience v=2015 or smth