Mathematical Reflections 2016, Issue 5 - Problem O385

Problem:
Let $f(x,y)=\frac{x^3-y^3}{6}+3xy+48$. Let $m$ and $n$ be odd integers such that $$|f(m,n)| \leq mn+37.$$ Evaluate $f(m,n)$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution:
Let $m$ and $b$ be two odd integers such that $$-mn-37 \leq \dfrac{m^3-n^3}{6}+3mn+48 \leq mn+37.$$ We get the two inequalities
$$m^3-n^3+64+12mn \leq -2$$ and $$m^3-n^3+512+24mn \geq 2.$$ Using the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ and setting $(a,b,c)=(m,-n,4)$ and $(a,b,c)=(m,-n,8)$, we obtain the two inequalities
$$
(m-n+4)(m^2+n^2+16+mn-4m+4n) \leq -2
$$
and
$$
(m-n+8)(m^2+n^2+64+mn-8m+8n) \geq 2
$$
Since $a^2+b^2+c^2-ab-bc-ca \geq 0$ with the equality if and only if $a=b=c$, then and as $m,n$ are odd integers, we obtain that $m^2+n^2+16+mn-4m+4n>0$ and $m^2+n^2+16+mn-4m+4n>0$. It follows that $m-n+4<0$ and $m-n+8>0$, i.e. $m-n+4 \leq -2$ and $m-n+8 \geq 2$, which gives $-6 \leq m-n \leq -6$, i.e. $m-n=-6$. Therefore, $$f(m,n)=-(m^2+mn+n^2)+3mn+48=-(m-n)^2+48=12.$$