Comparing $P(x,y)$ and $P(x+y,0)$, we get $f(x+y) + f(x) + f(y) = 2f(x+y) + f(0)$ or $f(x+y) + f(0) = f(x) + f(y)$ for all $x \neq y$. This also implies $f(x+y)^2 = 2f(x+y) + f(0)$, or $(f(x+y)-1)^2 = f(0) + 1$ for all $x \neq y$. Setting $x = - y \neq 0$, we get $f(0) = 0$ or $f(0) = 3$. Setting $y=0$ we just get $(f(x) - 1)^2 = f(0) + 1$ for all $x \neq 0$. If $f(0) = 0$, then $f(x) = 0,2$ for all $x$. Suppose there's an $x$ with $f(x) = 2$. Then $f(x+y) + f(0) = f(x+y) = f(x) + f(y)$ gives $f(y) = 0$ for all $y \neq x$, but plugging in original equation gives contradiction. $f(x) = 0 \forall x$ is a solution on the other hand If $f(0) = 3$, we get $f(x) = -1,3$ for all $x$. Suppose there's an $x$ with $f(x) = -1$. Using $f(x+y) + f(0) = f(x+y) + 3 = f(x) + f(y) = -1 + f(y)$, and changing $y$ with $y-x$, we get $f(y) + 4 = f(y-x)$ for all $x$, implying $f(y) = -1$ for all $y$ not equal to $x$, but in original equation we get contradiction. $f(x) = 3$ is a solution though.

straight wrote: Comparing $P(x,y)$ and $P(x+y,0)$, we get $f(x+y) + f(x) + f(y) = 2f(x+y) + f(0)$ or $f(x+y) + f(0) = f(x) + f(y)$ for all $x \neq y$. This also implies $f(x+y)^2 = 2f(x+y) + f(0)$, or $(f(x+y)-1)^2 = f(0) + 1$ for all $x \neq y$. Setting $x = - y \neq 0$, we get $f(0) = 0$ or $f(0) = 3$. Setting $y=0$ we just get $(f(x) - 1)^2 = f(0) + 1$ for all $x \neq 0$. If $f(0) = 0$, then $f(x) = 0,2$ for all $x$. Suppose there's an $x$ with $f(x) = 2$. Then $f(x+y) + f(0) = f(x+y) = f(x) + f(y)$ gives $f(y) = 0$ for all $y \neq x$, but plugging in original equation gives contradiction. $f(x) = 0 \forall x$ is a solution on the other hand If $f(0) = 3$, we get $f(x) = -1,3$ for all $x$. Suppose there's an $x$ with $f(x) = -1$. Using $f(x+y) + f(0) = f(x+y) + 3 = f(x) + f(y) = -1 + f(y)$, and changing $y$ with $y-x$, we get $f(y) + 4 = f(y-x)$ for all $x$, implying $f(y) = -1$ for all $y$ not equal to $x$, but in original equation we get contradiction. $f(x) = 3$ is a solution though. U give $P(x+y,0)$ then u give $P(x,-x)$ actually u give $P(0,0)$ however $x \neq y$ I think there is a mistake in this case.

@above Thank you! Didn't notice at first! That's the trick of this problem - not to accidently make a prohibited substitution

nAalniaOMliO wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any reals $x \neq y$ the following equality is true: $$f(x+y)^2=f(x+y)+f(x)+f(y)$$ Let $P(x,y)$ be the assertion $f(x+y)^2=f(x+y)+f(x)+f(y)$ $\forall x\ne y$ Let $a=f(0)$ $P(x,0)$ $\implies$ $(f(x)-1)^2=a+1$ $\forall x\ne 0$ and so $a\ge -1$ and $f(x)=1\pm\sqrt{a+1}$ $\forall x\ne 0$ $P(x,-x)$ $\implies$ $f(x)+f(-x)=a^2-a$ $\forall x\ne 0$ And since $LHS \in\{2-2\sqrt{a+1},2,2+2\sqrt{a+1}\}$, we easily get $a\in\{-1,0,2,3\}$ If $a=-1$, first result implies $f(x)=1$ $\forall x\ne 0$, which does not fit. If $a=0$, first result implies $f(x)\in\{0,2\}$ $\forall x\ne 0$ and second result is $f(x)+f(-x)=0$ $\forall x\ne 0$ and so : $\boxed{\text{S1 : }f(x)=0\quad\forall x}$, which indeed fits If $a=2$, first result implies $f(x)\in \{1-\sqrt 3,1+\sqrt 3\}$ $\forall x\ne 0$ Let $x,y,x-y,x+y\ne 0$ so that $f(x)=1+\epsilon_x\sqrt3$, $f(y)+\epsilon_y\sqrt 3$ and $f(x+y)=1+\epsilon_s\sqrt 3$ $P(x,y)$ becomes $(\epsilon_x+\epsilon_y-\epsilon_s)\sqrt 3=1$, which is impossible. If $a=3$, first result implies $f(x)\in\{-1,3\}$ $\forall x\ne 0$ and and second result is $f(x)+f(-x)=6$ $\forall x\ne 0$ and so : $\boxed{\text{S2 : }f(x)=3\quad\forall x}$, which indeed fits

nAalniaOMliO wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any reals $x \neq y$ the following equality is true: $$f(x+y)^2=f(x+y)+f(x)+f(y)$$D. Zmiaikou (f(x+y)-1)f(x+y)=f(x)+f(y) Then easy to assume that f(x) = 0 and f(x)=3 Assume f(x)= 3 as it only one number which accepts given problem

Let $f(0)^2-f(0)=c$. $P(x,-x): f(-x)=c-f(x)$ for $x \neq 0$ $P(-x,-y): (c-f(x+y))^2=c-f(x+y)+c-f(x)+c-f(y)=3c-f(x+y)^2$ for $x \neq y$ Thus, $2f(x+y)^2-2cf(x+y)+c^2-3c=0 \implies f(x+y)=\frac{2c \pm \sqrt{24c-4c^2}}{4}$, so $f(x)$ is constant for nonzero $x$. $P(x,0): f(0)=k^2-2k$ if $f(x)=k$ Thus, checking, only $f(x)=0$ and $f(x)=3$ work.

navier3072 wrote: $\implies f(x+y)=\frac{2c \pm \sqrt{24c-4c^2}}{4}$, so $f(x)$ is constant for nonzero $x$. Could you kindly explain how from $f(x)=$either $u$, either $v$, you immediately conclude that $f(x)$ is constant ? (why would be immediately impossible to have $f(x)=u$ for some $x$ and $f(x)=v$ for another $x$ ?)

$c=f(0)^2-f(0)$ so $\frac{2c \pm \sqrt{24c-4c^2}}{4}$ is either fixed or does not exist?

navier3072 wrote: $c=f(0)^2-f(0)$ so $\frac{2c \pm \sqrt{24c-4c^2}}{4}$ is either fixed or does not exist? It is not fixed, these are two distinct values $\frac{2c -\sqrt{24c-4c^2}}{4}$ and $\frac{2c +\sqrt{24c-4c^2}}{4}$

Oops, I forgot it was a quadratic, sorry. Suppose these roots are distinct roots $\alpha$ and $\beta$. Then, $f(x)=\alpha$ or $\beta$ at each $x \neq 0$. Since $\alpha + \beta = c$, if $f(x_0)=\alpha$ for some positive $x_0 \neq 0$, then $f(-x_0)=\beta$ Considering $P(x_0,0)$ and $P(-x_0,0)$, we get $\alpha ^2-2\alpha=\beta^2-2\beta \implies \alpha+\beta=c=2$ since $\alpha \neq \beta$ Since $\alpha \beta = \frac{c^2-3c}{2}$, WLOG $a= 1+\sqrt{2}$ and $b=1-\sqrt{2}$. Suppose $f(x_1)=\alpha$ and $f(y_1)=\beta$. Then, either $f(x_1+y_1)=\alpha$ or $\beta$ Thus, either $\alpha^2=2\alpha+\beta$ or $\beta^2=2\beta+\alpha$, which, substituting the numerical values, does not work. There is probably a faster way to deal with this pointwise (if this does work)

2 liner

I think, similar to my error above, this line lksb wrote: $f(x)^2=2f(x)+f(0)\implies f(x)=c$ is not correct

f(x)=0 by assuming and f(x)=3 by checking