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## Prize Puzzles concerning Primes and Composites of the form 20x^2 -1:

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• I offer a \$25 prize to the first person who can submit a verifiable counterexample or proof for conjectures 5) or 7) below, and also a \$25 prize for a proof of
Message 1 of 2 , Dec 10, 2008
I offer a \$25 prize to the first person who can submit a verifiable
counterexample or proof for conjectures 5) or 7) below, and also a
\$25 prize for a proof of 6) or 8). The deadline is 1/19.

x, A(x), k, T(k) : integers;

Let A(x) = 20x^2 - 1;
Let T(k) = 5*(2*k -1)^2 - 4*A(x) ;

5) If x > 1 , then A(x) will be prime if no k value
exists in the interval 2x < k < 3x such that T(k) is a
square of an integer.

Example: if x = 5, then A(x) = 499, 10< k < 15 and the
values of T(k) are 209, 649, 1129 and 1649 .
None of these is a square, therefore 499 is prime.

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

x, A(x),K,M,S,n,p : integers;

Let A(x) = 20x^2 -1. For any value of x let K1 = x, K2 = 2x+1
and let K2...Kn be consecutive integers with n denoting the
maximal possible number of terms in a sequence for S1 = K1
+K2...+ Kn within the interval 0 < n < x+1 and also for
S2 = K1 + K2...+ Kn within the interval 1 < n < x+2.
(note: 40*S1 + 9 and 40*S2 + 9 are always squares)

Example: If x = 4, then S1 = 4 + 9 + 10 + 11 = 34 and n = 4.
If x = 4, then S2 = 4 + 9 + 10 + 11 + 12 = 46 and n = 5.

Using the same S1 and S2 values found for each x, find any shorter
runs of the form S1 or S2 = M1 + M2...+ Mp, with p < n, M1 > x,
M2 = 2M1 +1, and M2...Mp consecutive integers with p denoting
the number of terms in the sequence.

Example: For S1 = 34 from above it is evident that 11 + 23
also equals 34, with M1 = 11 and p = 2.
For S1 = 46 from above it is evident that 15 + 31
also equals 46, with M1 = 15 and p = 2.

7) If any value for S1 > 1, S2 or M1 found by the processes
described above is substituted for x in A(x) = 20x^2 -1 then
A(x) is always composite.

Examples: if S1 = 34 then A(34) = 23119 = 61*379.
if S1 = 34 and M1 = 11 then A(11) = 2419 = 41*59.
if S2 = 46 then A(46) = 42319 = 101 * 419.
if S2 = 46 and M1 = 15 then A(15) = 4499 = 11*409.

8) If all possible composites from 7) above are removed from the
sequence 20x^2 -1 , then all the remaining values of A(x) are prime.

Aldrich Stevens
• 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
Message 2 of 2 , Dec 13, 2008
Taking number 6...
<snip>
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
<snap>

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

so, 20k^2a^2b^2-20kab-80a^2b^2+9.

we can rewrite this as follows...

(20k^2-80)(a^2b^2)-20k(ab)+9

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
work.

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
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