(In reply to
re: possible answer by Ady TZIDON)
Ady,
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.
Odd primes:
I M is odd: x^3=2(2n1)+1 or 4n1=x^3, a=2n1
II A cube x^3, is of the form 4n1 iff x is of the form (4k1)
III Now we have (4n1)=(4k1)^3; n = 16k^312k^2+3 k
IV 2n1 =32k^324k^2+6k1 = (2k1) (16k^24k+1)
So unless k=1, 2n1 is compound. If k=1, 164+1=13, the sole solution.
Edited on October 9, 2011, 11:55 pm

Posted by broll
on 20111009 23:51:53 