List all integers n, below 10000, such that 12*(n^2 + 2) is a perfect cube.
First, I'll assume you meant
positive integers.
I thought I'd try getting rid of the 12 = 2*2*3. We need n^2+2 to have a single factor of 2. Which will happen for any even n. So n=2m, the expression becomes
12((2m)^2+2) = 8*3(2m^2+1)
we don't need to worry about the 8.
Next we need 2m^2+1 to have two factors of 3. It turns out m=9p±2 and both work and simplify to
27(18p^2 ± 8p +1)
From here we still need (18p^2 ± 8p +1) to be a cube. I'm stuck here. It is easy to find two solutions for the '+' though:
p=0 and p=1
m=2 and m=11
n=4 and n=22
the cubes are 216 = 6^3 and 5832 = 18^3

Posted by Jer
on 20131001 14:15:42 