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Proth prime

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how do I add it to the Proth prime page ???

Proth theorem extended:

Let Q= k*2^n +1, where n=>3 is a odd natural number & k<= 2^n +1.
If for some 'a', a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime.

If 'm' is from the set of natural numbers, then every odd prime
divisor 'q' of a^(2^(m+1))+/-1 implies that q == +/-1(mod a^(m+2))
[concluded from generalized Fermat-number 'proofs' by Proth along
with my replacing 'm' with 'm+1'].

Now, if 'p' is any prime divisor of 'R', then a^((Q-1)/4) = (a^k)^
(2^(n-2)) == +/-1(mod p) implies that p == +/-1 (mod 2^n).

Thus, if 'R' is composite, 'R' will be the product of at least two
primes each of which has minimum value of (2^n +1), and it follows

k*2^n +1 >= (2^n +1)*(2^n +1) = (2^n)*(2^n) + 2*(2^n) +1; but the
1's cancel, so k*(2^n) >= (2^n)*(2^n) + 2*(2^n) and upon dividing
by 2^n... k>= 2^n +2.

(2^n -1)*2^n +1 = (2^n)*(2^n) - 2^n +1 >= (2^n -1)*(2^n -1)= (2^n)
*(2^n) - 2*(2^n) +1; but the 1's cancel again, so 1 >= 2 is a con-
tradiction, and the smallest product of at least two primes cannot
be derived using factors (2^n -1).

However, the first result is incompatible with our definition, so
if k<= 2^n +1 and a^((Q-1)/4) == +/-1(mod Q), then 'Q' is prime.

Just make it an article and attach it to the Proth prime page.

If leavemsg2 already knows how to use TeX, then this should be a snap. He would just make sure to remove the comment marker from the package for theorems and prooves, and it would be extremely straightforward for him to convert what he has written into proper TeX.

But if he doesn't know how to use TeX, then he should look for a tutorial. There's one here on PM, and there are others on the Web as well as in books.

It gets better... let Q= k*2^n +1; prime n >= 5;
let A=(((Q-3)/2-1)/2-1)2; 2^A == (Q+/-1)/2 mod Q
it's even quicker and only misses a few potential
primes; base 2 tests of this sort are also immune
to the pseudo-prime effect.
I tried to script it in PFGW but to no avail.
can you help me discover a large prime number ???

an 'absolute'-perfect cuboid CANNOT exist... (simple logic)
if you miss the logic, please re-read it ('til it sinks in)

i. it would have to appear before PT's exceed 2^3 < 3^2.

ii. a Pythagorean Triple's standard derivation:

m,n; m^2-n^2; 2mn; m^2+n^2

2,1; 3= 2^2-1^2 4=2*2*1 5= 2^2+1^2

3,2; 5= 3^2-2^2 12=2*3*2 13= 3^2+2^2

4,3; 7= 2^2-1^2 24=2*4*3 25= 4^2+3^2

iii. these 3 PT's also happen have a minimum expanse property:

2,1 3= 2+1 ... 5= 3*1+2

3,2 5= 3+2 ... 13= 5*2+3

4,3 7= 4+3 ... 25= 7*3+4

m,n = m+n = (m+n)n +m

no other PT's are of minimum expanse after 2^3 < 3^2,
so the idiosyncracy of finding an absolute perfect cube
becomes more absurd as the search is continued further.

logically, only a 'semi'-perfect cuboid could exist with
edges (3), (4), (12), and main diagonal (13); if it were
an absolute-perfect cuboid, then the other face diagonal
would have to be integral, as in (3), (4) & (5)'s case of
the first face; it's impossible; draw it out, if necessary.

iv. an absolute-perfect cuboid can't exist; it doesn't get
any simpler than that... enjoy.
think it through 'til you get it; no need to respond; an 8th
grader understood it... the first time!

What I don't understand is what this has to do with Proth primes. There's surely some connection, but I'm not sure what that connection is, much less whether it's something I need to include in the entry.

I tried to post and couldn't except to an existing post. The entry is out of place, but I wanted to be able to post it. sorry...

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