Re: [PrimeNumbers] Prize Puzzles concerning Primes and Composites of the form 20x^2 -1:
- Taking number 6...
6) If x > 1 , then A(x) will be composite if there exists at
least one k value in the interval 2x < k < 3x such that T(k) is a
square of an integer.
Example: if x = 4, then A(x) = 319, 8< k < 12 and the
values of T(k) are 169, 529 and 929 .
Two of these are squares, therefore 319 is composite
let x equal a composite number a * b
therefore, A(x) = 20a^2b^2-1
thus, T(k) = 5(2K-1)^2-80a^2b^2+4
Simplified, T(k) = 20k^2-20k-80a^2b^2+9
By your theory, there should be a value of k where 2ab<k<3ab where the
above is a perfect square
k can be redescribed as kab, where k is a number between 2 and 3
we can rewrite this as follows...
now, we can begin possibly factoring this into a square.
the 9 at the end, and the negative in the center, gives us part of the answer
( - 3)^2
we will throw in another variable to represent the missing portion, called w
(w-3)^2 will factor into w^2-6w+9
using comparison, w^2 = (20K^2-80)(a^2b^2) and -6w = -20k(ab)
using second comparison, -6w must equal -20k(ab), thus w = 10/3k(ab)
so our factor must be (10/3k(ab)-3)^2
we need to solve for k, to ensure k is between 2 and 3. if it is, then
the premise of number 6 can be proven
using first comparison, we know (10/3k(ab))^2=(20k^2-80)(a^2b^2)
100/9k^2a^2b^2 = (20k^2-80)(a^2b^2)
the a^2b^2 cancel out, giving us 100/9k^2=20k^2-80
subtract 20k^2 from both sides to give us -80/9k^2=-80 multiply both
sides by -9/80 gives us k^2=9, or k = 3
With your original theory, this would prove it false. I would surmise
that there will be some numbers that your rules as written will not
However, adjusted so that the interval is 2x<k<=3x and the mental
exersise above will prove it true.
Sorry if it isnt in long proof form, its been forever since ive done
one of those. I hope this was easy to follow