(i) Determine all possible prime(s) M such that 2*M + 1 is a perfect cube.

(II) Determine all possible prime(s) M such that 16*M + 1 is a perfect fourth power.

(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(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