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Wednesday, May 10, 2017

Mathematical Reflections 2016, Issue 6 - Problem S393

Problem:
If n is an integer such that n^2+11 is a prime, prove that n+4 is not a perfect cube.

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


Solution:
Assume that n+4 is a perfect cube. Then n+4=m^3, where m \in \mathbb{Z}, i.e. n=m^3-4. It follows that
\begin{array}{lll} n^2+11&=&(m^3-4)^2+11\\&=&m^6-8m^3+27\\&=&m^6+m^3+27-9m^3\\&=&(m^2)^3+(m)^3+(3)^3-9m^3\\&=&(m^2+m+3)(m^4-m^3-2m^2-3m+9). \end{array} As m^2+m+3 \geq 3 and m^4-m^3-2m^2-3m+9 \geq 3 for all m \in \mathbb{Z}, then  n^2+11 is not prime, contradiction.

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