On Jun 25, 3:25 pm, Martin Michael Musatov <marty.musa...@gmail.com> wrote: > > 2n * 2 (n+1) - 1 is prime > > > it's pretty easy to show if n = 1 or 5 (mod 7) then 7 > > will divide the > > whole thing > > I never wrote this:
you did, in words, in your very first post. this was the problem before your revisions. n is any integer, 2n any even integer, 2 (n+1) the next even integer.
but back to the main problem. it's true, and true that any number, times that number plus 2, take away 1, will not be a square number. it will always be 2 less than a square number. eg 8 * 10 - 1 = 79, 2 less than 81.
this is shown by "completing the square"
n (n+2) - 1 = (n+1)(n+1) - 2, in other words the number will always be two less than a square number.
by the way squares are distributed this can never be a square number.
for your expression [[N*N]*([N*N]+2)]-1 = (N*N+1)(N*N+1) - 2.
i liked the original problems you posted, but now everything has become trivial