Question 77052
<pre><font size = 5><b>We use the factor theorem, which is the identity: 
{{{x^n - y^n = (x - y)}}}({{{x^(n-1)+x^(n-2)y+x^(n-3)y^2}}}+ ··· + {{{x^(n-k)y^(k-1) }}} + ··· + {{{ xy^(n-2)+y^(n-1) }}})
Suppose p is divisible by 4, then there exists 
positive integer q such that p=4q, then 
{{{2^p-1 = 2(4q)-1 = (2^4)^q-1 = 16^q-1}}} =
{{{(16 - 1)}}}({{{16^(q-1)+16^(q-2)+16^(q-3))}}}+ ··· + {{{16^(q-k)}}} + ··· + {{{16 + 1}}}) =
15({{{16^(q-1)+16^(q-2)+16^(q-3))}}}+ ··· + {{{16^(q-k)}}} + ··· + {{{16 + 1}}}) so 
{{{2^p-1}}} is either 15 (when q=1) or divisible 
by 15, and in either case is not prime.

For the case when p is divisible by 3, then there exists 
positive integer q such that p=3q. Do the same as
above and you have 7 where the 15 is above and {{{2^p-1}}}
is not prime unless it equals 7, i.e., unless q=1, i.e.,
unless p=3, but that is ruled out in the hypothesis.
Edwin</pre>