Find all functions $f\colon \mathbb{R} \to \mathbb{R}$ such that $f(af(b) − b^2) = (b + 1)f(a − b)$ for all reals $a$ and $b$

Question

Find all functions $f\colon \mathbb{R} \to \mathbb{R}$ such that $$ f(af(b) − b^2) = (b + 1)f(a − b) $$ for all $a, b \in \mathbb{R}$.

Holds for f(x) = x and f(x) = 0 but I can't seem to prove its the only solution (if it is).

Edit: Sorry made a mistake, it doesn't hold for f(x) = x.

Answer 1

Consider $a=0$ which gives us $f(-b^2)=(b+1)f(-b)$ Making the substitution $b \rightarrow -b$ we arrive at $f(-b^2)=(1-b)f(b)$ and therefore $(b+1)f(-b)=(1-b)f(b)$. So $b=1$ gives us $2f(-1)=0$ and we have $f(-1)=0$. Similarly with $b=-1$ we have $2f(1)=0$ so $f(1)=0$.

Now setting $b=1$ in the original expression gives $f(af(1)-1)=2f(a-1)$ and so $f(af(1)-1)=f(-1)=0$ meaning $f(a-1)=0$ and so $f$ is the zero function.

Answer 2

Note that $b = -1$ yields the relation $$ f(af(-1)-1) = 0, $$ which holds for all $a \in \mathbb R$. Assume that $f(-1) \neq 0$, then this implies that $f(x) = 0$ for all $x$. A contradiction since we said $f(-1) \neq 0$. Therefore $f(-1) = 0$ is forced. Consider the line $a = b -1$ then we have $$ f((b-1)f(b)-b^2) = 0. $$ Suppose that $(b-1)f(b) - b^2 = c^2$, then we have $$ f(x) = \frac{x^2 + c^2}{x - 1} $$ but also we have $f(c^2) = 0$. Which is never true unless $c = 0$, but then note that $x = -1$ is not 0 so this is not a solution. So we instead look at $(b-1)f(b) - b^2 = -c^2$, which yields $$ f(x) = \frac{(x+c)(x-c)}{x-1} $$ again with the condition $f(-c^2) = 0$, which implies that the only solutions are $c = 1$.

If $c = 1$, then we see that $f(x) = x+1$ with $x \neq 1$. What happens at $x = 1$? Well, lets go back to the functional equation with $a-b = 1$: \begin{align} f((b+1)f(b) - b^2) & = (b+1)f(1) \\ f(2b+1) & = (b+1)f(1) \\ 2(b+1) & = (b+1)f(1) \\ 2 & = f(1). \end{align} The above equalities require $b\neq -1, 0, 1$ for all the operations to be valid. Hence $f(1) = 2$ and so there is no discontinuity. Therefore $f(x) = x + 1$ is the other solution.

Answer 3

I'm not sure $f(x)=x$ works. I think I solved your problem :

If I understood you well we have $f:\mathbb{R}\mapsto\mathbb{R}$ verifying, for any $a,b\in\mathbb{R}$ : $$f(af(b)-b^{2}) = (b+1)f(a-b)\label{eq0}\tag{0}$$

if you evaluate it for $a=0$, you get : $$f(-b^{2}) = (b+1)f(-b)\label{eq1}\tag{1}$$

if you evaluate \ref{eq1} by replacing $b$ by $-b$, seeing that $f(-b^{2}) =f( (-b)^{2})$ we see that :

$$(-b+1)f(b) = (b+1)f(-b)\label{eq2}\tag{2}$$

if you evaluate it in $b=1$ and $b=-1$ you get $$f(1)=f(-1)=0\label{eq3}\tag{3}$$

know if we evaluate \ref{eq0} with b=1, you get:

$$f(af(1)-1) = f(a-1)\label{eq4}\tag{4}$$

so if we remember \ref{eq2} we have, for any a :

$$f(a-1)=0\label{eq5}\tag{5}$$

so we can conclude that f is identically zero.

Answer 4

The solutions are $f(x) = 0$ and $f(x) = x + 1$. The following is my un-optimized deduction.

• Set $b=1$: $$ f(f(1)a-1)=2f(a-1). $$ This implies that $f(1) \neq 0$ otherwise $f(x)$ is constant, meaning $f(x) \equiv 0$ (one solution). So suppose $f(1) \neq 0$.

• Set $a=0$: $$ f(-b^2)=(b+1)f(-b)=(1-b)f(b), \tag{1}$$ where the latter equation follows from the former by replacing $b$ with $-b$. Setting $b=1$ in (1) gives us $f(-1) = 0$.

• Set $b=0$ then we have $f(f(0)a)=f(a)$. Setting $b=f(0)$ in (1) gives us $f(0) =1$, since $f(-1) = 0$ and $f(1) \neq 0$.

• Set $a=b$: $$f(b(f(b)-b))= b+1. \tag{2}$$

• Set $a=2b-f(b)$: $$f(-(b-f(b))^2)=(b+1)f(b-f(b)). \tag{3}$$ By (1), $$(1-b+f(b))f(b-f(b)) = (b+1)f(b-f(b)).$$ This equation implies that if $f(b-f(b)) \neq 0$ then $f(b) = 2b$. For such $b$, $f(b^2) = b+1$ by (2). Moreover by comparing the left-most side and the right-most side of (1), we have $f(-b^2)=2b(1-b)$. On the other hand, by replacing $b$ with $b^2$ in the latter equation in (1), we have $2b(1-b)(b^2+1) = (1-b^2)(b+1)$. In reals, $b=1$ is a unique solution to this equation. Summarizing the above, $f(b-f(b)) = 0$ for $b \neq 1$.

• If $f(1-f(1)) \neq 0$ then $f(1)=2$. However now $f(1-f(1))=f(-1)=0$. Therefore $f(b-f(b)) = 0$ for all $b$.

• Finally, replace $b$ with $b-f(b)$. By (3) we have $$(b-f(b)+1)f(a-b+f(b))=0.$$ If $f(b) \neq b+1$ for some $b$, then $f(x)$ is constant and zero. Therefore $f(b) = b+1$ (which is a solution).