quacksaysduck wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that \[f(x)^2+f(2y+1)=x^2+f(y)+y+1\]for all reals $x$, $y$. Let $P(x,y)$ be the assertion $f(x)^2+f(2y+1)=x^2+f(y)+y+1$ $P(x,-1)$ $\implies$ $f(x)^2=x^2$ and so $\forall x$, either $f(x)=x$, either $f(x)=-x$ If $f(x)=-x$ for some $x\ne 0$, $P(0,x)$ implies $f(2x+1)=1$ and so : Either $1=f(2x+1)=2x+1$ and so $x=0$ (and so $f(x)=x$ too) Either $1=f(2x+1)=-(2x+1)$ and so $x=-1$ Anc the two solutions : $\boxed{\text{S1 : }f(x)=x\quad\forall x}$, which indeed fits $\boxed{\text{S2 : }f(-1)=1\text{ and }f(x)=x\quad\forall x\ne -1}$, which indeed fits

One solution is $f(x) = x$. The other one is $f(x) = x \forall x \ne -1$ and $f(-1) = 1$. These clearly work. Now we show they are the only ones. Let $P(x,y)$ be the given assertion. $P(x,-1): f(x)^2 = x^2$, which implies $f(x)$ is $x$ or $-x$ for each $x$. Thus, \[ f(2y + 1) = f(y) + y + 1\] If $f(y) \ne y$, then $f(y) = -y$ so $f(2y + 1) = 1$, so $2y + 1$ is $1$ or $-1$. The former implies $y = 0$, so $f(0) = -0 = 0$, absurd. Thus, $2y + 1 = -1$, so $y = -1$. This implies $f(x) =x $ for all $x \ne -1$, and $f(-1) \in \{-1,1\}$, as desired.

megarnie wrote: One solution is $f(x) = x$. The other one is $f(x) = x \forall x \ne -1$ and $f(-1) = 1$. These clearly work. Now we show they are the only ones. Let $P(x,y)$ be the given assertion. $P(x,-1): f(x)^2 = x^2$, which implies $f(x)$ is $x$ or $-x$ for each $x$. Thus, \[ f(2y + 1) = f(y) + y + 1\] If $f(y) \ne y$, then $f(y) = -y$ so $f(2y + 1) = 1$, so $2y + 1$ is $1$ or $-1$. The former implies $y = 0$, so $f(0) = -0 = 0$, absurd. Thus, $2y + 1 = -1$, so $y = -1$. This implies $f(x) =x $ for all $x \ne -1$, and $f(-1) \in \{-1,1\}$, as desired. wow! solution!