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Tuesday, December 20, 2016

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.

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