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June 21st, 2018, 08:58 AM  #1 
Member Joined: Aug 2017 From: India Posts: 54 Thanks: 2  Prime number solution
I have come across the following question Find all integers n such that $\displaystyle 2n^36n^2+4n+3$ is prime My attempt is this is the maximum I could reach $\displaystyle 2n^36n^2+4n+3 => 2n^2(n3)+4n+3 => 2n^2(n3) +(n3)+ (3n+6) => (2n^2+1)(n3)+3(n+2) $ How should I proceed further? Is it that I have to find the roots of the equation? Please help. Last edited by skipjack; June 21st, 2018 at 09:12 PM. 
June 21st, 2018, 09:34 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,647 Thanks: 1476  

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