amparvardi wrote: Let f be a function that satisfies the following conditions: $(i)$ If $x > y$ and $f(y) - y \geq v \geq f(x) - x$, then $f(z) = v + z$, for some number $z$ between $x$ and $y$. $(ii)$ The equation $f(x) = 0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions; $(iii)$ $f(0) = 1$. $(iv)$ $f(1987) \leq 1988$. $(v)$ $f(x)f(y) = f(xf(y) + yf(x) - xy)$. Find $f(1987)$. Proposed by Australia. According to me, such a problem means that : either there is a unique function matching all the requirements, either many functions match all the requirements but they all have the same value for $x=1987$ Using this fact, and since $f(x)=x+1$ matches all the requirements, the answer is $\boxed{1988}$

pco wrote: amparvardi wrote: Let f be a function that satisfies the following conditions: $(i)$ If $x > y$ and $f(y) - y \geq v \geq f(x) - x$, then $f(z) = v + z$, for some number $z$ between $x$ and $y$. $(ii)$ The equation $f(x) = 0$ has at least one solution, and among the solutions of this equation, there is one that is not smaller than all the other solutions; $(iii)$ $f(0) = 1$. $(iv)$ $f(1987) \leq 1988$. $(v)$ $f(x)f(y) = f(xf(y) + yf(x) - xy)$. Find $f(1987)$. Proposed by Australia. According to me, such a problem means that : either there is a unique function matching all the requirements, either many functions match all the requirements but they all have the same value for $x=1987$ Using this fact, and since $f(x)=x+1$ matches all the requirements, the answer is $\boxed{1988}$ : Why there couldn't exist classes of functions, which yield several possible values of $f(1987)$? I think the above is very flawed reasoning (unless proven right) - based on "what would be a pattern for the problems proposed at an IMO", which has nothing to do with mathematics.

Farenhajt wrote: : Why there couldn't exist classes of functions, which yield several possible values of $f(1987)$? I think the above is very flawed reasoning (unless proven right) - based on "what would be a pattern for the problems proposed at an IMO", which has nothing to do with mathematics. Sorry, Farenhajt. This was just a joke since the problem was quite easy in a direct way. Here is a more convenient proof (I hope you'll agree it) : Let $P(x,y)$ be the assertion $(v)$ : $f(x)f(y)=f(xf(y)+yf(x)-xy)$ Using $(ii)$, let $a$ the greatest solution of $f(x)=0$ $f(0)\ne 0$ and so $a\ne 0$ If $a>0$ : $f(0)-0\ge 0 \ge f(a)-a$ and so, using $(i)$ : $\exists z$ such that $f(z)=z$ but then $P(a,z)$ $\implies$ $0=f(0)$, in contradiction with $(iii)$ So $a<0$ and $P(1987,a)$ $\implies$ $f(a(f(1987)-1987))=0$ and so $a(f(1987)-1987)\le a$ and so $f(1987)\ge 1988$ and so, using $(iv)$, $f(1987)=1988$. (and this mandatory value may be reached using for example $f(x)=x+1$) And, btw, clever rating (congrats)

pco wrote: And, btw, clever rating (congrats) Well, for the distinguished and undisputably capable Mathlinkers, which are expected to know better at any given time of day - especially when it comes to their narrowest area of expertise (functional equations in this case) - some tough love must be applied. At least the rating got you to work a little, which can't be considered bad

Farenhajt wrote: At least the rating got you to work a little, which can't be considered bad Yes, you're right. I'm very well known on this forum as a quite lazy guy who very rarely works. And thanks to you, i'll try to change and begin to work a little. Thanks

pco wrote: Farenhajt wrote: At least the rating got you to work a little, which can't be considered bad Yes, you're right. I'm very well known on this forum as a quite lazy guy who very rarely works. And thanks to you, i'll try to change and begin to work a little. Thanks "To work a little" = "to work a little on the given problem", and not just dismiss it on the grounds of a loophole in wording. And please, don't be so bittersweet. You're dangerously close to the "offended diva" attitude, and in spite of it being very fashionable at the beginning of the 21st century, you're too clever to need it.

Let $P(x,y)$ be the assertion of (v). And let $z$ be the greatest solution of $f(x)=0.$ Note that $z\neq 0.$ If $z<0,$ then $P(x,z)$ gives $z(f(x)-x)\leq z.$ And so $f(x)\geq x+1.$ And if $z>0,$ then $f(z)-z<0<f(0)-0,$ so by (i), $\exists z:f(z)=z.$ Then $0=1$ by $P(a,z),$ absurd. So $f(1987)\geq 1988.$ Combine with (iv) to get $f(1987)=1988.$