Let's put down $a = -b$, then we have that $f(b^2) = f(b) + f(-b) - f(b)f(-b)$. If we need that f would be nondecreasing, it must be held that $f(b^2) >= f(b)$, so $f(-b)[1-f(b)] >= 0$ too. From that we have only two possibilities to discuss: CASE 1: $f(b) = 1$ and $f(-b) = 0$ (this implies that for each $a<0: f(a)=0$), then we can put down this into the formula (ii) from the task and obtain that $1 = f(a+b-ab)$. It can be easily seen that for each convenient $a, b$ we have $a+b-ab = x$ where $x$ can be each real number more than or equal $x$ because of the equation $(b-1)(1-a) = x-1$ has always a solution $b>=1$ for fixed $a, x$. So the function f generated in case 1 must satisfy (and it's enough): $ a<=0 ==> f(a) = 0,$ $0<a<1 ==> f: a ---> f(a)$, such that $0<f(a)>1$ and for each$0<a_1<a_2<1: f(a_1)<=f(a_2),$ $b>=1 ==> f(b) = 1,$ CASE 2: we have only $f(-b) = 0$ (this implies that for each $a<0: f(a)=0$), then we can put down this into the formula (ii) in which we only consider that $a<0$, then we obtain that $f(b) = f(a+b-ab)$. We can easily see that for each $a<=0$ and convenient $b: a+b-ab>=b$ and for each $b$ we can find any $a<0$ such that $a+b-ab = x$, where $x$ is fixed real number more or equal 1. This signifies that for each $b: f(b) =$ some constant $C>=1$. Now let's take remaining $0<a<1$. From formula (ii) $f(a) + C - f(a).C = C$, so $f(a).(1-C) = 0$. Now there are 2 possibilities: $C = 1$, then we obtain the same set of functions f as in case 1, or $C > 1$, then $f(a) = 0$ and we have another set of convenient function f: $ a<1 ==> f(a) = 0,$ $f(1)=1,$ $b>1 ==> f(b) = $some constant$ C > 1,$ I think that's all... It may contain some mistake, but I think that the inicial substitution $a = -b$ to obtain $f(b^2) = f(b) + f(-b) - f(b)f(-b)$ is good way to solve this problem...

I think this solution is broken, not least because it misses f(x) = x. What happened (I think) was that the inequality $f(-b)[1-f(b)]\geq0$ turned into the equality $f(-b)[1-f(b)]=0$. Does anyone have a link to a correct solution to this question on the forum? Jack

Please check the IMO ShortList 2003 solutions booklet!

In which part of Mathlinks can I download this booklet?

Thanks, orl.

Couldn't anyone solve this problem? It seems the above solution is not correct.

(i) $f(0) = 0, f(1) = 1$ (ii) $f(a)+f(b) = f(a)f(b)+f(a+b-ab)$ for all $a < 1 < b$ $g(x): =1-f(1-x)$ is a nondecreasing function $\mathbb{R}\rightarrow\mathbb{R}$ with $g(0)=0$ and $g(1)=1$ and (ii) gives $g((1-a)(1-b))=g(1-a)g(1-b)$ for all $a < 1 < b$, or (*) $g(ab)=g(a)g(b)$ for all $a>0>b$. Then for all $a,a'>0>b$ we have $g(aa')g(b)=g(aa'b)=g(a)g(a'b)=g(a)g(a')g(b)$. So either $g\equiv 0$ on $\mathbb{R}_{-}$, or $g$ is a nondecreasing solution of the Cauchy equation (**) $g(aa')=g(a)g(a')$ for all $a,a'\in\mathbb{R}_{+}$. Also $c: =-g(-1)\geq-g(0)=0$, then $g(x)=-cg(|x|)$ for all $x<0$. The rest is easy. It leads to $2$ types of solutions. Type I: $f: \mathbb{R}\to\mathbb{R}$ arbitrary nondecreasing with $f(0)=0$ and $f\equiv 1$ on $[1,\infty)$. So $h: (-\infty,0)\to (-\infty,0]$ and $k: (0,1)\to [0,1]$ arbitrary nondecreasing with $f(x)=\{\begin{array}{cl}h(x)&\text{for }x<0\\ 0 &\text{for }x=0\\ k(x) &\text{for }0<x<1\\ 1&\text{for }x\geq 1 \end{array}$ Type II: $f(x)=\{\begin{array}{cl}1&\text{for }x=1\\ 1-(1-x)^\alpha &\text{for }x<1\\ 1+c(x-1)^\alpha&\text{for }x>1 \end{array}$ for some $\alpha,c\in\mathbb{R},\alpha,c\geq 0$.

rope0811 wrote: Proposed by A. Di Pisquale & D. Matthews, Australia Correction: A. Di Pisquale = Angelo Di Pasquale http://www.imo-official.org/year_reg_team.aspx?year=2013&code=AUS Comment: D. Matthews = Daniel Matthews

I claim the solution set is all functions $f$ such that $f(0) = 0$ $f(x) = 1$ for all $x \geq 1$ $f$ is non-decreasing and all functions such that $$f(x) = \begin{cases} -(1-x)^c + 1 &\text{ if $x < 1$} \\ 1 &\text{ if $x = 1$} \\ k(x-1)^c + 1 &\text{ if $x > 1$} \end{cases}$$for some constants $c \geq 0, k > 0$. First, we show they all work: Note that as $a < 1 < b$, \begin{align*} 0 &> (a-1)(b-1) \\ 0 &> ab - a - b +1 \\ a+b-ab &> 1\\ \end{align*} We first show all functions $f$ such that $f(0) = 0$ $f(x) = 1$ for all $x \geq 1$ $f$ is non-decreasing works. Clearly, condition 1 is fulfilled and $f$ is increasing, so we only need to check condition 2. Thus, for all $a < 1 < b$, \begin{align*} f(a) + f(b) &= f(a) \cdot 1 + 1\\ &= f(a)f(b) + f(a+b-ab) \end{align*}as desired. Next we show $$f(x) = \begin{cases} -(1-x)^c + 1 &\text{ if $x < 1$} \\ 1 &\text{ if $x = 1$} \\ k(x-1)^c + 1 &\text{ if $x > 1$} \end{cases}$$for some constants $c \geq 0, k > 0$ works as well. Since $c \geq 0$, $f$ is non-decreasing. Next, $f(0) = -(1-0)^c + 1 = 0$ and $f(1) = 1$ so condition 1 is met. Finally, \begin{align*} f(a)f(b) + f(a+b-ab) =& (-(1-a)^c + 1)(k(b-1)^c + 1) + k(a+b-ab-1)^c + 1 \\ =& -k\left[(1-a)(b-1)\right]^c -(1-a)^c + k(b-1)^c + 1\\ & +k\left[(1-a)(b-1)\right]^c + 1\\ =& (-(1-a)^c + 1) + (k(b-1)^c + 1)\\ =& f(a) + f(b) \end{align*}so condition 2 is met, as desired. Next, we show they are the only possible. Let $a = x+1, b = y+1, u = -x$, with $x < 0 < u,y$. Define $g(x) = f(x+1)-1$. Then \begin{align*} f(a) + f(b) &= f(a)f(b) + f(a+b-ab) \\ 1-f(x+1+y+1-xy-x-y-1) &= (f(x+1) + 1)(f(y+1) + 1) \\ -g(-xy) &= g(x)g(y) \\ -g(uy) &= g(-u)g(y) \text{ for }u,y > 0\\ \end{align*} Since $f(0) = 0$ and $f(1) = 1$, this implies $g(-1) = -1$ and $g(0) = 0$ respectively. Let $y=1$ to get $-g(u) = g(-u)g(1)$. Then \begin{align*} -g(uy) &= g(-u)g(y) \\ -g(uy) &= \frac{-g(u)}{g(1)}g(y) \\ g(uy)g(1) &= g(u)g(y) \end{align*} We then split into cases. Note that since $f$ is non-decreasing and $g(0) = 0$, $g$ is non-decreasing and $g(x) \geq 0$ for $x \geq 0$ Case 1: $g(x) = 0$ for some $x > 0$. Then let $y=x$ to get \begin{align*} -g(ux) &= g(-u)g(x) \\ g(ux) &= 0 \text{ for }u,y > 0 \end{align*} Thus $g(x) = 0$ for all $x > 0$. This implies that $f$ is any function such $f(0) = 0$ $f(x) = 1$ for all $x \geq 1$ $f$ is non-decreasing We showed this worked earlier, so we are done with this case. Case 2: $g(x) > 0$ for all $x > 0$. Then let $h(x) = g(x)/g(1)$, so then for positive $u,v$, $h(uv) = h(u)h(v)$, and $h(1) = 1$. Also note $h(x) > 0$ for $x > 0$ and $h$ is non-decreasing because $g$ holds those properties and $g(1) > 0$. Then define $j(x)$ taking $\mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $j(x) = \ln \left[ h(e^x)\right]$. ($j$ is undefined for negatives). Then as \begin{align*} h(uv) &= h(u)h(v) \\ h(e^x e^y) &= h(e^x)h(e^y) \\ \ln \left[ h(e^{x+y})\right] &= \ln \left[ h(e^x)\right] + \ln \left[ h(e^y)\right], \end{align*}we get that $j(x+y) = j(x) + j(y)$. Also, as $h$ is non-decreasing, $j$ is non-decreasing. Thus by Cauchy we get that $j(x) = cx$ for some constant $c$. Then over the positives $h(x) = x^c$. Let $g(1) = k$, with $k > 0$, then $g(x) = kx^c$ over the positives. Then by substitution, \begin{align*} -g(uy) &= g(-u)g(y) \\ -k(uy)^c &= g(-u)k(y)^c \\ g(-u) &= -u^c \end{align*} Thus, $$g(x) =\begin{cases} -(-x)^c &\text{ if $x < 0$} \\ 0 &\text{ if $x = 0$} \\ kx^c &\text{ if $x > 0$} \end{cases} $$ In order to have $g$ non-decreasing, we must have $c \geq 0$. So ultimately, as $f(x) = g(x-1)+1$, we get that $$f(x) = \begin{cases} -(1-x)^c + 1 &\text{ if $x < 1$} \\ 1 &\text{ if $x = 1$} \\ k(x-1)^c + 1 &\text{ if $x > 1$} \end{cases}$$ for some real constants $c \geq 0$, $k > 0$ in this case, which we showed earlier indeed works. As we have exhausted all cases and showed all remaining functions worked, we are done. $\blacksquare$ Notes: Solved with MOP 2020. Motivations: We introduce the substitution $g(x) = f(x)$ because after factoring the insides and outsides, it seems like a useful substitution. This reduces it down to a very Cauchy-like equation, and perhaps with logs and substitution and the like we can force it into Cauchy. So we do, and all that's left is being extremely careful, which is not easy, but it is straight forward to finish.

Essentially the same as the solution above with similar motivation - posting anyways for storage.

ISL marabot solve Let $P(a,b)$ denote the given assertion. \[P(-a+1,b+1): f(-a+1)+f(b+1)=f(-a+1)f(b+1)+f(-a+b+2+ab+a-b-1)=f(-a+1)f(b+1)+f(ab+1)\implies (f(-a+1)-1)(f(b+1)-1)=1-f(ab+1).\] Let $g(x)=f(x+1)-1$. Note $g(-1)=-1$ and $g(0)=0$. So \[g(-x)g(y)=-g(xy)\forall x,y>0.\] Setting $y=1$ here gives $g(-x)g(1)=-g(x)$. Thus, $g(-x)g(y)=\frac{-g(x)}{g(1)}g(y)=-g(xy)$. So let the new assertion $Q(x,y)$ denote \[g(x)g(y)=g(xy)g(1)\forall x,y>0\] Case 1: There exists a positive real number $k$ so that $g(k)=0$. If $g(1)=0$, then $Q(x,x)$ gives $g(x)= 0\forall x\ge 0$. If $g(1)\ne0$, then $Q(k,x): g(kx)=0$, so $g(x)=0\forall x\ge0$. So \[g(x)=0\forall x\ge 0\implies f(x+1)=1\forall x\ge 0\implies f(x)=1\forall x\ge 1.\] Thus, the solution here for $f$ is $\boxed{\text{anything that is nondecreasing}, f(0)=0, \text{ and } f(x)=1\forall x\ge 1}$. We will now check that this works. Note that $f(b)=1$. So we need $f(a)+1=f(a)+f(a+b-ab)\iff f(a+b-ab)=1$. Note if\[a+b-ab<1\iff ab-a-b+1<0\iff (a-1)(b-1)>0\iff a,b>1\text{ or }a,b<1,\]a contradiction. Thus, $f(a+b-ab)=1$, so this function works. Case 2: $g(x)>0\forall x>0$. Then define $h(x)=\frac{g(x)}{g(1)}$. So \[h(x)h(y)=h(xy)\forall x,y>0\]and $h(x)>0\forall x>0$. Now define $s\colon \mathbb{R}\to \mathbb{R}$ that satisfies $s(x)=\ln(h(e^x))$. Note that $s(x)+s(y)=s(x+y)$, so $s$ is additive. We have $g$ is non decreasing, so $h$ is nondecreasing, so $s$ is nondecreasing. Since $s$ is Cauchy and nondecreasing, $s(x)=kx$ for some real constant $k$. Thus, $h(x)=x^k\forall x>0$. Since $h$ is nondecreasing, $k\ge 0$. So $g(x)=cx^k\forall x>0$. Note $g(0)=0$ and \[g(-x)c=-g(x)=-cx^k\implies g(-x)=-x^k\forall x>0\implies g(x)=-(-x)^k\forall x<0\]Thus, \[g(x)=\begin{cases} -(-x)^k &\text{ if $x < 0$} \\ 0 &\text{ if $x = 0$} \\ cx^k &\text{ if $x > 0$} \end{cases} \] Note $f(x)-1=g(x-1)\implies f(x)=g(x-1)+1$. Thus, \[\boxed{f(x)=\begin{cases} -(1-x)^k+1 &\text{ if $x < 1$} \\ 1 &\text{ if $x = 1$} \\ c(x-1)^k+1 &\text{ if $x > 1$} \end{cases}} \]for some constants $c>0$ and $k\ge 0$ for this case. Now we will show that this works. Since $c>0$, this is nondecreasing. Also, it's obvious $f(0)=0$ and $f(1)=1$. Note $f(a)=-(1-a)^k+1$ and $f(b)=c(b-1)^k+1$. Since $a,b\ne1$, we had shown earlier that $a+b-ab>1$. So $f(a+b-ab)=c(-ab+a+b-1)^k+1$. We have \[f(a)f(b)+f(a+b-ab)=-c((1-a)(b-1))^k-(1-a)^k+c(b-1)^k+1+c((1-a)(b-1))^k+1=-(1-a)^k+1+c(b-1)^k+1=f(a)+f(b),\]as desired.

(i) is already solutions, now we find rest of them. Setting $g(x)=f(x+1)-1$ and $b-1=x, a-1=y$ we get $-g(x)g(y)=g(-xy)~~\forall y<0<x.$ $x\to 1\implies g(y)=-g(-y)g(1)=-kg(-y) \implies g(x)g(y)=kg(xy).$ As like $f,$ the function $g$ is non-decreasing and $g(0)=0, g(-1)=-1.$ We do case-work on the value of $k:$ Case 1. $k=0:$ Then we have $g(x)= 0~\forall x\geq 0 \implies f\equiv 1~\forall x\geq 1,$ which indeed fits. Case 2. $k\neq 0:$ Then we can easily get, for some $r$ we have $g(x)=kx^r~ \forall x>0.$ Also $g(-x)=-x^r.$ Thus for $x>1, f(x)=k(x-1)^r$ and $x<1, f(x)=1-(1-x)^r.$ Clearly both fit.

Substitute $f(x)=g(x-1)+1$ to obtain $g(-1)=-1$, $g(0)=0$, and $-g(-xy)=g(x)g(y)$ for positive $x$ and negative $y$. Now suppose that $g(1)=0$. Then we obtain $-g(-y)=0$ so that for all positive $x$ we have $g(x)=0$. Then we can set the other half of $g$ to be any nondecreasing function such that $g(-1)=-1$. This corresponds to $f(x)=1$ for all $x\ge 1$ and $f(x)=p(x)$ for all $x<1$ for any nondecreasing function $p$ such that $p\le 1$ and $p(0)=0$. Otherwise, set $h(x)=\frac{g(x)}{g(1)}$. Since $-g(-y)=g(1)g(y)$ we see that we must have $h(x)h(y)=h(xy)$ for all positive $x,y$. We also see that $h$ is nondecreasing and that $h(1)=1$. Set $h(2)=k$ and note that for all $x\in Q$ we have that $h(2^x)=k^x$ which immediately implies that for all positive $x$ we have that $h(x)=x^a$ for some constant $a\ge 0$. This means that for all $x>0$ we have $g(x)=cx^a$ for constants $c>0$ and $a\ge 0$. Then for $x\le 0$ we have $g(x)=-x^a$ (where $x^a$ is positive). Finally, we have that for $x>1$, $f(x)=c(x-1)^a+1$, and for $x\le 1$, we have $f(x)=-(x-1)^a+1$. Because of the way we constructed this function, it is evident that it works. We are done. $\blacksquare$

Let $g(a)=f(a+1)-1$ then $f(a)=g(a-1)+1$. The conditions become $g$ is nondecreasing, $g(0)=0$, $g(-1)=-1$, and \[0=g(a-1)g(b-1)+g(-(a-1)(b-1))\]for $a<1<b$ so this becomes $-g(a)g(b)=g(-ab)$ for all $a<0<b$. Setting $b=1$ gives $-g(a)g(1)=g(-a)$. If $g(1)=0$ then $g(a)=0$ for all positive $a$. Then, all restrictions on the function's value on negative numbers become lifted, and all that is left is that $g(x)$ is nondecreasing, but $g(-1)=-1$ and $g(a)=0$ for all $a\ge 0$. If $g(1)\neq 0$ then we get $g(-a)g(b)=g(-ab)g(1)$ for positive $-a$ and $b$. This becomes $g(a)=g(1)a^k$ for some positive $k$. $g(-a)$, then is $-a^k$. The result follows by simply substituting this into $f$. I'm simply too lazy to verify.