Problem 1

The function

f(n)

is defined on the positive integers and takes non-negative integer values.

f(2)=0

f(3)>0

f(9999)=3333

, and for all

m,n

f(m+n)-f(m)-f(n)=0 \text{ or } 1.

Determine

f(1982)

Step 3 of 7: Squeezing

f(3k)

between a floor and

f(9999)=3333

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In plain words

This is the key squeeze of the whole proof: a lower bound built purely from superadditivity meets an upper bound forced by the one exact large value we are given, pinning down infinitely many values of $f$ at once.

f(3(k{+}1))\ge f(3k)+f(3)=f(3k)+1 \quad\text{for all } k\ge1

Detailed analysis

Superadditivity with $m=3k,\,n=3$ shows the sequence $a_k=f(3k)$ increases by at least $1$ at every step, so $a_k\ge a_1+(k-1)=k$ for all $k\ge1$ . But $a_{3333}=f(9999)=3333$ exactly, matching the lower bound with equality. If any single step increased by $2$ or more, $a_{3333}$ would exceed $3333$ , a contradiction. Hence every step increases by exactly $1$ , so $f(3k)=k$ for every $1\le k\le3333$ .