Hello,
I was thinking over the following problem. Let be any composite number divisible by . Let :
for and .
So, . Furthermore, an additional property is required, which is that , .
Finally, the last property is that when the polynomial is factorized into , neither of the factors equal when .
I got to the before-last part, because I noticed that :
and
So
But each time I apply this formula to find , and , it turns out that the factorized form of the polynomial yields a factor equal to when . Why is that ? Why doesn't it yield a nontrivial factor of ? Does it come from the formula ?
This might not be very clear ... If you need me to make it clearer don't hesitate. Thanks all.
Well, this still gives a factor of 3 (because 2x + 1 with x = 1), and which is a composite divisible by 3.
In fact, I'm looking for a systematic way to find , and such that where is a composite number, and satisfying the condition is a perfect square. Then, when factored, the polynomial should yield the factorization of with .
I made it up to the condition, because the only way I can think of that guarantees it to be a perfect square requires n to be a multiple of 3 (and strangely enough, the factorization of the polynomial keeps giving me 3).
So I'm asking if there are any other, and more efficient, methods of finding the values of the coefficients that satisfy both properties.
No, what I conjecture is that :
(don't know if that's how one formally writes it but I think the idea is pretty straightforward)
And I was wondering if there was a (tractable) way of finding values , and that satisfy (1) and (3), but that do not imply that (or any variant). This question of course assumes that it is (2) that is causing to be divisible by (I only conjecture it, again, but I think it can be shown quite easily. If not, we have a new Clay problem anyway xD)
PS : sorry if this is not very clear, I know I really need to work on my communication and logical skills