Question -
The function $f(n)$ is defined for all positive integers $n$ and takes on non-negative integer values. Also, for all $m, n$ $$ \begin{array}{c} f(m+n)-f(m)-f(n)=0 \text { or } 1 \\ f(2)=0, f(3)>0, \text { and } f(9999)=3333 \end{array} $$ Find $f(1982)$ $$ (\mathrm{IMO}-1982) $$
My doubt -
by putting $m=n=1$ we easily get $f(1)=0$..
now putting $n=1$ in given FE we get $f(m+1)=f(m)+0$ or $1$
now it can't be $0$ because then $f(3)=0$ contradicting the given fact that $f(3)>0$... so hence $f(m+1)=f(m)+1$......(1)
now putting $n=2$ in given FE we get $f(m+2)=f(m)+0$ or $1$ (since $f(2)=0$ given )
but it can't be $0$ because then $f(3)=0$ contradicting the given fact that $f(3)>0$ again hence $f(m+2)=f(m)+1$ ......(2)
now combining $(1)$ and $(2)$ we get $f(m+1)=f(m+2)$ which is again contradicting, by putting $m=1$ that $f(3)=0$ !!!!
how this can be true ...because i have taken all cases into considerations but then also getting contradiction ...where is the flaw ???
thankyou
In the functional equation, the term "$0$ or $1$" means that for each pair of integers $m, n$, the quantity $f(m + n) - f(m) - f(n)$ belongs to the set $\{0, 1\}$.
In particular, the value of $f(m + n) - f(m) - f(n)$ depends on $m, n$. It can be $0$ for some values of $m, n$, while being $1$ for others.
In your argument, when you come to $f(m + 1) = f(m) + 0$ or $1$, you seem to assume that this "$0$ or $1$" does not depend on $m$, which is an incorrect assumption.