$f(x+y)=f(x)+f(y)+2547$ Put $x=y=0$,we have $2547=-f(0)$, so we obtain $f(x+y)-f(x)=f(y) -f(0)$. $\lim_{y\to 0}\frac{f(x+y)-f(x)}{y}=\lim_{y\to 0}\frac{f(y)-f(0)}{y}$ $f'(x)=f'(0)\Longleftrightarrow f(x)=f'(0)x+C$.From $f(2004)=2547$,we have $C=2547-2004f'(0)$. Therefore by $f(2547)=2547f'(0)+C$, the answer is $f(2547)=543f'(0)+2547$.

There is a really easy solution Kunny Pierre.

f(0)=-2547 you can show that f(n)=nf(1)+(n-1)2547; from rec using f(2004)=2004f(1)+2003*2547, we get f(1)= -(2002/2004)*2547; f(n) therefore =2547*((n/1002)-1); f(2547)=2547*(2547/1002-1)=2547*1545/1002

Note that $g(x)=f(x)+2547$ does satisfy the Cauchy equation. Pierre.

yes, the knowledge of cauchy's makes quite a few problems quite obvious

it's been a while since the last time ! ok? Can we talk about a derivation int the Q set of rationals????

pbornsztein wrote: Note that $g(x)=f(x)+2547$ does satisfy the Cauchy equation. Pierre. what is the cauchy-equation? Is this pre-olympiad?

Peter VDD wrote: pbornsztein wrote: Note that $g(x)=f(x)+2547$ does satisfy the Cauchy equation. Pierre. what is the cauchy-equation? Is this pre-olympiad? f(x)+f(y)=f(x+y) only has f(x)=cx as a solution where c is a constant for f a continuous function on R or a function on Q, Z, N.

as a side note, anything which stops the function being dense in the whole plane (like boundedness in an interval) would rule out the mad functions in R.