The first step is to establish that $f(0) = 0$. Putting $x = y = 0$, and $f(0) = t$, we get $f(t) = t^2$. Also, $f(x^2+t) = f(x)^2$, and $f(f(x)) = x + t^2$. We now evaluate $f(t^2+f(1)^2)$ two ways. First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$. Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$. So $t = 0$, as required. It follows immediately that $f(f(x)) = x$, and $f(x^2) = f(x)^2$. Given any $y$, let $z = f(y)$. Then $y = f(z)$, so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$. Now given any positive $x$, take $z$ so that $x = z^2$. Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$. Putting $y = -x$, we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$. Hence $f(-x) = - f(x)$. It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$, $y$. Take any $x$. Let $f(x) = y$. If $y > x$, then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$. If $y < x$, then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$. In either case we get some $z > 0$ with $f(z) = -z < 0$. But now take $w$ so that $w^2 = z$, then $f(z) = f(w^2) = f(w)^2 \ge 0$. Contradiction. So we must have $f(x) = x$

Set x = 0 to get $f(f(y)) = y+f(0)^{2}$. We'll let $c = f(0)^{2}$, so $f(f(y)) = y+c$. Then $f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$ $f(a^{2}+b+c) = f(b)+f(a)^{2}$ $f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$ $a^{2}+b+2c = b+(a+c)^{2}$ $2c = c^{2}+2ac$ Since this holds for all $a$, it follows that $c = 0$. Now we have $f(0) = 0 \implies f(f(y)) = y$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$, then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$. Finally, if $u > v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x.

$k=x_{1}+x_{2}+...+x_{n}$

orl wrote: Let $ \,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $ \,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2} + f(y)\right) = y + \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.\] Denote with $ P(x,y)$ the property $ f(x^2+f(y))=y+f^{2}(x)$ Let $ c=f^{2}(0)$. So we have $ P(0,x) \Rightarrow f^{(2)}(x)=x+c\;\;\;x\in \mathbb{R} \: (*)$ where $ f^{(2)}(x)=f(f(x)).$ Consider $ a,b\in \mathbb{R}$ with $ a\ne b$ and $ f(a)=f(b)$. So we have $ f^{(2)}(a)=f^{(2)}(b) \Rightarrow a+c=b+c \Rightarrow a=b.$ This is a contraddiction. So it must be $ f(a)\ne f(b)$ when $ a \ne b$ and $ f(x)$ must be injective. Consider now $ a\in \mathbb{R}$. From $ (*)$ we have $ f^{(2)}(a-c)=a.$ Therefore, $ \forall a\in \mathbb{R} \;\;\; \exists b\in \mathbb{R}, b=f(a-c),$ such that $ f(b)=a.$ So $ f(x)$ must be surjective. Therefore $ f(x)$ must be bijective. Then there must be $ u\in \mathbb{R}$ such that $ f(u)=0$ and this point is unique. So we have $ (1)$ $ P(u,u) \Rightarrow f(u^2)=u$ $ (2)$ $ P(-u,u) \Rightarrow f(u^2)=u+f^{2}(-u)$ From $ (1)$ and $ (2)$ we have $ u=u+f^{2}(-u) \Rightarrow f(-u)=0$ Since $ f(x)$ must be bijective we have $ u=0$. Therefore $ f(0)=0, \ c=0$ The equation $ (*)$ becomes $ f^{(2)}(x)=x\;\;\;\forall x\in \mathbb{R}\: (**)$ From $ (**)$ we derive that $ f^{(2)}(x)$ is strictly increasing. So consider $ a,b\in \mathbb{R}$ with $ f(a)>f(b)$. Then $ f(f(a))>f(f(b))$ and this implies $ a>b.$ Consider now $ a,b\in \mathbb{R}$ with $ f(a)<f(b)$. Then $ f(f(a))<f(f(b))$ and this implies $ a<b.$ Therefore we derive that $ f(x)$ is also strictly increasing. Consider $ a \in \mathbb{R}\;\;a\ne 0$. If $ f(a)>a$ then $ f(f(a))>f(a) \Rightarrow a>f(a).$ This is a contraddiction. If $ f(a)<a$ then $ f(f(a))<f(a) \Rightarrow a<f(a).$ This is again a contraddiction. The last two results say that $ \forall x\in \mathbb{R}, \;\;x \ne 0,$ it must be $ f(x)=x.$ Since $ f(0)=0$ we have that $ \forall x \in \mathbb{R}$ it must be $ f(x)=x.$ Therefore $ \boxed{f(x)=x,\ \forall x \in \mathbb{R}}$ is a solution of the initial equation and it is the unique solution.

prester wrote: So consider $ a,b\in \mathbb{R}$ with $ f(a) > f(b)$. Then $ f(f(a)) > f(f(b))$ and this implies $ a > b.$ Consider now $ a,b\in \mathbb{R}$ with $ f(a) < f(b)$. Then $ f(f(a)) < f(f(b))$ and this implies $ a < b.$ Therefore we derive that $ f(x)$ is also strictly increasing. The above steps are obviously wrong. We cannot derive that $ f(f(a)) > f(f(b))$ because we do not know that $ a > b$ but we do know that $ f(a) > f(b)$ and then the hypothesys uses the assert we want to demonstrate. It is non-sense. To demonstrate that $ f(x)$ is strictly increasing, I did not find a best way than previous post (by probability1.01). Let $ a > b\;\;\; a,b \in \mathbb{R}.$ So $ \exists u \in \mathbb{R}$ such that $ a = u^2 + b$. So $ f(a) = f(u^2 + b) = f^2(u) + f^{ - 1}(b) > f^{ - 1}(b) = f^{(2)}(f^{ - 1}(b)) = f(b)$ and therefore $ f(x)$ is strictly increasing. Then the demonstration can proceed as before. Sorry for mistake.

probability1.01 wrote: Set x = 0 to get $f(f(y)) = y+f(0)^{2}$. We'll let $c = f(0)^{2}$, so $f(f(y)) = y+c$. Then $f(a^{2}+f(f(b))) = f(b)+f(a)^{2}$ $f(a^{2}+b+c) = f(b)+f(a)^{2}$ $f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2}$ $a^{2}+b+2c = b+(a+c)^{2}$ $2c = c^{2}+2ac$ Since this holds for all $a$, it follows that $c = 0$. Now we have $f(0) = 0 \implies f(f(y)) = y$. Note that f must be surjective since we may let y vary among all reals, and f must be injective since if $f(a) = f(b)$, then $a+f(x)^{2}= f(x^{2}+f(a)) = f(x^{2}+f(b)) = b+f(x)^{2}$. Finally, if $u > v$, then there is some $t$ s.t. $u = t^{2}+v$, and so $f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)$. Hence f is strictly increasing. It is now clear that since $f(f(y)) = y$, we must have $f(x) = x$ for all x. heloo, can you explain me how you get $ f(f(a^{2}+b+c)) = f(f(b)+f(a)^{2}) = b+f(f(a))^{2} $??

I think it's easy to chek that $f(0)=0$ thereby $f(f(x))=x$ So , consider the sequence defined by : $f(f(f(....)))=x_{n} $ and $x_{0}=x$ So $f(f(x))=x \implies x_{n+2}=x_{n}$ so for all $n \in \mathbb N $ there existe $a;b \in \mathbb R $ such that $x_{n}=a+b(-1)^{n} $ for $n=1$ we have $f(x)=a-b=a+b-2b=x-2b$ because $a+b=x$ replacing this values in the equation $ f(f(x))=x$ and we will merely find that $b=0$ then $ ( \forall x \in \mathbb R ) f(x)=x$ Wich is indeed a solution for the initial fonctioanl equation . Redwane.

R.Maths wrote: I think it's easy to chek that $f(0)=0$ thereby $f(f(x))=x$ So , consider the sequence defined by : $f(f(f(....)))=x_{n} $ and $x_{0}=x$ So $f(f(x))=x \implies x_{n+2}=x_{n}$ so for all $n \in \mathbb N $ there existe $a;b \in \mathbb R $ such that $x_{n}=a+b(-1)^{n} $ for $n=1$ we have $f(x)=a-b=a+b-2b=x-2b$ because $a+b=x$ replacing this values in the equation $ f(f(x))=x$ and we will merely find that $b=0$ then $ ( \forall x \in \mathbb R ) f(x)=x$ Wich is indeed a solution for the initial fonctioanl equation . Redwane. helooo, van you explian me why So f(f(x))=x \implies x_{n+2}=x_{n} so for all n \in \mathbb N there existe???

e.lopes wrote: The first step is to establish that $f(0) = 0$. Putting $x = y = 0$, and $f(0) = t$, we get $f(t) = t^2$. Also, $f(x^2+t) = f(x)^2$, and $f(f(x)) = x + t^2$. We now evaluate $f(t^2+f(1)^2)$ two ways. First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$. Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$. So $t = 0$, as required. It follows immediately that $f(f(x)) = x$, and $f(x^2) = f(x)^2$. Given any $y$, let $z = f(y)$. Then $y = f(z)$, so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$. Now given any positive $x$, take $z$ so that $x = z^2$. Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$. Putting $y = -x$, we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$. Hence $f(-x) = - f(x)$. It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$, $y$. Take any $x$. Let $f(x) = y$. If $y > x$, then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$. If $y < x$, then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$. In either case we get some $z > 0$ with $f(z) = -z < 0$. But now take $w$ so that $w^2 = z$, then $f(z) = f(w^2) = f(w)^2 \ge 0$. Contradiction. So we must have $f(x) = x$ helooo, can you explain me why Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$

arshakus wrote: e.lopes wrote: The first step is to establish that $f(0) = 0$. Putting $x = y = 0$, and $f(0) = t$, we get $f(t) = t^2$. Also, $f(x^2+t) = f(x)^2$, and $f(f(x)) = x + t^2$. We now evaluate $f(t^2+f(1)^2)$ two ways. First, it is $f(f(1)^2 + f(t)) = t + f(f(1))^2 = t + (1 + t^2)^2 = 1 + t + 2t^2 + t^4$. Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$. So $t = 0$, as required. It follows immediately that $f(f(x)) = x$, and $f(x^2) = f(x)^2$. Given any $y$, let $z = f(y)$. Then $y = f(z)$, so $f(x^2 + y) = z + f(x)^2 = f(y) + f(x)^2$. Now given any positive $x$, take $z$ so that $x = z^2$. Then $f(x + y) = f(z^2 + y) = f(y) + f(z)^2 = f(y) + f(z^2) = f(x) + f(y)$. Putting $y = -x$, we get $0 = f(0) = f(x + -x) = f(x) + f(-x)$. Hence $f(-x) = - f(x)$. It follows that $f(x + y) = f(x) + f(y)$ and $f(x - y) = f(x) - f(y)$ hold for all $x$, $y$. Take any $x$. Let $f(x) = y$. If $y > x$, then let $z = y - x. f(z) = f(y - x) = f(y) - f(x) = x - y = -z$. If $y < x$, then let $z = x - y$ and $f(z) = f(x - y) = f(x) - f(y) = y - x$. In either case we get some $z > 0$ with $f(z) = -z < 0$. But now take $w$ so that $w^2 = z$, then $f(z) = f(w^2) = f(w)^2 \ge 0$. Contradiction. So we must have $f(x) = x$ helooo, can you explain me why Second, it is $f(t^2 + f(1 + t)) = 1 + t + f(t)^2 = 1 + t + t^4$ yes yes, I have just understand it

Here's my solution: Taking $y = -f(x)^2$ we get $f(x^2 + f(y)) = 0$ i.e., $\exists b$ such that $f(b)=0$ Then $f(b^2+f(y))=y$ and so $f$ is bijective. Then $f(b)^2= f(b^2+f(0))=f((-b)^2 +f(0)) = f(-b)^2$ so $f(-b)^2 = 0$ Hence $f(-b)=0=f(b) $ and so $-b=b$ since $f$ is injective, giving us $b=0$ So $f(0)=0$ Then $f(f(x))=x$ Suppose $x>y$ for some $x$ and $y$. Let $x=y+z^2$ Then $f(x)=f(z^2+f(f(y)))=f(y)+f(z)^2>f(y)$ so $f$ is strictly increasing. It is then well known that the only increasing solutions to $f(f(x))=x$ is $f(x)=x$ So $f(x)=x$ which works.

hi guys, I found this solution in shortlist, but I think that it is not full, can you look at this solution and explain me dear reader!!! It is easy to see that f is injective and surjective. From $f(x^2 + f(y)) =$ $f((-x)^2 +f(y))$ it follows that $f(x)^2 = (f(-x))^2$, which implies $f(-x) = -f(x)$ because f is injective. Furthermore, there exists $z∈R$ such that $f(z) = 0.$ From $f(-z) = -f(z) = 0$ we deduce that $z = 0$. Now we have $f(x^2) = f(x^2 + f(0)) = 0 + (f(x))^2 = f(x)^2$, and consequently $f(x) = f(sqrt(x)^2) > 0$ for all $x > 0$. It also follows that $f(x) < 0$ for $x < 0$. In other words, f preserves sign. Now setting $x > 0$ and $y = -f(x)$ in the given functional equation we obtain $f(x - f(x)) = f(sqrt(x)^2+ f(-x)) = -x + f(√x)2 = -(x - f(x)).$ But since f preserves sign, this implies that $f(x) = x for x > 0. $Moreover, since $f(-x) = -f(x),$ it follows that $f(x) = x$ for all x. It is easily verified that this is indeed a solution.

orl wrote: Let $ \,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $ \,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2} + f(y)\right) = y + \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.\] Denote by $ P(x,y)$ the assertion $ f(x^2+f(y))=y+f^{2}(x)$. [hide="1) $f(0) = 0$"]\[P(0,0) \implies f(f(0)) = f(0)^2 \\ \\ P(0,x) \implies f(f(x)) = x+f(0)^2 \implies f(x) + f(0)^2 = f(f(f(x))) = f(x+f(0)^2) \] \[P(x,0) \implies f(x^2+f(0)) = f(x)^2 \\ \\ P(-x,0) \implies f(x^2+f(0)) = f(-x)^2 \\ \\ \implies f(-x) = \pm f(x)\] But, \[f(-x) = f(x) \implies x+f(0)^2 = f(f(x)) = f(f(-x)) = -x+f(0)^2\] But this is true only for $x=0$. Now, take any non-zero real $a$. We have, \[-a +f(0)^2 = f(f(-a)) = f(-f(a)) = -f(f(a)) = -(a+f(0)^2) \\ \\ \implies f(0) = 0\][/hide]

[hide="3) $f(x+1) = f(x) +1$"] We have, \[P(1,0) \implies f(1) = f(1)^2 \implies f(1) = 0 \ \text{or } f(1) = 1.\] But, if $f(1) =0$, then \[P(x,1) \implies f(x^2) = 1+f(x)^2\] This is clearly not possible, because \[P(x,0) \implies f(x^2) = f(x)^2 \not = 1+f(x)^2\] Therefore, $f(1) =1$. Now, \[P(1,x) \implies f(f(x) + 1) = x+1 \implies f(x+1) = f(f(f(x) +1)) =f(x) +1\] since $f(f(x)) =x$.[/hide] [hide="4) $f(x)=x \ \forall x \in \mathbb{R}$"] We have, \[P(x,1-x^2) \implies f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2)\] Now, \[f(1-x^2) = 1+f(-x^2) = 1-f(x^2) \implies f(x^2+ f(1-x^2)) = f(1+(x^2 - f(x^2))) = 1+ f(x^2 - f(x^2))\] Also, using $f(x)^2 = f(x^2)$, which has been proved earlier, we have \[1+ f(x^2 - f(x^2)) = f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2) = 1 + (f(x^2) - x^2)\] Therefore, we have, \[f(x^2 - f(x^2)) = f(x^2) - x^2\] Therefore, from 2nd lemma, \[x^2 - f(x^2) = 0 \iff f(y) = y\ \forall y \in \mathbb{R}^+ \] Combining this with $f(-x) = -f(x)$, we get the desired.[/hide]

Let $P(x,y)$ be the assertion that $f(x^2 + f(y))=y+f(x)^2$. By $P(0,y)$, we have that $f(f(y))=y+f(0)^2$ which implies that $f$ is a surjective function. Furthermore, if $f(x)=f(y)$, then it follows that $x+f(0)^2=f(f(x))=f(f(y))=y+f(0)^2$ which implies that $x=y$ and hence that $f$ is injective. Now let $a \in \mathbb{R}$ be such that $f(a)=0$. By $P(a,f(y))$, it follows that $f(a^2+f(f(y)))=f(y)+f(a)^2=f(y)$ which implies that $y=f(f(y))+a^2$ since $f$ is injective. However, as previously established, $f(f(y))=y+f(0)^2$ which implies that $a^2 + f(0)^2=0$ and hence that $a=f(0)=0$ and $f(f(y))=y$. Now by $P(x,0)$, it follows that $f(x^2)=f(x)^2$ and furthermore that $f(x)^2=f(x^2)=f((-x)^2)=f(-x)^2$ which implies that $f(-x)=-f(x)$ since $f$ is injective. Now by $P(x,f(y))$, it follows that $f(x^2+y)=f(x^2+f(f(y)))=f(y)+f(x)^2=f(y)+f(x^2)$. Combining this with the fact that $f(-x)=-f(x)$ yields that $f(x+y)=f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Now note that for all $a \ge 0$, $f(a)=f((\sqrt{a})^2)=f(\sqrt{a})^2 \ge 0$. Hence $f$ is bounded on an interval which implies by the Cauchy functional equation that $f(x)=cx$ for some constant $c \in \mathbb{R}$. Substituting this into the equation yields that $f(x)=x$.

So that is Alex Song's solution from http://www.artofproblemsolving.com/Forum/viewtopic.php?f=320&t=422503.

Yes. Sorry, I forgot to mention that the solution shown above is due to Alex Song.

Rijul saini wrote: orl wrote: Let $ \,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $ \,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2} + f(y)\right) = y + \left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}.\] Denote by $ P(x,y)$ the assertion $ f(x^2+f(y))=y+f^{2}(x)$. [hide="1) $f(0) = 0$"]\[P(0,0) \implies f(f(0)) = f(0)^2 \\ \\ P(0,x) \implies f(f(x)) = x+f(0)^2 \implies f(x) + f(0)^2 = f(f(f(x))) = f(x+f(0)^2) \] \[P(x,0) \implies f(x^2+f(0)) = f(x)^2 \\ \\ P(-x,0) \implies f(x^2+f(0)) = f(-x)^2 \\ \\ \implies f(-x) = \pm f(x)\] But, \[f(-x) = f(x) \implies x+f(0)^2 = f(f(x)) = f(f(-x)) = -x+f(0)^2\] But this is true only for $x=0$. Now, take any non-zero real $a$. We have, \[-a +f(0)^2 = f(f(-a)) = f(-f(a)) = -f(f(a)) = -(a+f(0)^2) \\ \\ \implies f(0) = 0\][/hide]

[hide="3) $f(x+1) = f(x) +1$"] We have, \[P(1,0) \implies f(1) = f(1)^2 \implies f(1) = 0 \ \text{or } f(1) = 1.\] But, if $f(1) =0$, then \[P(x,1) \implies f(x^2) = 1+f(x)^2\] This is clearly not possible, because \[P(x,0) \implies f(x^2) = f(x)^2 \not = 1+f(x)^2\] Therefore, $f(1) =1$. Now, \[P(1,x) \implies f(f(x) + 1) = x+1 \implies f(x+1) = f(f(f(x) +1)) =f(x) +1\] since $f(f(x)) =x$.[/hide] [hide="4) $f(x)=x \ \forall x \in \mathbb{R}$"] We have, \[P(x,1-x^2) \implies f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2)\] Now, \[f(1-x^2) = 1+f(-x^2) = 1-f(x^2) \implies f(x^2+ f(1-x^2)) = f(1+(x^2 - f(x^2))) = 1+ f(x^2 - f(x^2))\] Also, using $f(x)^2 = f(x^2)$, which has been proved earlier, we have \[1+ f(x^2 - f(x^2)) = f(x^2+ f(1-x^2)) = 1 + (f(x)^2 - x^2) = 1 + (f(x^2) - x^2)\] Therefore, we have, \[f(x^2 - f(x^2)) = f(x^2) - x^2\] Therefore, from 2nd lemma, \[x^2 - f(x^2) = 0 \iff f(y) = y\ \forall y \in \mathbb{R}^+ \] Combining this with $f(-x) = -f(x)$, we get the desired.[/hide] Hi,Rijul.After showing that $f(x+1)=f(x)+1$ one can also argue in another way.This implies $f(z)=z \forall z \in \mathbb{Z}$ using induction.Then one can show that $f(r)=r \forall r \in \mathbb{Q}$.Finally use the density of $\mathbb{Q}$ in $\mathbb{R}$ to conclude that $f(x)=x \forall x \in \mathbb{R}$.

R.Maths wrote: I think it's easy to chek that $f(0)=0$ thereby $f(f(x))=x$ So , consider the sequence defined by : $f(f(f(....)))=x_{n} $ and $x_{0}=x$ So $f(f(x))=x \implies x_{n+2}=x_{n}$ so for all $n \in \mathbb N $ there existe $a;b \in \mathbb R $ such that $x_{n}=a+b(-1)^{n} $ for $n=1$ we have $f(x)=a-b=a+b-2b=x-2b$ because $a+b=x$ replacing this values in the equation $ f(f(x))=x$ and we will merely find that $b=0$ then $ ( \forall x \in \mathbb R ) f(x)=x$ Wich is indeed a solution for the initial fonctioanl equation . Redwane. Your solution doesn't work: You started off with all this stuff by taking a fixed $x$. For this $x$, you are right to get some constants $a$ and $b$ and you'll have $f(x)=x-2b$ for this particular $x$. But from that you cannot derive that $f(x)=x+c$ for every $x$ because for different starting values $x$ your $b$-value might be different.

Let $c=f(0)$. Then $P(0,x)\implies f(f(x))=x+c^2$ so $f$ is surjective. Also \[f(a)=f(b)\Longleftrightarrow a+c^2=f(f(a))=f(f(b))=b+c^2\Longleftrightarrow a=b\] hence $f$ is injective, thus bijective. Choose $k$ so that $f(k)=0$. So \[P(k,f(x))\implies f\left(k^2+x+c^2\right)=f(x)\implies k^2+c^2=0\implies k=c=0.\] So $f(f(x))=x$. Consider $m>n$, let $m=n+t$. Choose $s$ so that $f(s)=n$. Then we have \[f(m)=f(n+t)=\sqrt t + s> s=f(f(s))=f(n)\] so strictly increasing. Now $f(x)>x\implies x=f(f(x))>f(x)$, similar for $f(x)<x$ leads to contradiction. Therefore $f(x)=x~\forall x\in\mathbb R$.

orl wrote: Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \] Let $P(x,y)$ the assertion of the given F.E. $P(0,x)$ $$f(f(x))=x+f(0)^2 \implies f \; \text{bijective}$$Let $c$ be a real such that $f(c)=0$ (it exists becuase $f$ is bijective) $P(c,f(x))$ and injectivity $$y+f(0)^2=f(f(y))=y+c^2 \implies c=\pm f(0) \implies f(\pm f(0))=0$$On the last equation if $y=-c^2$ then $f(-c^2)=c$ but by $P(c,c)$ we have $f(c^2)=c$ thus by injectivity $c=0$. $P(0,x)$ $$f(f(y))=y \implies f \; \text{involution}$$$P(x,0)$ $$f(x^2)=f(x)^2 \implies f(f(x)^2)=x^2$$And now by replacing this on the original F.E. $$f(x^2+f(y))=y+f(x^2)$$Let $Q(x,y)$ the assertion of this F.E. Choose $x>y$ so that $x=y+a^2$ where $a$ is a real number. $Q(a,y)$ $$f(x)=f(y)+f(a^2)>f(y) \implies f \; \text{increasing}$$Thus $f$ involutive+increasing means that $f$ is the identity. Thus the only sol is $f(x)=x$ for every real $x$. Thus we are done

$f$ is bijective. $P(0,y) \implies f(f(y))=y+f(0)^2$. $f(a)=0, P(x,y-f(x)^2) \implies f(f(y)-f(x)^2)=y-x^2 \implies f(f(y))=y-a^2$ and $a-a^2=f(0)$. $-a^2=f(0)^2=(a-a^2)^2 \implies a=0$ or $a^2-2a+2 \implies a=0 \implies f(f(x))=x$. $P(x,0) \implies f(x^2)=f(x)^2 \implies f(x^2+f(y))=y+f(x^2) \implies f(x^2+y)=f(x^2)+f(y)$ for $y \rightarrow f(y)$. $f(x^2)=f(x)^2 \implies f(x)=\pm f(-x)$. If $f(c)=f(-c), P(x,c)$ and $P(x,-c)$ together implies $c=0$. Thus $f(x)=-f(-x)$ and $f(x^2+y)=f(x^2)+f(y) \implies f(-x^2+y)=-f(x^2-y)=-(f(x^2)+f(-y))=f(-x^2)+f(y) \implies f(x+y)=f(x)+f(y)$ for all $x,y \in R$. $f(x)^2+2f(x)f(y)+f(y)^2=(f(x)+f(y))^2=f(x+y)^2=f(x^2+2xy+y^2)=f(x^2)+f(2xy)+f(y^2) \implies 2f(x)f(y)=f(2xy) \implies 2f(x)=f(2x) \implies f(x)f(y)=f(xy)$ for $x,y \in R$. With $f(x)+f(y)=f(x+y)$ we conclude $f(x)=x$.

We'll use $P(x,y)$ for the assertion $(x,y)$ into the FE: $P(0,y)$ gives $f(f(y))=y+f(0)^2\Longrightarrow f$ is bijective (as $y$ runs from $-\infty$ to $+\infty$, $y+f(0)^2$ covers $\mathbb{R}$, so $f$ in surjective and if we assume that $\exists a\neq b\in\mathbb{R}$ such that $f(a)=f(b)$ then $b+f(0)^2=f(f(b))=f(f(a))=a+f(0)^2$, contradiction$\implies$ $f$ is bijective) Therefore $\exists$ a unique $ z\in\mathbb{R}$ such that $f(z)=0$ since $f$ is bijective: $P(0,0)\Longrightarrow f(f(0))=f(0)^2$ $(1)$ $P(z,f(0))\Longrightarrow f(z^2+f(f(0)))=f(0)+f(z)^2\overset{(1)}{\Longrightarrow} f(z^2+f(0)^2)= f(0)\Longrightarrow z^2+f(0)^2=0\Longrightarrow z=0, \boxed{f(0)=0}$ Now $P(0,y)$ gives $f(f(y))=y$ and $P(x,0)$ gives $f(x^2)=f(x)^2\implies f(x)=-f(-x)$. Using these two results we have that $P(x,f(y)):\quad f(x^2+y)=f(y)+f(x^2)$. This is the Cauchy equation and as $f(x^2)=f(x)^2\geq 0$ this means that $f$ is monotonic, which implies that $f(x)=cx$ for some constant $c\in \mathbb{R}$. As $c\times 1^2=f(1^2)=f(1)^2=c^2\times 1^2\Longrightarrow c\in\{0,1\}$. The function $f\equiv 0$ is not a solution as $f$ is bijective, so the only possible solution is $f(x)=x$. And indeed: $$f(x^2+f(y))=f(x^2+y)=x^2+y=y+f(x)^2$$Therefore $f(x)=x$ is the only solution to the FE.

orl wrote: Let $\,{\mathbb{R}}\,$ denote the set of all real numbers. Find all functions $\,f: {\mathbb{R}}\rightarrow {\mathbb{R}}\,$ such that \[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \] $P(0;y)$ we have $f(f(y))=y+(f(0))^2, (1)$ so f is bijective Exist a such that $f(a)=0$ $P(a;0):$ $f(a^2+f(0))=0=f(a)$ -> $a^2+f(0)=a, (2)$ Let $y=a$ in $(1):$ $f(0)=a+f(0)^2, (3)$ From $(2)$ and $(3)$ we have $f(0)=a=0$ We get $f(x^2)=f(x)^2, (*)$ and $f(f(y))=y$ $P(x;f(y)):$ $f(x+y)=f(x)+f(y), (4)$ for all $x \geq 0$ and all real $y$ $Q(x;y)$ in $(4)$ Let $x>0$ then from $(4)$ we have $0=f(0)=f(x)+f(-x)$ so f is odd From $(*)$ we have $f(x) \geq 0$ for all $x \geq 0, (5)$ From $(4)$ and $(5)$ we have $f(x)=x$ , which is a familiar problem Q.E.D

A rather lengthy solution IG ;-;(Pls correct if I am wrong at some point) We claim that only $f(x)=x$ which can be easily checked. Let the assertion be $P(x,y)$. It is easy to see $f(x)$ is injective. $P(x,-f(x)^2) \Rightarrow f(x^2+f(-f(x)^2)=0$ Since $f(x)$ is injective it follows that $ x^2+f(-f(x)^2)=t ...(1) $ for some unique $t$ Now, Plugging $x=t$ in $(1)$ we get $t^2+f(0)=t$ Again, $P(x,t) \Rightarrow f(x^2)=f(x)^2 +t \Rightarrow f(0)=f(0)^2+t$ Combining the above two expressions we get $f(0)=0$ Plugging $x=f(x)$ in $(1)$ we get $f(-x)=-f(x)$ Again,$P(x,-x^2) \Rightarrow f(x^2-f(x)^2)=f(x)^2-x^2$ Now lets assume $f(a)=-a$ for some $a \neq 0$ also assume WLOG that $a>0$ since $f(-x)=-f(x)$ Then $P(\sqrt{a},a) \Rightarrow f(\sqrt{a})^2=-a$ which is a contradiction. So, $a=0 \Rightarrow f(x)^2=x^2$ Now we fix the point wise error. assume for $f(a)=a,f(b)=-b$ $P(a,b) \Rightarrow f(a^2 -b)=a^2+b$ Contradiction. $f(x)=x$ is the only solution(since $f(x)=-x$ don't work).

$P(0,x): f(f(x))=x+f(0)^2$, so $f$ is bijective. Claim: $f(0)=0$ $P(x,f(y)): f(x^2+y+f(0)^2)=f(x)^2+f(y)$. $P(f(x),y): f(f(x)^2+f(y))=(x+f(0)^2)^2+y$. Note that $f(f(x)^2+f(y))=x^2+y+2f(0)^2$. Suppose FTSOC that $f(0)\ne 0$. So we get \begin{align*} x^2+y+2f(0)^2=x^2+2xf(0)^2+f(0)^4+y \\ 2f(0)^2=2xf(0)^2+f(0)^4 \\ 2=2x+f(0)^2\forall x \\ \end{align*} This is a contradiction because it implies that $f(0)^2+2\cdot 332.5=f(0)^2+665=2\implies f(0)^2=-663$ and also $f(0)^2+2\cdot 93.5=f(0)^2+187=2\implies f(0)^2=-185$. So $f(0)=0$, which implies $f(f(x))=x$. $P(x,f(y)): f(x^2+y)=f(x)^2+f(y)$. $P(x,0): f(x^2)=f(x)^2$. So $f(x^2+y)=f(x^2)+f(y)$. Claim: $f$ is odd. Proof: We have $f(x^2)=f(-x)^2=f(x)^2$. If $f(x)=f(-x)$ for some $x$, then we have $f(f(x))=f(f(-x))\implies x=-x\implies x=0$. So $f(x)=-f(-x)$ for all $x$ since $f(0)=0$. $\blacksquare$ For all reals $x$ and $y$ that are not both negative, we get $f(x+y)=f(x)+f(y)$. Now if $u$ and $v$ are negative, we get $f((-u)+(-v))=-f(u+v)=f(-u)+f(-v)=-(f(u)+f(v))$, so $f(u+v)=f(u)+f(v)$. Thus, $f$ is additive. From $f(x^2)=f(x)^2$, $f$ is bounded over the interval (always nonegative) $[0,\infty]$, so $f(x)=kx$ for some constant $k$. From involution we get $k=1$ or $k=-1$. From $f(x^2)=f(x)^2$, we get $k=1$. So $\boxed{f(x)=x}$ is the only solution.

Edit June 10: added a crucial word. Claim: $f(0)=0.$ Proof. Let $P(x,y)$ denote the original equation. $P(0,y)\implies f(f(y))=y+(f(0))^2.$ $P(0,0)\implies f(f(0))=(f(0))^2.$ $P(0,1)\implies f(f(1))=1+(f(0))^2.$ Applying $f$ to both sides of $P(x,y)$ yields, $x^2+f(y)+(f(0))^2=f(y+(f(x))^2).$ $P(1,f(0))\implies f(f(0)+(f(1))^2)=1+2(f(0))^2.$ $P(f(1),0)\implies f(f(0)+(f(1))^2)=(1+(f(0))^2)^2.$ From this the claim follows. And $f(f(x))=x.$ $\blacksquare$ Claim: $f$ is strictly increasing. Proof. Let for any reals $a,b$, $a>b.$ Note that $f(f(\sqrt{a-b})) =\sqrt{a-b} \implies f(\sqrt{a-b}) \neq 0.$ $P(\sqrt{a-b}, b) \implies (f(\sqrt{a-b}))^2+f(b) =f(a).$ And thus our claim follows. $\blacksquare$ Claim: $f(x)=x,$ which in fact holds. Proof. It's pretty well known and easy to show that the solution of any strictly increasing and involutive function is an identity function. This completes the proof. $\blacksquare$

Video Solution: https://youtu.be/ZZaIPN-ILdk

Cool problem! I think I understand Cauchy Functional Equations a little bit now. Solution. The answer is $f \equiv \mathrm{id}$ which clearly works. We will proceed to show that this is unique. Denote $P(x,y)$ as the assertion to the functional equation. One can quickly see that $f$ is bijective from $P(0,x)$. Let $\alpha$ be "the zero" of $f$. Now we show that $\alpha = 0$. For this consider the following substitutions. $P(\alpha, 0)$ gives us $f(\alpha^2 + f(0)) = 0 = f(\alpha)$. By injectivity, we have that $\alpha^2 + f(0) = \alpha$. $P(0, \alpha)$ gives us $f(0) = f(0)^2 + \alpha$. Now solve the simultaneous equations to get $f(0) = 0$ as desired. $P(x,0)$ reveals $f(x^2) = f(x)^2 \ge 0$. So $f$ is positive over positive domain values. $P(0,x)$ gives that $f$ is an involution. Finally observe that $P(x,f(y))$ gives the following partial "Cauchy Condition" \[f(x^2 + y) = f(x^2) + f(y) \implies f(a+b) = f(a) + f(b)\]for all reals $a$ and all $y \ge 0$. One can extend this partial condition to a full "Cauchy" easily. Observe that for some arbitrary $r \in \mathbb{R}$ \[f(0) = f(r-r) = f(r) + f(-r) \iff f(r) = -f(-r)\]which shows $f$ is odd. With the odd nature of $f$, we can now say that $f$ is completely additive over $\mathbb{R}$. We will now present two finishes. Since $f(x) > 0$ for $x > 0$, $f$ must be strictly increasing. For strictly increasing additive function $f$, it is well-known that $f(x) = ax$. Plugging this in would finish the solution. The second finish uses the involution. Since $f$ is an involution and strictly increasing, $f$ must be the identity function. So we're done. $\blacksquare$

$f(x^2+f(y))=y+f(x)^2$ Plug in $y=-f(x)^2$ to get $f(x^2+f(-f(x)^2))=0$ Let $u=x^2+f(-f(x)^2)$ $f(u^2+f(y))=y$ so $f(y)=f(f(u^2+f(y)))=u^2+f(y)$ $u^2=0$ $u=0$ $f(0)=0$ Plug in $x=0$ to get $f(f(y))=y+f(0)^2=y$, so $f$ is injective and surjective $f(x^2+f(-f(x)^2))=0=f(0)$, so $x^2+f(-f(x)^2)=0$ $f(-f(x)^2)=-x^2$ $f(f(-f(x)^2))=f(-x^2)=-f(x)^2$ Plug in $y=0$ to get $f(x^2+f(0))=f(x^2)=f(x)^2$, so $f$ is odd Plug in $x=1$ to get $f(1)=f(1)^2$, so $f(1)=0$ or $1$, but beacuse $f$ is injective, $f(1)=1$ Plug in $y=-y$ to get $f(x^2-f(y))=f(x)^2-y$ $f(x^2+f(y))=y+f(x)^2=f(x)^2+f(f(y))$, so $f$ is additive Because $f(x^2)=f(x)^2>0$, $f$ is bounded over the positives, so $f(x)=kx$ Because $f(x)^2=f(x^2)$, $k^2x^2=kx^2$, so $k=0$ or $1$. Plugging $f(x)=0$ and $f(x)=x$ in, get that only $\boxed{f(x)=x}$ works

\[f(x^2+f(y))=y+f(x)^2\]First, by looking at $y$ we see that $f$ is injective, also surjective. We compare $P(x,y)$ and $P(-x,y)$ to get $y+f(x)^2=y+f(-x)^2$ or $|f(x)|=|f(-x)|$ and since $f$ is injective, $f(x)=-f(-x)$ thus, $f$ is odd. \[f(x^2-f(y))+f(x^2+f(y))=-y+f(x)^2+y+f(x)^2=2f(x)^2\]$f$ is surjective hence $f(x^2-y)+f(x^2+y)=2f(x)^2$ and $x=0$ implies $f(0)=0$. $x=0$ in the original equation gives $f(f(y))=y$ so $f$ is involution. $y=0$ yields $f(x^2)=f(x)^2$. \[f(x^2+y)=f(y)+f(x)^2=f(y)+f(x^2)\]Let's show that $f(a)+f(b)=f(a+b)$ for all reals. If one of $a,b$ is nonnegative, then $f(x^2)+f(y)=f(x^2+y)$ gives the result. If both of them are negative, then since $f$ is odd $f(-x^2)+f(-y)=f(-x^2-y)$ gives it. Since $f$ is both additive and bounded below on positive values by $f(x^2)=f(x)^2\geq 0$, this is a Cauchy function with solution $f(x)=cx$ and subsituting back implies $f(x)=x$ as desired.$\blacksquare$

Old-Classic: Here is the proof: Let $f(0)=c$. Then: $P(0, 0): f(c)=c^2$. $P(x, f(y))$: $f(x^2+f(f(y))) = f(y)+f(x)^2 \implies f(x^2+y+c) = f(y)+f(x)^2$ $$\implies f(f(x^2+y+c)) = f(f(y)+f(x)^2) \implies x^2+y+2c = y+c^2+(x+c^2)^2$$$$\implies 2c = 2c^2 x+c^4+c^2$$Inorder for the above equation to hold for all real $x$, we have $c=0$. Thus: $$f(f(x))=x, f(x^2)=f(x)^2.$$The latter implies $f(x) \ge 0$ for all $x \ge 0$. Thus, plugging back $P(x, f(y))$: $$f(x^2+y) = f(x^2)+f(y).$$Note that $f$ is surjective thus, $f(y)$ covers all reals. Thus: $f$ is additive along with $f(x) \ge 0$ for all $x \ge 0$. Therefore: $f(x)=ax$ and plugging back $f(x)=x$ is the only solution,

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