(In reply to re: possible answer
by Ady TZIDON)
I agree (II) is rather easy.
For (I) there is a pretty approach:
Even prime: 2:2*2+1= 5, which is not a cube.
I M is odd: x^3=2(2n-1)+1 or 4n-1=x^3, a=2n-1
II A cube x^3, is of the form 4n-1 iff x is of the form (4k-1)
III Now we have (4n-1)=(4k-1)^3; n = 16k^3-12k^2+3 k
IV 2n-1 =32k^3-24k^2+6k-1 = (2k-1) (16k^2-4k+1)
So unless k=1, 2n-1 is compound. If k=1, 16-4+1=13, the sole solution.
Edited on October 9, 2011, 11:55 pm
Posted by broll
on 2011-10-09 23:51:53