Guest Posted December 29, 2009 Report Share Posted December 29, 2009 Prove that at least one of the numbers x + 3 and x2 + 3x + 3 is not a perfect cube, where x is a nonnegative integer. Quote Link to comment Share on other sites More sharing options...
0 unreality Posted December 29, 2009 Report Share Posted December 29, 2009 (edited) edit: nvm. Edited December 29, 2009 by unreality Quote Link to comment Share on other sites More sharing options...
0 Izzy Posted December 31, 2009 Report Share Posted December 31, 2009 Your edit scares me.. Am I missing something? x=1. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 31, 2009 Report Share Posted December 31, 2009 (edited) Your edit scares me.. Am I missing something? x=1. The OP asks for a proof, that for every value of a nonnegative integer x, at least one of the numbers amongst x + 3 and x^2 + 3x + 3 is not a perfect cube. Edited December 31, 2009 by K Sengupta Quote Link to comment Share on other sites More sharing options...
0 Guest Posted December 31, 2009 Report Share Posted December 31, 2009 Prove that at least one of the numbers x + 3 and x2 + 3x + 3 is not a perfect cube, where x is a nonnegative integer. if x+3 and x^2+3x+3 are both perfect cubes, then their product must be a perfect cube. thus x^3+6x^2+18x+9 is a perfect cube (x^3+6x^2+18x+9)^(1/3) can never be an integer at an integer value of x. As a result, x+3 and x^2+3x+3 cannot both be perfect cubes. Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted December 31, 2009 Report Share Posted December 31, 2009 if x+3 and x^2+3x+3 are both perfect cubes, then their product must be a perfect cube. thus x^3+6x^2+18x+9 is a perfect cube (x^3+6x^2+18x+9)^(1/3) can never be an integer at an integer value of x. As a result, x+3 and x^2+3x+3 cannot both be perfect cubes. You state that "(x^3+6x^2+18x+9)^(1/3) can never be an integer at an integer value of x." Why is that? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 1, 2010 Report Share Posted January 1, 2010 You state that "(x^3+6x^2+18x+9)^(1/3) can never be an integer at an integer value of x." Why is that? (x+2) cubed is x^3+6x^2+18x+8. this will always be a perfect cube. x^3+6x^2+18x+9 is one greater. the only cubes that differ by one are -1,0,and 1. if x is a nonnegative integer, then x is not any of those cubes, let alone three less than any of those cubes. sorry I forgot to mention that part of the proof. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted January 2, 2010 Report Share Posted January 2, 2010 Then (X+3)(X^2 + 3X +3) = Y^3 +1 which cann't be product of 2cubic numbers unless if Y= 0 i.e. X= (-2) but given that X is non negative. So no solution. Quote Link to comment Share on other sites More sharing options...
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Prove that at least one of the numbers x + 3 and x2 + 3x + 3 is not a perfect cube, where x is a nonnegative integer.
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