emregirgin35 wrote: Find all $f$ functions from real numbers to itself such that for all real numbers $x,y$ the equation $f(f(y)+x^2+1)+2x=y+(f(x+1))^2\quad (1)$ holds. It is easy to see that $f$ is a bijective. Let $a=f(0)$ and $b=f(1).$ Replacing $x=0$ in $(1),$ we get \[y=f\big(f(y)+1\big)-b^2,\quad \forall y \in \mathbb R.\quad (2)\] Plugging this into $(1),$ we get \[f\big(f(y)+1+x^2\big)+2x+b^2=f\big(f(y)+1\big)+f^2(x+1),\quad \forall x,\, y\in \mathbb R.\quad (3)\] Since $f $ is a bijective, it follows that \[f(x^2+y)+2x+b^2=f(y)+f^2(x+1),\quad \forall x,\,y \in \mathbb R.\quad (4)\] Now, replacing $y=0$ in $(4),$ we get \[f^2(x+1)=f(x^2)+2x+b^2-a,\quad \forall x \in \mathbb R.\quad (5)\] Plugging this into $(4),$ we get \[f(x^2+y)=f(x^2)+f(y)-a,\] from which it follows that \[f(x+y)=f(x)+f(y)-a,\quad \forall x,\,y \in \mathbb R,\, x \ge 0.\quad (6)\] From here, we can also show that $(6)$ still holds in the case $x<0$ (this is an easy job), so we have \[f(x+y)=f(x)+f(y)-a,\quad \forall x,\,y \in \mathbb R.\quad (7)\] Taking $x=y=1$ in $(7),$ we get $f(2)=2b-a.$ Taking $x=1$ in $(5),$ we also have $f^2(2)=b^2+b+2-a.$ So we have \[(2b-a)^2=b^2+b+2-a.\quad (8)\] Also, taking $x=-1$ in $(5),$ we have \[a^2+a=b^2+b-2.\quad (9)\] Solving the system $(8)$ and $(9),$ we get $a=0$ and $b=1.$ Therefore, the equations $(5)$ and $(7)$ can be rewritten as \[f^2(x+1)=f(x^2)+2x+1,\quad \forall x \in \mathbb R\quad (5')\] and \[f(x+y)=f(x)+f(y),\quad \forall x,\,y \in \mathbb R.\quad (7')\] Using $(7')$ and $f(1)=1,$ we can rewrite $(5')$ as \[f(x^2)=f^2(x)+2\cdot f(x)-2x,\quad \forall x \in \mathbb R.\quad (10)\] Replacing $x$ by $-x$ in $(10)$ with notice that $f$ is odd (because $f$ is additive), we also have \[f(x^2)=f^2(x)+2x-2\cdot f(x),\quad \forall x\in \mathbb R.\quad (11)\] Comparing $(10)$ and $(11),$ we get $f(x)=x,\, \forall x \in \mathbb R.$

here my solution: Obviously, $f$- bijective. Then there exist $a,b\in R$ such that $f(a)=0$ and $f(b)=1$. (1) $P(0,y)$ $ \Rightarrow $ $ f(f(y)+1)=f(1)^2+y $ and $ a=f(1)-f(1)^2$. (2) $P(f(1),y)$ $ \Rightarrow $ $f(f(y)+f(1)^2+1)+2f(1)=y+f(1)^4+2f(1)^2+1$ (since (1)). (3) since (1), $ f(f(1)^2+y+1)=f(f(f(y)+1)+1)=f(1)^2+f(y)+1 $ $ \Rightarrow $ $ f(f(y)+f(1)^2+1)=f(1)^2+f(f(1))+1 $. So (2) and (3) $ \Rightarrow $ $ f(f(y))=y+f(1)^4+f(1)^2-2f(1) $ (*). It gives us $ f(0)=f(1)^4-f(1) $ and $f(1)=b+f(1)^4+f(1)^2-2f(1)$. Since (1), we have that $f(2)=f(1)^2+b$. \[ f(f(y)+2)+2=y+(f(1)^2+b)^2 \] and \[ f(f(y)+2)=y+2+f(0)^2 \] $ \Rightarrow $ \[ (f(1)^2+b)^2=f(0)^2+4 \] and \[ (f(1)-1)(f(1)^4+f(1)^3+f(1)^2-f(1)-1)=0 (**) .\] We also have $ f(f(y)+2)=y+2+f(0)^2 $ $ \Rightarrow $ $ f(1)^2+b=f(2)=a+2+f(0) $ and \[ (f(1)-1)(f(1)^7+f(1)^6+f(1)^5-f(1)^4-2)=0 (***) .\] Since (**) and (***) we get easily $ f(1)=1 $ is unique root. Hence $f(1)=1$, $f(0)=f(1)^4-f(1)=0$, $f(2)=2$ and $f(-1)=-1$. Since (*) we have $f(f(y))=y$ and we can find $f$- additive and $f(x^2)+2x=f(x)^2+2f(x) $. So $f(f(x)^2)+2f(x)=x^2+2x$ and $x^2+2f(x)=f(f(x)^2)+2x$ $ \Rightarrow $ $f(x)=x$. Answer is $f(x)=x$ for all $x \in R$.

Let the original equation be $(1)$. Notice the function is clearly bijective, so $f^{-1}$ is well defined. Now let $a=f^{-1}(-1)$. In $(1)$ we see that $f(x+1)^2=f(x^2)+2x-a$. Let this equation be $(2)$. Together, we get $f(X+Y+1)=f(X)+f^{-1}(Y)-a$ whenever $X \ge 0$. Let this equation be $(3)$. Now suppose we have numbers $x_1$ and $x_2$ such that $f(x_1+1)+f(x_2+1)=0$. By $(2)$ we get $f(x_1^2)+2x_1 = f(x_2^2)+2x_2$. In (3) with $X=x_1^2$ and $X=x_2^2$ we get $f(Y+x_1^2)+2x_1=f(Y+x_2^2)+2x_2$ for all $Y$. So if $C=x_1^2-x_2^2$ and $D=2(x_1-x_2)$ we get that $A-B=C \Rightarrow f(A)-f(B)=D$. $(4)$ I will prove that ($4$) implies $C=D$. Notice that in $(3)$ if we plug in $Y=y$ and $Y=y+C$, we get that $f^{-1}(y+C)-f^{-1}(y)=D$ and so $A-B=D \Rightarrow f(A)-f(B)=C$. Together with $(4)$, this implies that, if $Z=C-D$, $A-B = Z \Rightarrow f(A)-f(B)=-Z$. It is easy to see $-Z$ also satisfies that relationship, so WLOG $Z \textgreater 0$ (if $Z=0$ then $C=D$). Notice $(1) \Rightarrow f(f(y)+x^2+1)+2x \ge y$. But notice that in $(1)$, we can plug in $x=x_0$ a very big (and positive) number and $x=x_0+k$ an even bigger number such that $(k^2+2x_0k)/Z$ is an integer. We have $y \le f(f(y)+(x_0+k)^2+1) + 2(x_0+k) = f(f(y)+x_0^2+1) - (k^2+2x_0k) + 2(x_0+k)$. But $(k^2+2x_0k) - 2(x_0+k)$, with $x_0$ fixed, can be as big as we want, and therefore eventually we will contradict this inequality. So we have $Z=0$, and so $C=D$. So we have $f(a)=-f(b) \Rightarrow a=-b$ or $a=b=f^{-1}(0)$. But if $f(0) \neq 0$ we see that if $f(q)=-f(0)$ ($q$ exists because $f$ is bijective) then $q=-0$ or $f(q)=0$. From this we get that $f(0)=0$ and so $f$ is odd. Now take $(1)$ and take $z=f(y)+X$, we get that $z-y=f(f(z)-f(y))-f(0)$ if $f(z) \ge f(y)$. Now suppose $X$ is negative, and take $f(z)=f(y)+X$. Applying $(1)$ we see $f(X+f(y)+1) = f(f(z)+1)=f(0)-a+z=f(X)-a+y$. This is because $z-y=f(f(z)-f(y))-f(0)$ using the above result (since $f(y) \ge f(z)$) and using that $f$ is odd. Therefore $(3)$ is valid for all $X,Y$. Using symmetry we get that $f(X)-f^{-1}(X)$ is constant. Using $X=Y+1$, $Y=X-1$ we get that $f(X)-f^{-1}(X-1)$ is constant. So combining these we get that $f(X)-f(X-1)=C$ is constant. Notice this is another statement like $(4)$, and so $1=C$. Therefore $f(x)-f(x-1)=1$ always. $(5)$. Using $(2)$ with $x=0$ and $x=-1$, we can solve and get that $f(-1)=-1$. From this, $a=-1$, and using $(5)$ we get that $f(n)=n$ for all $n \in \mathbb{Z}$. Using the fact that $f(X)-f^{-1}(X)$ is constant, we get that $f(f(x))=x$ for all $x$. And using this in $(3)$, we get that $f$ satisfies the Cauchy equation $f(x)+f(y)=f(x+y)$ (plugging $X=x-1$ and remembering that $f(X)=f(x)-1 = f(x)+a$. So we have $f(x)+f(y)=f(x+y)$ and $f(f(x))=x$. Assume $f(w) \neq w$ for a real $w$. Then we easily get that, if $C=w-f(w)$, that $A-B = C \Rightarrow f(A)-f(B)=-C$. We had seen that this implied $C=0$. So no such $w$ exists. Therefore $f(x)=x$ for all $x$.

Suppose $P(x,y)$ denotes the original equation and $f(1)=c$. $P(0,y)\Longrightarrow f(f(y)+1)=c^2+y\quad (1)$ $\Longrightarrow f$ is bijective. Also $ c^2+f(y)+1=f(f(f(y)+1)+1)=f(c^2+y+1)$ Hence $P(c,y)\Longrightarrow f(f(y))=\text { constant }+y\quad (2)$ From (1) & (2), we see $f(f(y)+1)-f(f(y))=d$ for some constant $d$. And since $f$ is surjective, we further derive $f(x+1)-f(x)=d\forall x\in \mathbb R$. Comparing $P(x-1,y)$ and $P(-x+1,y)$, we get \[4(x-1)=f(x)^2-f(-x+2)^2=f(x)^2-f(-x)^2-4f(-x)d-4d^2\quad (3)\]Substituting $x$ by $-x$ we get \[4(-x-1)=f(-x)^2-f(x)^2-4f(x)d-4d^2\quad (4)\]Adding (3) & (4) will give $f(x)+f(-x)=e$ for some constant $e$. But substituting $f(-x)$ by $e-f(x)$ in (3) reveals that $f$ is a linear function. So $f(x)=ax+b$. Now checking in the main equation shows that $a=1,b=0$. In other words, $f(x)=x\forall x$

Mathuz I think that in the seventh line must be:$f(f(y)+f(1)^{2}+1)=f(1)^{2}+f(f(y))+1$

baopbc wrote: Mathuz I think that in the seventh line must be:$f(f(y)+f(1)^{2}+1)=f(1)^{2}+f(f(y))+1$ yes it is

Who can explain to me the solution of CanVQ please Quote: from which it follows that \[f(x+y)=f(x)+f(y)-a,\quad \forall x,\,y \in \mathbb R,\, x \ge 0.\quad (6)\]From here, we can also show that $(6)$ still holds in the case $x<0$ (this is an easy job), so we have \[f(x+y)=f(x)+f(y)-a,\quad \forall x,\,y \in \mathbb R.\quad (7)\]

retre wrote: Who can explain to me the solution of CanVQ please Quote: from which it follows that \[f(x+y)=f(x)+f(y)-a,\quad \forall x,\,y \in \mathbb R,\, x \ge 0.\quad (6)\]From here, we can also show that $(6)$ still holds in the case $x<0$ (this is an easy job), so we have \[f(x+y)=f(x)+f(y)-a,\quad \forall x,\,y \in \mathbb R.\quad (7)\] Me too Thank you!

@above (now I can do it) pluggin $y\implies -x^2,y\implies -x^2+y$ Combine this two equalities we get the desired result.

emregirgin35 wrote: Find all $f$ functions from real numbers to itself such that for all real numbers $x,y$ the equation \[f(f(y)+x^2+1)+2x=y+(f(x+1))^2\]holds. here is how i did it. . let $$p(x,y)=f(f(y)+x^2+1)=f(x+1)^2+y-2x$$now we have from $p(z,f(y)+x^2+1)$ that $$f(y+(f(x+1)^2-2x+1+z^2))=f(y)+(f(z+1)^2-2z+x^2+1)$$so let $g(x,z)=f(x+1)^2-2x+1+z^2$ so $$f(y+g(x,z))=f(y)+g(z,x)$$also $$f(y+g(z,x))=f(y)+g(x,z)$$so if $Q(x,z)=f(x,z)+f(z,x)$ we get $$f(y+Q(x,z))=f(y)+Q(x,z)$$now we have from $$p(x,y)-p(-x,y)=f(x+1)^2=f(1-x)^2+4x=Z(x)$$now we have from $Z(x+g(a,b))$ that. $$f(1+x+g(a,b))^2=4(x+g(a,b))+f(1-x-g(a,b))^2$$so $$(f(1+x)+g(b,a))^2=4x+4g(a,b)+(f(1-x)-g(b,a))^2$$so $$g(b,a)f(1,x)=2g(a,b)-g(b,a)f(1-x)$$so either $g(b,a)=0$ then $g(a,b)=g(b,a)$ or $$\frac{g(a,b)}{g(b,a)}$$is a constant function changing $a,b$ gives $g(a,b)^2=g(b,a)^2$ so if $g(a,b)=g(b,a)$ is not true then $g(a,b)+g(b,a)=0$ but $g(a,b)+g(b,a)=f(a+1)^2+f(b+1)^2+(a-1)^2+(b-1)^2$ so $a=b$ and $g(a,b)=g(b,a)$ so in all these cases we have $g(a,b)=g(b,a)$. so we are left with $f(x+1)^2-(x+1)^2=f(z+1)^2-(z+1)^2$ so $f(x)^2=x^2+c$ but as we know if $c=0$ is not true then there is a non_zero $a$ such that $f(x+a)=f(x)+a$ so $$x^2+a^2+2ax+c=f(x+a)^2=f(x)^2+a^2+2af(x)$$so $f(x)=x$ and $c=0$ . so $f(x)=x$ for all real $x$ which finishes the problem. . . really liked this one it was super fun to solve.

I think your $P(x,f(y)+1)$ is incorrect, so is the next step.

Let $P(x,y)$ be the assertion. Comparing $P(x,a)$ and $P(x,b)$ shows us that $f$ is injective. Also, fixing $x$ and varying $y$ shows us that $f$ is surjective. Subtracting $P(0,y)$ from $P(-1,y)$ gives us $$f(f(y)+2)-f(f(y)+1)=2+f(0)^2-f(1)^2\text{ (which is constant)}$$Let $a=2+f(0)^2-f(1)^2$. Since $f$ surjective, we find that $f(x+1)-f(x)=a$ .......(1) Then, we can easily show that $f(x+k)-f(x)=ka$ for all $k\in\mathbb{Z}$ .....(2) By (1), $f$ is linear over the integers. Let $f(n)=an+b$ for all $n\in\mathbb{Z}$. (By (1), we know that the coefficient of $n$ should be $a$). Now, $P(n,y)$ (where $n\in\mathbb{Z}$) gives us $$f(f(y)+n^2+1)+2n=y+f(n+1)^2\Rightarrow f(f(y))+(n^2+1)a+2n=y+(an+a+b)^2\Rightarrow$$$$f(f(y))-y=n^2(a^2-a)+2n(a^2+ab-1)+(a+b)^2-a\text{ ......(3)}$$Fix $y$ in (3). Then, RHS is constant for all integers $n$. Therefore, we need to have $a^2-a=a^2+ab-1=0$. This gives us $(a,b)=(1,0)$. Hence, $f(n)=n$ for all $n\in\mathbb{Z}$. By (3), we find that $f(f(y))-y=0\Rightarrow f(f(y))=y$ for all $y\in\mathbb{R}$. By (2), we have $f(x+k)-f(x)=k$ for all $k\in\mathbb{Z}$. $P(x,0)$ gives us $f(x+1)^2=f(x^2+1)+2x=f(x^2)+2x+1$ .......(4) Then, $y+f(x^2)+2x+1=f(f(y)+x^2+1)+2x=f(f(y)+x^2)+2x+1\Rightarrow y+f(x^2)=f(f(y)+x^2)$. Since $f$ is surjective and $f(f(y))=y$, we find that $f(u)+f(v)=f(u+v)$ if at least one of $u,v$ is non-negative. Hence, $f(x)+f(-x)=f(0)=0\Rightarrow -f(x)=f(-x)$. By (4), we find that $f(x^2)+2x+1=f(x+1)^2=[f(x)+1]^2=f(x)^2+2f(x)+1$. Hence, $f(x^2)+2x=f(x)^2+2f(x)$ ....(5) Hence, $f(x^2)-2x=f(-x)^2+2f(-x)=f(x)^2-2f(x)$ ........(6) Subtracting (6) from (5) gives us that $4x=4f(x)\Rightarrow f(x)=x$, done.

This solution does not use injectivity or surjectivity. $P(x,y), P(-x,y) \implies (*)4x=f(x+1)^2-f(1-x)^2$ $P(0,x) \implies f(f(x)+1)=x+f(1)^2 \implies f(x+f(1)^2+1)=f(x)+1+f(1)^2 \implies (**)f(x+c)=f(x)+c$ where $c=f(1)^2+1 \neq 0$. $x \rightarrow x+c$ in $(*)$ implies $$4(x+c)=(f(x+1)+c)^2-(f(1-x)-c)^2=4x+2c(f(x+1)+f(1-x)) \implies f(x+1)+f(1-x)=2$$Using the last one in $(*)$ gives $f(x+1)^2-(2-f(x+1))^2=4x \implies f(x)=x$ for all $x$.

Nice problem.

Redacted

This solution is so disgusting it has a certain kind of beauty to it. The answer is $f \equiv x$ only, which works. Opening Observations: By fixing $x$ and letting $y$ vary, $f$ must be surjective. Furthermore, if $f(y_1) = f(y_2)$ and $x$ is fixed, then $y_1=y_2$, so $f$ must also be injective. First Part: Set $x = 0$ to obtain \[f(f(y) + 1) = y+f(1)^2. \ \ (1)\]Set $g(y) = f(y) + 1$, so $g(g(y)) = y+1+f(1)^2$. Using the so-called ``$fff$ trick", \[g\left(y+f(1)^2+1\right) = g(g(g(y))) = g(y) + 1 + f(1)^2.\]Equivalently, \[f\left(y+f(1)^2+1\right) = f(y) + f(1)^2+1. \ \ (2)\]Setting $x=f(x)$ in the original and using statement $(1)$ yields \[f\left(f(y)+f(x)^2+1\right) + 2f(x) = y + x^2 + 2f(1)^2 x + f(1)^4.\]Now setting $x = 1$ here and equating with $(2)$ yields \[f(f(y)) + f(1)^2 + 1 = f\left(f(y) + f(1)^2 + 1\right) + 2f(1) = y+2f(1)+f(1)^4+1.\]Thus \[f(f(y)) = y+2f(1)+f(1)^4 - f(1)^2 \ \ (3)\]but equating this with $(1)$ and noting that $f$ is surjective, it follows that $f(y+1) = f(y) + c$ for some constant $c$; in other words, $f \equiv ax+b$ on $\mathbb Z$. Equating in the original equation, \[a\left(ay+b + x^2 + 1\right) + b = y + (ax+a+b)^2\]yields $a=1$ and $b=0$, so $f \equiv x$ on $\mathbb Z$. Second Part: Now we close in for the kill. Equation $(3)$ now reads $f(f(y)) = y$, and setting $y=0$ in the original and using $f(x+1) = f(x)+1$ yields \[f\left(x^2\right) + 2x = f(x)^2 + 2f(x) \ \ (4).\] Combining this with the original equation, \[f\left(f(y) + x^2\right) + 2x = y + f(x)^2 + 2f(x) = y + f\left(x^2\right) + 2x.\]In other words, \[f\left(f(y) + x^2\right) = y + f\left(x^2\right).\]As $f(f(y)) = y$, it follows that $f$ is additive, and letting $f(y) = -x^2$ yields that $f$ is odd. Thus setting $-x$ in $(4)$ yields \[f\left(x^2\right) - 2x = f(x)^2 - 2f(x) \ \ (5).\]Subtracting $(5)$ from $(4)$ yields $f(x) = x$, as needed.

The only solution is $\boxed{f(x) = x}$, which clearly works. Now we show it's the only one. Let $P(x,y)$ denote the given assertion. Claim: $f$ is bijective. Proof: To show $f$ is injective, note that if $f(a) = f(b)$, comparing $P(x,a)$ and $P(x,b)$ gives $a=b$. To show $f$ is surjective, setting $y = -f(x+1)^2$ gives that $-2x$ is in the image of $f$, so $f$ is surjective. $\square$ $P(0,y): f(f(y) + 1) = y + f(1)^2$. Adding both sides by $1$ gives \[f(f(y) + 1) + 1 = y + f(1)^2 + 1 \ \ \ \ (1) \] $P(0, f(y) + 1): f( f(f(y) + 1) + 1) = f(y) + f(1)^2 + 1$. Thus from applying $f$ to both sides of $(1)$, $f(y + c) = f(y) + c$, where $c = f(1)^2 + 1$. $P(x + c, f^{-1}(y)) - P(x, f^{-1}(y)): f(y + x^2 + 1 + 2xc + c^2 ) - f(y + x^2 + 1) + 2c = 2f(x+1) c + c^2$. Firstly, setting $y = -1 -x^2 $ gives that $f(2xc + c^2) = f(0) + 2f(x+1) c + c^2 - 2c$. Hence $f(y + x^2 + 1 + 2xc + c^2) = f(y + x^2 + 1) + f(2xc + c^2) - f(0)$. For any reals $a,b$, setting $y = a - (x^2 + 1)$ and $x = \frac{b - c^2}{2c}$ (which is possible since $c > 0$) gives that $f(a+b) = f(a) + f(b) - f(0)$ for any reals $a,b$. Claim: For any $t \ge f(0) + 1$, we have $f(t) \ge 0$. Proof: Set $y = 0$ and $x$ be the non-positive number so that $f(0) + 1 + x^2 = t$. We have\[f(t) = f(x+1)^2 - 2x \ge f(x+1)^2 \ge 0 \ \ \ \ \square\]Now let $g(x) = f(x) - f(0)$. We get that $g(a+b) + f(0) = g(a) + f(0) + g(b) + f(0) - f(0) $, which implies $g(a+b) = g(a) + g(b) \forall a,b \in \mathbb R$. Therefore, $g$ is additive, and since we have for all $t \ge f(0) + 1$, $g(t) \ge -f(0)$, $g$ is bounded on a nontrivial interval ($[f(0) + 1, \infty)$) and thus is linear, so $f$ is also linear. Let $f(x) = ax + b$ for real constants $a,b$. $P(0,y): f(ay + 1 + b) = y + f(1)^2$. Thus, $a^2 y + a + ab + b = y + f(1)^2$. If $a^2 \ne 1$, we can vary $y$ so that this is not true, so $a^2 = 1$ must hold. If $a = -1$, the above equation becomes $-1 = f(1)^2$, which is false. Thus, $a = 1$ and $f(x) = x + b$. The above equation becomes $2b + 1 = (b+1)^2$, so $b^2 = 0 \implies b = 0$, and therefore $f$ is the identity. edit: oops i just realized you can get $f$ is additive + constant by $P(x+1,f^{-1}(y)) - P(x,f^{-1}(y))$ and the whole thing about $c$ was unnecessary