Croatian Mathematical Society Competitions

Plugging $(a,b)\leftarrow (1,1)$ into the first given condition, we get $f(1,1)=1$.

On the other hand, by plugging $(a,b)\leftarrow (a,b+1)$ we get $$f(a,b+1)+a+b+1=f(a,1)+f(1,b+1)+ab+a.$$ Using the first given condition in its original form, it follows that $$f(a,b+1)-f(a,b)=f(1,b+1)-f(1,b)+a-1.$$ Let $p>2$ be a fixed prime number such that $p\mid a+b$, meaning that $p\mid f(a,b)$. Since $p\mid a+(b+1)-1$, we also have $p\mid f(a,b+1)$. Therefore, $p\mid f(1,b+1)-f(1,b)+a-1$ and $$p\mid f(1,b+1)-f(1,b)-b-1(\ast )$$ for all positive integers $a$ and $b$, and primes $p>2$ such that $p\mid a+b$. Note that the right-hand side in $(\ast )$ does not depend on $a$, hence fixing $b$ and varying $a$ would yield infinitely many odd prime divisors of $f(1,b+1)-f(1,b)-b-1$, which is possible only if the latter equals 0, i.e. $$f(1,b+1)=f(1,b)+b+1$$ for every positive integer $b$. Therefore, $f(1,2)=f(1,1)+2=1+2$, $f(1,3)=f(1,2)+3=1+2+3$, i.e. $f(1,n)=1+\cdots +n=\frac{n(n+1)}{2}$ (by mathematical induction). In a similar fashion, starting with $(a,b)\leftarrow (a+1,b)$, we infer that $f(n,1)=\frac{n(n+1)}{2}$. Finally, from $$\begin{array}{rl}f(a,b)+a+b& =f(a,1)+f(1,b)+ab\\ & =\frac{a(a+1)}{2}+\frac{b(b+1)}{2}+ab,\end{array}$$ it follows that $f(a,b)=\left(\genfrac{}{}{0ex}{}{a+b}{2}\right)$. This function clearly satisfies the second given condition, hence it is the only solution.