List all integers n, below 10000, such that 12*(n^2 + 2) is a perfect cube.

I Add 3 to both sides:  a^3+3 = 3(4n^2+9)  
II Multiply both sides by 3:  3(a^3+3) = 9(4n^2+9)  
III Expand and switch constants:  3a^3-81 = 36n^2-9  
IV Factor:  3(a-3)(a^2+3a+9) = 9(2n-1)(2n+1)  
V Divide by 3: (a-3)(a^2+3a+9) = 3(2n-1)(2n+1) 

Now 3 divides a, so substitute 3b = a

VI ((3b)-3)((3b)^2+3(3b)+9) = 3(2n-1)(2n+1)  
VII Factor:  27(b-1)(b^2+b+1) = 3(2n-1)(2n+1)  
VIII Divide by 3:  9(b-1)(b^2+b+1) = (2n-1)(2n+1)  

IXa Either: 9 divides (2n-1), when n = (9k+5)  
9b^3-9 = 4(9k+5)^2-1  
9b^3-9 = 9(2k+1)(18k+11)  
(b-1)(b^2+b+1) = (2k+1)(18k+11)  
b = (2k+2)  
4k^2+10k+7 = 18k+11  
and k = (1-rt2), (1+rt2), so n is not an integer.
  
IXb Or: 9 divides  (2n+1), when n = (9k+4)  
9b^3-9 = 4(9k+4)^2-1  
9b^3-9 = 9(2k+1)(18k+7)  
(b-1) (b^2+b+1) = (2k+1)(18k+7)  
b = (2k+2)  
4k^2+10k+7 = 18k+7  
and k = {0,2}, when b ={2,6}, so a ={6,18} and n ={4,22}  
are the only solutions, whether below 10000 or not.

Edited on October 2, 2013, 2:09 am

Posted by broll on 2013-10-02 01:49:50