[Guest]

Determine all possible value(s) of a positive integer x such that (x³-1)/5 is a prime number.

Edited January 30, 2010 by K Sengupta

[Guest]

x = 6 is the only solution.

Since x³- 1 must be divisible by 5, x³-1 must end in 0 or 5.

But if it ends in 0,(and x is clearly >= 3) then dividing by 5 would yield even number ( > 2 ) which is not prime.

So it ends in 5, which means x ends in 6.

So x can be written in the form 10y + 6 for some integer y, so x-1 = 10y+5

Factoring (x³ - 1)/5 = (x - 1) ( x² + x + 1)/5

Substituting for x-1, this becomes (10y + 5)(x² + x + 1)/5

= (2y+1)(x² + x + 1)

The only way this expression can be prime is if y = 0, (otherwise 2y+1 is one of two factors)

so x = 10*0 + 6 = 6 is the only solution [ (x² + x + 1) = 43 is the prime number]