Functional equation

The only solution is $f(x) = 1 - \frac{x^2}{2}$ . This problem is IMO 1999/6, and it's actually pretty hard to proof rigorously.

Note: Most solutions make the error of substituting in $f(y) = C$ for some value $C$ , but provide no justification for why this is possible. For example, to calculate $f(2016)$ , some solutions use "Let $f(y) = 2016$ , then ... ". However, if observe the function $f$ , you will see that 2016 is not in the range, and thus there is no corresponding $y$ value that we can use.

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On to the solution. I've broken it up into 6 parts, so that it's clearer to see why I took those steps.

Let $M$ be the image $f( \mathbb{R} )$ . Let $c = f(0)$ .

1) We will show that for $m \in M$ , $f(m) = \frac{ 1 + c - m^2 } { 2 }$
Since there is a real value $p$ such that $f(p) = m$ , we can set $x = m, y = p$ :
$f ( m - m) = f( m) + m^2 + m - 1 \Rightarrow f(m) = \frac{ 1 + c - m^2 } { 2 } \quad (1)$

2) We will show that $c \neq 0$ .
Proof by contradiction. Suppose not, then with $x = x, y = 0$ , $f(x) = 0 + x \times 0 + f(x) - 1 = f(x) - 1$ which is a contradiction.

3) We will show that $M - M = \mathbb{R}$ . What this means it that given any real value $r$ , there exists $a, b \in M$ such that $a - b = r$ .
Set $x = x, y = 0$ : $f(x-c) = f(c) + xc + f(x) - 1$ .
Hence $f(x-c) - f(x) = xc + f(c) - 1$
Observe that the RHS is a linear (non-constant) function in $x$ , and thus the range is all real numbers.

4) We will show that $f(x) = c - \frac{r^2}{2}$ for all real $x$ .
Now, given any $r \in \mathbb{R}$ , we have $a, b \in M$ such that $r = a - b$ , with $a = f(p), b = f(q)$ .
Set $x = a, y = q$ , we get $f( a- b ) = f(b) + ab + f(a) - 1$ .
Using equation 1, we get $f(r) = \frac{ 1 + c - b^2 } { 2} + ab + \frac{1+c - a^2 } { 2} - 1 = c - \frac{ r^2}{2} \quad (4)$

5) We will find the value of $c$ .
In equation (1), set $m = c$ to get $f(c) = \frac{ 1 + c - c^2 } { 2 }$ .
In equation (4), set $r = c$ to get $f(c) = c - \frac{ c^2 } { 2 }$ .
Equating these 2 expressions, $c = 1$ .

6) In conclusion, $f(x) = 1 - \frac{ x^2}{2}$

Now, it remains to show that this satisfies the original functional equation, which can be manually done.

$\begin{aligned} f(x-f(y)) & = f(f(y))+xf(y)+f(x) - 1 & \small \color{#3D99F6}{\text{Putting }x=f(y)} \\ f(0) & = f(f(y))+(f(y))^2+f(f(y)) - 1 \\ \implies 2f(f(y)) & = f(0) + 1 - (f(y))^2 & \small \color{#3D99F6}{\text{Replacing }f(y) \text{ with }x} \\ 2f(x) & = f(0) + 1 - x^2 & \small \color{#3D99F6}{\text{Putting }x=0} \\ 2f(0) & = f(0) + 1 \\ \implies f(0) & = 1 \\ \implies f(x) & = 1 - \frac {x^2}2 \\ \implies f(2016) & = 1 - \frac {2016^2}2 = \boxed{-2032127} \end{aligned}$

Putting $x=1$ and $f\left( y \right) =1$ .
We get $f\left( 0 \right) =2\times f\left( 1 \right)$ $\longrightarrow (1)$

Now, let's put $x=2$ and $f\left( y \right) =1$ .
We get $f\left( 2 \right) =-1$

Now we put $x=2$ and $f\left( y \right) =0$ which gives $f\left( 0 \right) =1$ .
If we put this value into the 1st equation we will get $f\left( 1 \right)$ = $\frac{1}{2}$

Till now, We have
$f\left( 0 \right) =1$
$f\left( 1 \right)$ = $\frac{1}{2}$
$f\left( 2 \right) =-1$

Now, let's try and find a pattern in the equation. If we put $f\left( y \right) =1$ in the given equation,

$f\left( x-1 \right) =f\left( 1 \right) +x+f\left( x \right)-1$

i.e. $f\left( x \right) =f\left( x-1 \right) -x+\frac { 1 }{ 2 }$ $\longrightarrow (2)$

If we put $x=2016$ in 2nd equation,

$f\left( 2016 \right) = f\left( 2015 \right)-2016+\frac{1}{2}$ $\longrightarrow (3)$

Similarly, if $x=2015$ , $f\left( 2015 \right) = f\left( 2014 \right)-2015+\frac{1}{2}$

We can put back the value of $f\left( 2015 \right)$ in the 3rd equation
We will drill down to $f\left( 2 \right)$ similarly.

Therefore, $f\left( 2016 \right)= f\left( 2 \right)+ 2014 \times \frac{1}{2} - (2016+2015......+4+3)$

Putting the value of $f\left( 2 \right) = -1$ .

We will get $f\left( 2016 \right)$ as -2032127.

So the correct answer is $\boxed{-2032127}$

I suggest a slightly faster solution where we don't need to solve for $f(0)$ :

Rearranging,

$f(x)-f(x-f(y)) = 1-f(f(y)) - xf(y)$

From this we can see all real $c$ can be expressed as $f(m)-f(n)$ for some $m, n$ .

(To see this, we choose y such that $f(y)$ is not 0 (and this exists as $f(x)=0$ $\forall x$ does not satisfy the functional equation). Then we can find a value of $x$ such that $1-f(f(y)) - xf(y) = c$ . So in this case $m = x, n=x-f(y)$ .)

For all real $m, n$ , letting $x = f(m), y=n$ we see that $f(f(m)-f(n)) = f(f(n)) + f(f(m)) + f(m)f(n) - 1$ which is symmetric about $m, n$ . Thus $f(f(m)-f(n)) = f(f(n)-f(m))$ . But by the above fact, for all $c$ , we have $f(c)=f(-c)$ , so the function is even.

Now substitute $x = 0.5f(y)$ . We can do this as $0.5f(y)$ is real. Then $f(x)=f(x-f(y))$ since the function is even, hence $1-f(f(y)) - xf(y) = 0$ .

Hence we have $f(f(y)) = -0.5(f(y))^2 + 1$ for all $y$ .

Let $c = f(m)-f(n)$ , then

$f(c) = f(f(m)-f(n)) = f(f(m)) + f(f(n)) + f(m)f(n) - 1$

$= (-0.5[f(m)]^2 + 1) + (-0.5[f(n)]^2 + 1) + f(m)f(n) - 1 = -0.5(f(m)-f(n))^2 + 1 = -0.5c^2 + 1$

Thus that's our solution! (Checking can be done easily.)