WHERE:       4k + 1, 4k + 3 = Odd, and 4k, 4k + 2 = Even

IF:                   Odd + Even = Odd

AND:             (4k + 1) + (4k + 2) = (4 * 2k + 3)

SO:                (4k + 3) + (4k + 0) = (4 * 2k + 3)

THEN:           (4k + 1) + (4k + 0) = (4 * 2k + 1)

OR:                (4k + 3) + (4k + 2) = (4 (2k + 1) + 1)

IF:                   (4k + 0)^2 = (4n)

THEN:           (4k + 2)^2 = (4n)

AND:             (4k + 1)^2 = (4n + 1)

SO:                (4k + 3)^2 = (4n + 1)

IF:                  (4k + 1)^2 + (4k + 0)^2 = (4n + 1)

AND:             (4k + 1)^2 + (4k + 2)^2 = (4n + 1)

THEN:           (4k + 3)^2 + (4k + 0)^2 = (4n + 1)

SO:                (4k + 3)^2 + (4k + 2)^2 = (4n + 1)

THEN:            4k + 1 + 4k = 4 (2k) + 1

HENCE:        (4a + 1) + 4b = 4 (a + b) +1

In the last example above, the addition of the first two terms of the equation which corresponds to the sum of two squares to equal a 4n + 1 number is reduced to its most simplified general case.  That the properties of the terms always default to the example given, leads to the omission rule for the 4n + 3 series integers in the case of the sum of two squares.  Since this is always going to be the case, a general law or axiom may be inferred from this which is based upon the sum of the properties of the squaring of the individual terms and not necessarily upon the law of areas alone.  In the case of certain 4n + 1 series integers never equaling the sum of two perfect squares, it must be further implied that this is to be based upon a law of areas as concerns the value of the third term alone, since the 4n + 1 series integers may also be the "Only Once Sum of Two Perfect Squares" as well as more than once.  This means that the Law of Omission as applied to the yield for an Odd^2 + Even^2 never equaling a 4n + 3 series number, is different than the Law of Areas as applied to the yield as a 4n + 1 series number, even though the individual properties for the yields themselves may be derived from their own axiom.  Again, I say this because the properties of the individual terms are dependent upon the squaring of those terms prior to their summation, whereas the yields for the sum of those terms are strictly based upon the values as they might correspond to a given area.

To further complicate things, the only once sum of two squares as equal to an even integer is also allowed.  It matters not whether a^2 + b^2 is the sum of two Odd integers squared or the sum of two Even integers squared, since every Odd integer squared defaults to a 4n + 1 number, and every Even integer squared defaults to a 4n number.  If each of the two Odd terms squared both default to a 4n + 1 number, regardless of whether they are 4n + 3 numbered terms, then for the sum of the squares of those terms we will always get a 4n + 2 number as the result.  Likewise for the sum of two 4n or 4n + 2 numbers squared, we will always get the sum of two 4n numbers.  For example:

WHERE:       Odd + Odd = Even, and Even + Even = Even

IF:                  (4k + 1)^2 + (4k + 1)^2 = 4n + 2

AND:             (4k + 1)^2 + (4k + 1)^2 = 4^2 k (2k +1) + 2

THEN:           (4k + 3)^2 + (4k + 3)^2 = 4n + 2

SO:                (4k + 3)^2 + (4k + 3)^2 = 4^2 (k + 1) (2k + 1) + 2

IF:                  (4k + 2)^2 + (4k + 2)^2 = 4n

AND:             (4k + 0)^2 + (4k + 0)^2 = 4n

THEN:           (4k + 0)^2 + (4k + 2)^2 = 4n

SO:                  4k + 4k = 4n

Now comes the properties for the individual terms as they might be raised to a an Odd power.  This particular solution may be applied as a rule of omission for the problem known as "The Last Theorem of Fermat".  This rests upon the fact that while certain 4n + 1 series numbers may be root-able to the square... can 4n + 1 + 1 = 4n + 2 series numbers be root-able to the square or the cube, or to any power for that matter?  For example:

WHERE:        Odd + Even = Odd, Odd + Odd = Even, or Even + Even = Even

IF:                   (4k + 1)^even = 4n + 1

AND:              (4k + 1)^odd   = 4n + 1

AND:              (4k + 3)^even = 4n + 1

THEN:            (4k + 3)^odd   = 4n + 3

OR:                 (4k + 0)^even = 4n

SO:                  (4k + 0)^odd   = 4n

AND:               (4k + 2)^even = 4n

THEN:             (4k + 2)^odd   = 4n

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We have seen above, that when we attempt to yield a (4n + 2) integer by any and all of the combinations of possible bases with their corresponding exponents, the reverse of the operation is also prohibited.  So for three out of the four possible Odd integer combinations shown below, a (4n + 2) integer is the result.  We may now eliminate those three combinations from our mathematical choices.

IF:                     (4a + 1)^x + (4b + 1)^x = 4n + 2

AND:                (4a + 3)^x + (4b + 3)^x = 4n + 2

THEN:              (4a + 1)^even + (4b + 3)^even = 4n + 2

SO:                    (4a + 1)^odd  +  (4b + 3)^odd  = 4n

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IF:                     (4k + 1)^x + (4k + 1)^x = 4n + 2

AND:                (4k + 3)^x + (4k + 3)^x = 4n + 2

THEN:              (4a + 1)^x + (4b + 1)^x = 4n + 2

SO:                   (4a + 3)^x + (4b + 3)^x = 4n + 2

IF:                       4k + 1 + 4k + 1 = 4n + 2

AND:                  4k + 3 + 4k + 3 = 4n + 2

THEN:                4a + 1 + 4b + 1 = 4n + 2

SO:                     4a + 3 + 4b + 3 = 4n + 2

IF:                       4k + 1 + 4k + 3 = 4n

AND:                 (4k + 1)^odd = 4n + 1

AND:                 (4k + 3)^odd = 4n + 3

THEN:               (4k + 1)^odd + (4k + 3)^odd = 4n

SO:                    (4a + 1)^odd + (4b + 3)^odd = 4n

For any odd exponent greater than 2 and odd variable integer a or b:

IF:                      ((4*odd + 1)^odd + (4*odd + 3)^odd) /4 = odd integer

Since we may show that the sum of the terms of the (4*odd + 1) + (4*odd + 3) variety will always be divisible by 2^2 and only 2^2 to yield an odd integer, then by omission we have proven that the sum of those two terms raised to any odd power greater than 2 may never be root-able by the same power to yield the deuce.  At this point I must caution that this portion of the proof is only for a single series using odd precursors within the variable positions of the terms, however it is the general case for the entire series of those sums.  It will, therefore, be easy to continue to move toward a more complete comprehensive proof on the basis of elimination.  It should be interesting to note, that the sum of any two terms (4a + 1)^1 + (4b + 3)^1 = 4n = (4a + 2)^1 + (4b + 2)^1...

I don't even have to say at this point, that since the Even + Odd term combination listed below will always yield a 4n + 3 integer when both are raised to an Odd power, we may now dispense with the third term yield as ever being root-able by an Even Integer as well.  This is because a 4n + 3 number may not have any square root at all.  Since a (4k + 2)^x term will always default to a 4n integer, then these two terms may be considered synonymous.  Now you might be thinking what does this mean, several things in fact concerning the secondary properties of the integers along with their etymology.  In the case of three Odd exponents which are greater than 2, it has been historically shown that those exponents may never be entirely equivalent, and at the same time a common factor must be present for this to be so… it’s just that no one has ever been able to prove it in a lean or economical way.  For example: 

IF:                     (4k + 2)^odd + (4k + 3)^odd = 4n + 3

THEN:               (4a + 2)^odd + (4b + 3)^odd = 4n + 3

___________________________________________________

We may see from the examples immediately below that the properties of the yield may be traced cleanly through the first default an right into the secondary properties of the variables contained within (4 (4k + 1) + 1) term by simply changing it to a (4 (4k + 3) + 1) integer.  Since I have found that the (4 (4k + 1) + 2) terms are apparently unaffected by this, (probably the result of defaulting to a 4n number for all exponents) then we may now discover the properties of the internal variable of a single term by knowing only the powers of the terms in combination with the secondary denumeration sequence of the yield.  You may also see at this point the possible applications for configuring terms as well as Prime Numbers.  Thank you. 

IF:                    (4 (4k + 1) + 2)^3 + (4 (4k + 1) + 1)^3 = 4n + 1

AND:               ((4 (4k + 1) + 2)^3 + (4 (4k + 1) + 1)^3 – 1) /4 = 4n + 1

THEN:              (4 (4k + 1) + 2)^3 + (4 (4k + 3) + 1)^3 = 4n + 1

SO:                 ((4 (4k + 1) + 2)^3 + (4 (4k + 3) + 1)^3 – 1) /4 = 4n + 3

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IF:                    (4k + 2) + (4k +2) = 4n

AND:               (4k + 2)^2 = 4n

AND:               (4k + 2)^2 + (4k + 2)^2 = 4n + 4n

THEN:             (4k + 2)^2 + (4k + 2)^2 = 2^3 (2k + 1)^2

SO:                  (4k + 4k) = 2^3 * k

HENCE:         ((4a + 2)^2 + (4b + 2)^2) /2^3 = integer

IF:                     (4a + 2)^2 = 4n

AND:               ((4a + 2)^2 /4)^(1/2) = 4a /2 + 1

AND:               ((4b + 2)^2 /4)^(1/2) = 4b /2 + 1

THEN:             ((4a + 2)^2 + (4b + 2)^2 = 4 (2a + 1)^2 + 4 (2b + 1)^2

OR:                    4 (2a + 1)^2 + 4 (2b + 1)^2 = 4n + 4n

HENCE:           (2 (2a + 1))^2 + (2 (2b + 1))^2 = 4n + 4n

IF:                     ((4a + 2)^x + (4b + 2)^x = 2^x ((2a + 1)^x +

(2b + 1)^x)

THEN:             ((4a + 2)^x + (4b + 2)^x) /2^(x+1) = integer

HENCE:           (4a + 2)^x + (4b + 2)^x = 4n

IF:                     ((4a + 2)^2 + (4a + 2)^2) /2^2 - 2) /4 = 2a (a + 1)

THEN:               (4a + 2)^2 + (4a + 2)^2 = 4 (2^3 * a (a + 1) + 2)

HENCE:            (4a + 2)^2 + (4a + 2)^2 = 4 (4k + 2) = 4n

IF:                       (4a + 2)^2 + (4a + 0)^2 = 4n

AND:                  (4a + 2)^2 + (4a + 0)^2 = 4 (4a (2a + 1) + 1) = 4n

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GOLDBACH TWIN PRIME REDUCTION SERIES (unfinished)

IF                    ((72^2 - 2^2) + (65^2 + 2^2))^(1/2) = 97

AND:              ((11^2 + 4^2) + (60^2 - 4^2))^(1/2) = 61

Fermat-Mersenne Prime Sequence to Base 2

by Brian S. McMillan

PS.  After I thoroughly distribute this work, I will send out a similar dissertation involving the Wilson Theorem.  This is truly an interesting piece of work.  Even for me. 

Foreword:  Because the following artifact has become known, this is that 4n +1 numbers other than Primes may only once be the sum of two perfect squares, then what makes Prime Numbers unique in this regard.  Furthermore, it may mean that there are discrepancies in the 4n +3 test as well, including the other 4n generators, for instance, some of the Odd 4n +1 numbers have no perfect squares summand at all just like ALL of the 4n +3 series numbers.  Since this would mean that the sum of two squares could never equal a 4n +3 number, and some of the 4n +1 series numbers also share in this property, it must be for a different reason.  This result concerning Non-Prime 4n +1 numbers could not have been known to Leonard Euler, since he would not have used it in his method of reduction for 4n +1 series Fermat Numbers, at least without mentioning it.  (This in no way diminishes the strategy employed by Euler to ultimately reduce Fermat Numbers since his method incorporates multiples of 4n series factors as the primary mode of denumeration).  However, knowledge of this result at this point in mathematical history may prompt a serious review of a significant number of proofs which rely heavily on the "Only Once Sum of Two Perfect Squares Rule" as exclusive to Prime Numbers or their inferred role within the framework of those proofs.  Lack of mention for Non-Prime candidates in any proof related to this "Rule" is an almost automatic disqualification for that line of reasoning or any inferred property assigned to the same.  I have seen no mention of this in all of the books on prime numbers, or any discussion of it on the internet... at all.  So if you see any revision to Prime Number Theory concerning this in the near future, you will know for a certainty where it came from.  It may be that 4n +1 series numbers that are generated in the Fermat Numbers sequence are devoid of this discrepancy in their denumeration and in fact the Fermat Numbers may produce an inordinately high number of Strong Pseudo-Primes.  This has far reaching consequences for number theory.  At any rate, Primality cannot be absolutely determined by the only once sum of two perfect squares. 

You may view part of the original work at: 

http://www.pumpraser.com/fermatmersenne.html

With what I have found, I am a little confused about the following web site. 

http://www.math.rutgers.edu/~cherlin/History/Papers1999/chellani.html

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It may be interesting to note that 2^(4n) +1 always ends in 7.  Likewise, all 2^(4n) -1 always end in 5.  So for Mersenne SSequence Numbers, since all yields for 2^P -1 are either Prime or Pseudo-Prime, then it may be stated that since all yields for numbers of the form 2^(P-1) -1 = Non-Prime for P greater than 3, lest the Fermat test fail.  Then are the yields for 2^(P+1) -1 also Non-Primes, are they always divisible by 3.  Are the yields for 2^(Odd+1) -1 always divisible by 3.. or (2^(Even) -1) /3 = integer, If so then are all 2^NP -1 also Non-Prime?  Furthermore, some odd integers, while not Prime are themselves only once the sum of two perfect squares, such as: 

THEN:      25 = 3^2 + 4^2

Although it is obvious that 25 is not Prime. 

Numbers which may be less obvious are: 

IF:           117 = 6^2 + 9^2

THEN:     117 /3^2 = 13 Prime

IF:           2597 = 14^2 + 49^2 

THEN:     2597 /7^2 = 53 Prime

I had noticed that 6 = 2 * 3, and 9 = 3 * 3, so I decided to create a number out of 14 = 2 * 7, and 49 = 7 * 7, last example above, to see if it was only once the sum of two perfect squares and voila!  They are both 4n +1 numbers and are NOT Prime.  So while Fermat's assertion that 4n +1 Primes are only once the sum of two perfect squares, this may only be used to prove a known Prime and not its reverse.  This means that there are an infinite number of Odd integers which are NOT Prime that are only once the sum of two perfect squares.  For example: 

IF:           13 +1 = 14

AND:       53 +1 = 54

THEN:    (6 * 3^2)^2 + (9 * 3^2)^2 = 9477  Not Prime

HENCE:  9477 /3^6 = 13 Prime

SO:        (9477 -1) /4 = 2369

THEN:   ((9477 -1) /4 -1) /4 /4 /4 = 37 Prime 

So I decided to take this further to see if I could do it with a Prime number as one of the factors.  And I got a surprising result. 

IF:           37^2 + (4 * 37)^2 = 23273

THEN:     23273 - (4 * 38)^2 = 13^2 

HENCE:   4 * 13 +1 = 53 and we are back where we started!  WOW! WOW! WOW! 

This time almost at random, I was able to pick a 4n +1 Odd number that was Not Prime and yet only once the sum of two perfect squares.  For example:

IF:           153 = 3^2 + 12^2

THEN:     153 /3^2 = 17 Prime 

___________________________________________________

To continue expanding on the original sequence: 

IF:           (4 * 11)^2 + (11 * 11)^2 = 16577  Not Prime

THEN:      16577 /11^2 = 137 Prime

SO:         (16577 -1) /4 = 4144

HENCE:  ((16577 -1) /4 /4 /4 -3) /4 /4 /4 = 4

OR:         (16577 -3) /2 = 8287 Prime  (alternate sequence)

THEN:      8287 = 4n +3  (No Square Summands)

SO:          8289 = 4k +1  (No Square Summands)

This final example above is NOT Prime, however as a 4k +1 number it behaves as its 4k +3 counterpart with regard to its lack of any perfect squares summand. 

As we have seen previously, the result of factoring the fabricated example to yield Prime as the simple remainder is not a determining element.  P^2 = 4n +1 is likewise in many case only once the sum of two perfect squares.  Again, these are only a limited set of strategies which may in actuality be much more diverse, and in reality the only once sum of two perfect squares as applied to Non-Primes (may), like their Prime counterparts, diminish as the Natural Integers ascend.  For example: 

IF:            (4 * 19)^2 + (19 * 19)^2 = 136097  Not Prime

THEN:       136097 /19^2 = 377  Not Prime

SO:           136097 - (19 * 11)^2 = (19 * 16)^2  (Second Square Summand)

IF:            (4 * 29)^2 + (29 * 29)^2 = 720737  Not Prime

THEN:       720737 /29^2 = 857  Prime

SO:           720737 - (31 * 16)^2 = (53 * 13)^2 (Second Square Summand) 

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http://www-lipn.univ-paris13.fr/~banderier/Recipro/node35.html

I have not been able to find the format below... anywhere else. 

IF:              2^(2-1) + 1 = 3

AND:          2^(3-1) + 1 = 5

AND:          2^(5-1) + 1 = 17

ALSO:        2^(17-1) + 1 = 65537

HENCE:     (2^((65537-1) /2^11) - 1) /65537 /5 /3 /17 = 257 Prime

THEN:        2^(65537-1) + 1 = Prime or Pseudo-Prime

ALSO:        2^(P-1) + 1  etc...

http://mathworld.wolfram.com/Catalan-MersenneNumber.html

If the Wolfram MathWorld page titled:  Catalan-Mersenne Number, lists the sequence:  2^2 - 1 = 3, 2^3 - 1 = 7, 2^7 -1 = 127, and 2^127 - 1 = 170141183469231731687303715884105727, as the numbers defined by the series equation which is posted on the web page, then would the above sequence that I have found be titled the McMillan-Fermat Numbers?  Heh, heh... Just Kidding.

At the very least this proves that an infinite series of pseudo-primes exist, that represent compounded values which are only once the product of certain primes and never square, cube, or repeat those primes in their composition for the composite numbers which pass the Fermat primality test.  I believe that I've found a very economical way of proving or disproving Primality in the case of 2^(2^n) numbers.  Also, please remember that any odd number squared, becomes a 4n+1 number.  At any rate I believe that this takes the issue just a bit farther than Euler did, at least from a provable computation perspective.  A pattern between the Fermat Numbers and the Mersenne Number sequence reveals a definite relationship between those sequences, and the The Prime Number Theorem which may be illuminated by the condensed series method in relation to the Goldbach Conjecture as well as the Twin Prime Conjecture followed by the Polignac and therefore the Opperman Conjecture, of which I have expanded upon at the bottom, titled The Wilson-Opperman Bridge. 

Also, all Mersenne Prime yields are of the 4n+3 type... which means that:

IF:              2^P -1 = Prime or Pseudo-Prime

AND:          (P or PSP -3) /4 = odd integer, Prime, or Pseudo-Prime

THEN:        (2^P -4) /4 +0 = 2^(P-2) -1 ... Twin-Prime Theorem

HENCE:     (((((((((2^P -4)/4 -3)/4 -3)/4 -3)/4 -3)/4 -3)/4 -3)/4 -3)/4 ... -3)/4 = Mersenne Sequence

THEN:        ((2^P -4) /4 +2 = 2^(P-2) +1 ... Fermat Sequence (bridge) 

IF:              ((2^(2^n) +1) * 2 - 3 = 2^(2^n+1) -1

THEN:        ((2^(P-1) +1) * 2 - 3 = 2^P -1... Fermat-Mersenne (bridge)

For Some Prime P, and n = Some P, and Some number of divisors of 4:

HENCE:     ((((((2^(2^(P-1)/2^n) +1 -1) /4 +0)/4 +0)/4 +0)/4 +0)/4... +0)/4 = Fermat Sequence

OR:            2^(2^(P-1)/2^n) /2^(P-n) +1 = odd integer, Prime, or Pseudo-Prime

If, for example, it may be that 2^P -3 is a Prime number... then it would be a 4n +1 Prime.  It matters not that it is Prime, for it is still a 4n +1 number.  However, with the exception of certain small Primes below 100 or so, this number may only once be a 4P +1 Prime.  That is it may only once have a 1 subtracted from it and then divided by 4 to reach Prime where this number is not again a 4P +1 Prime.  If the sequence is reversed... that is pro-engineered using the Fermat Rule, then a number may only once be a 4P +1 Prime.  If we try to iterate the equation again as 4P +1, then the resulting yield cannot be Prime without first moving through a Non-Prime 4n +1 iteration.  No exception here is: 4 * 18 +1 = 73 Prime, 4 * 73 +1 = 293 Prime, and 4 * 293 = 1173 Not Prime, because our first number 18 is NOT Prime.  A minor exception as an example would be:  4 * 3 +1 = 13 Prime, 4 * 13 +1 = 53 Prime, and 4 * 53 +1 = 213 Not Prime, it survived two iterations beginning with 3, because 3 is Prime.  The reversed sequence for all Primes belonging to the Fermat-Mersenne series, that is yielded in base 2, should default to the 4n +3 Prime series.  Therefore, the remaining subtractions and divisions will be indicative of a 4n +3 number.  This could be considered proof for the 4n +1 Primes or some odd integers only being once the sum of two perfect squares, for example: 

IF:             (2^P -3) = 4n +1

THEN:       ((2^P -3) -1)/4 = 4n +3

OR:           (((((((((2^P -3) -1)/4 -3)/4 -3)/4 -3)/4) -3)/4... -3)/4 = odd integer, Prime, or Pseudo-Prime, Mersenne Sequence.  Or the general case defaults to the Mersenne Sequence. 

IF:             ((2^n -3) -1)/4 = 2^(n-2) -1

THEN:       ((2^n -1) -3)/4 = 2^(n-2) -1

HENCE:    (((2^n -1) -3)/4 -3)/4 = 2^(n-4) -1... etc. 

Note:  Keep in mind that the two Sequential Masters represented here by the Fermat and Mersenne Series, have been traditionally devised to Base 2.  Which means that 2 raised to any power at all will automatically be subject to the Fermat 4n reduction series.  That ALL Natural Numbers may be subject to the 4n denumeration is without question, but not all Natural Numbers must be reduced according to this method.  However, also keep in mind that the Fermat and Mersenne Series is assumed to proceed along entirely different lines than the Wilson Theorem, which as far as we know does NOT generate Pseudo-Primes.  This is not to mean that the Fermat 4n series reduction method is totally without merit.  For a workable strategy may be devised to overcome this limitation... obviously, since the default rule that I have found can be demonstrated to be employed to this advantage, but like Pseudo-Primes themselves it is only a work-around to the base 2 limitation, which coincidentally proceeds along the magnitude of 2 separating all odd and even numbers alike, thus Twin-Primes. 

Unless this is accomplished, then ultimately the series would be a trap.  With this said, a superior methodology (may) be to consider the 2n, 2n +1, 2n +2, 2n +3 Series reduction as a way to reach a more Wilson Theorem-like result.  Since the Wilson Theorem uses Even Natural Numbers as ascending products of 2 combined with intermediate Primes in the Euclid Format, of which prime products may have a greater value than the next employment of the 2 raised to its next power, it is this which makes it difficult to separate a workable pattern from the Prime Theorem Frequency using the Fermat-Mersenne Method.  However, because the reverse reduction rule when applied to Fermat 4n +1 Primes defaults to the Mersenne Sequence in base 2 after only a single iteration, then we may automatically assume that the Twin-Prime progression will always leave the remaining Prime seperated by + or - 2 as a 4n +1 Prime.  Thank You. 

Exceptions for 4n +1 Primes when further reducible only by 4n numbers.  When a number may be reduced by using the Fermat strategy, such as (97 -1) /4 /4 before changing it, and it does not default to the Mersenne Sequence, then it will be considered subject to a third Sequential Master, that is the sequence which must include all other Prime Numbers.  This will be indicative of a 2n, 2n +1, 2n +2 or 2n +3 alternate sequence quickly leading to Prime since all even numbers are divisible by 2.  There are also Prime Numbers which apparently follow no straightforward denumeration.  With the exception of the number 3, the Master Sequence for Fermat Numbers are obviously 4n reductions, so I will elaborate on this down the road since this is crucial to further Primality in that sequence.  For now I will give only one default Fermat to Mersenne sequence example, since listing more of them would be entirely rendundant, because we already know that any further examples would end up as even reductions to base 2.  Again, this method of reverse sequencing may be applied directly to Primes and Pseudo-Primes which are generated within both the Fermat and Mersenne series, since the only values to be generated are ultimately either Prime or Pseudo-Prime.  For example: 

IF:             (65537 -1)/4 /4 /4... /4 = 2^(2^n)  Fermat Sequence

THEN:       ((65537 -3)/2 -3)/2 /2 = 8191  Mersenne Sequence (alternate)

___________________________________________________

IF:             ((97 -1)/4 /4 /2 = 3  Prime

THEN:       ((97 -1)/2 -2) /2 = 23  Prime

HENCE:     ((97 -3)/2 = 47  Prime

When reducible by 2n or 4n +2 numbers.  This introduces the 2n value as a possible 2P number in some if not a significant amount cases.  This method may represent The Wilson-Opperman Bridge. 

SO:            (89 -1)/4 = 22 = 2 * 11

AND:          (73 -1)/4 = 18 = 2 * 3^2

OR:            (41 -1)/4 = 10 = 2 * 5... etc.

I believe that while Euler's reasoning and method of proof for the break-down of the Fermat Numbers Sequence is quite useful and far reaching, there are still unanswered questions within the 2^(2^n) +1 series with regard to the possible discovery of single prime numbers appearing at some further position in that sequence.  There is also some degree of illumination concerning the nature as well as the configuration of the Pseudo-Primes that are revealed within the Fermat Numbers Sequence that have not been fully investigated.  Simply finding ever larger composite Pseudo-Primes in order to dissect the larger Prime numbers that evolve within that series is not enough for serious mathematics, as fun as it may be... it is not enough.  It must be understood exactly why these Pseudo-Primes are generated in the way that they are.  For example:  Are the Pseudo-Primes which are generated within the Fermat Numbers Sequence always a compound of two Prime numbers and two only, or will they be proven to be composed of more than two, and if so... why?  If it is found that the Pseudo-Primes in the Fermat Numbers Sequence are always composed of two Primes, then a mathematical strategy may be devised that goes far beyond the trial and error reduction method employed to dissect those composites into their respective Primes.  For example: 

http://www.godkings.com/pseudoprime.txt

If the dual compound or Two-Prime Composite Pseudo-Primes obey the same rule as the general case listed above, then we may have the beginning of a Twin-Prime Theorem.  Thank You. 

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It may be interesting to note that 2^(4n) +1 always ends in 7.  Likewise, all 2^(4n) -1 always end in 5.  So for Mersenne SSequence Numbers, since all yields for 2^P -1 are either Prime or Pseudo-Prime, then it may be stated that since all yields for numbers of the form 2^(P-1) -1 = Non-Prime for P greater than 3, lest the Fermat test fail.  Then are the yields for 2^(P+1) -1 also Non-Primes, are they always divisible by 3.  Are the yields for 2^(Odd+1) -1 always divisible by 3.  If so then are all 2^NP -1 also Non-Prime?  Has a more general omission rule been found here for the Levy's Conjecture? 

http://mathworld.wolfram.com/LevysConjecture.html

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PROOF to Infinity for 2^(2^n) + 1 PRIMES or PSEUDO-PRIMES with Conjectures Review by Brian S McMillan

For some Prime P and Variables (n, k)

IF:              2^(2^n) + 1 = Prime or Pseudo-Prime

THEN:        2^(2^n) + 1 = 4k + 1

WHERE:     4k+1 = Prime or Pseudo-Prime

SO:            (2^(4k+1) - 2) /(4k+1) = Integer

AND:          (2^((P-1)/2) + or -1) /P = N Integer

HENCE:     (2^(P-1) - 1) /P = N (PN + or - 2)   Fermat

For some Prime P and number of 2's

OR:            (2*2*2... *2 + or - 1)/ P = Integer   Euclid-Fermat

The last equation above completes the proof.

FERMAT ADDITIONS:

IF:              2^x /2^y = x

WHERE:    (x, y) are Integers

THEN:        2^x + 1  = Prime or Pseudo-Prime

OR:           (2^(2^x/2^y) + 1) /(2^x + 1) = 1

HENCE:     2^((P-1)/2^y) + 1 = Prime or Pseudo-Prime

Now the above approach may not be entirely unique except for one interesting relationship, and that is that the sequence which is represented (may) be related to the Prime-Gap problem, and therefore to the Goldbach Conjecture.  Thank You.

WHERE:     x is variable and n = Integer

IF:              x = n /(x+n) + n =  sqrt n(n+1)

AND:          x = n /(x-n) - n  = -sqrt n(n+1)

ALSO:

IF:              x = n /(x+1) + 1 = sqrt (n+1)

AND:          x = n /(x-1) - 1 = -sqrt (n+1)

WHERE:     n = (P-1)

THEN:        x = sqrt P

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For Mersenne 2^P - 1 =  Prime or Pseudo-Prime, I have ffound great agreement to the Pseudo-Prime but not much Prime agreement.  I would have to wonder if this trend continues.  For example:  2^11 -1 = 2047, and 2^23 -1 = 8388607 PSP which are not in the Fermat Number Series except as part of the Mersenne Pseudo-Primes in the Super Exponent, but missed it by only 2... I just thought it was interesting that so many Pseudo-Primes are generated.  I wonder if any more 2^(P+1) +1 Pseudo-Primes are generated besides 2^(31+1) +1 = 4294967297 PSP... (127 +1), (8191 +1), (131071 +1), (524287 +1), (2147483647 +1),  . Which brings us to an interesting argument.  If we periodically divide Mersenne Pseudo-Primes+1 or Mersenne Primes +1  into the Fermat Number Sequence as 2^y,

Keep in mind before I continue, that 2^(2^37) +1 = Prime or Pseudo-Prime, is an extraordinarily gargantuan number, however, it is this number that will pass the Fermat Primality Test, whether it is Prime or Composite.  This allows extremely large numbers to be tested for primality without actually computing the full values of the numbers.  This means that the proper form of the Fermat Equation incorporating this specific example would be:

      (2^(2^(2^37)) -1) /(2^(2^37) +1) = Integer

IF:  (2^(2^(2^37)+1) -2) /2/(2^(2^37) +1) = Integer, and I know for certain that it does!

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IF:   2^3 -1

=     7  Mersenne Number

AND:

       7 * 2 + 3

=    17

THEN:

        2^4 +1

=     17  Fermat Number

HENCE:  2^4 /2 -1

=     7

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IF:   2^7 -1

=     127  Mersenne Number

AND:

       127 * 2 + 3

=     257

THEN:

       2^8 +1

=     257  Fermat Number

HENCE:  2^8 /2 -1

=     127

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IF:   2^13 -1

=     8191  Mersenne Number

AND:

       8191 * 8 + 9

=     65537

THEN:  2^16 +1

=     65537  Fermat Number

HENCE:  2^16 /2^3 -1

=     8191

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IF:   2^31 -1

=     2147483647  Mersenne Number

AND:

      2147483647 * 2 + 3

=    4294967297

THEN:  2^32 +1

=     4294967297  Fermat Number (PSP)

HENCE:  2^32 /2 -1

=     2147483647

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IF:  2^61 -1

=    2305843009213693951  Mersenne Number

AND:

      2305843009213693951 * 8 + 9

=   18446744073709551617

THEN:  2^64 +1

=   18446744073709551617  Fermat Number (PSP)

HENCE:  2^64 /2^3 -1

=    2305843009213693951

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IF:  2^127 -1

=    170141183460469231731687303715884105727  Mersenne Number

AND: 

      170141183460469231731687303715884105727 * 2 + 3

=    340282366920938463463374607431768211457

THEN:  2^128 +1

=    340282366920938463463374607431768211457  Fermat Number (PSP)

HENCE:  2^128 /2 -1

=    170141183460469231731687303715884105727

COPYRIGHT 1996-2007 Brian S. McMillan

ALSO:

http://www.pumpraser.com/

http://www.godkings.com/physics.txt

http://www.godkings.com/quasar.txt

http://www.godkings.com/gravity.txt

http://www.godkings.com/raser.txt

http://www.godkings.com/fermat_theorem.txt

http://www.godkings.com/pseudoprime.txt

http://www.godkings.com/radiogravity2.txt

http://www.godkings.com/fermatlast.txt

http://www.godkings.com/pyramidindusnuke.html

___________________________________________________

Introductory note:

Since the following conjectures are all proposed in algebraic terms, would it not follow that the solutions could also be represented algebraically. 

The Euler-Goldbach Conjecture is this: 

Is every even number the sum of two primes? 

The Opperman Conjecture is this: 

Does a prime always fall between n^2 and (n +1)^2... ? 

The Twin Primes Conjecture is this: 

Are there an infinite number of twin primes? 

The Polignac Conjecture is this: 

For every even number 2n there are infinitely many pairs of consecutive primes which differ by 2n. 

I am referencing these conjectures from the internet "Prime Pages' resources" page titled: 

"Prime Conjectures and Open Questions" 

The click-on is listed immediately below. 

http://primes.utm.edu/notes/conjectures/ 

THE WILSON-OPPERMAN BRIDGE

When one takes the frequency for the occurrence in primes set forth in the Opperman Conjecture as that frequency represented by the difference between n^2 and (n +1)^2... that is:  (n +1)^2 - n^2 = (2n +1), then this is virtually identical to the conditions which are set forth in the solution to the Polignac conjecture.  Since it is stated on "The Prime Pages" resource, in the click-on above, that the Polignac Conjecture is identical to the Twin Prime Conjecture, when n = 1, and because all primes greater than 2 are odd, then beginning with n = 1, the frequency is only shifted by 1 to infinity, since we are only representing the difference as (2n +1) and not this value squared.  Now the popular interpretation for the progression of twin primes seems to favor a starting point without regard to the possibility, that while the expansion of twin prime sets may be separated by 2n, because all twin primes themselves are separated by 2, that the series for twin primes proceeds from infinitely many starting points, and therefore, all twin primes series may not share their progression with each other, and may even be devoid of such parallels.  That is the frequency of twin primes, like prime numbers themselves, may expand simply because there is a greater and greater pool of larger numbers with which a factorization may be performed.  And we see gaps in those series, because this is where composites are repeated as more frequent compounds of those smaller primary numbers.  While this is obvious, to say the least... it is the obvious that is often overlooked. 

If any even number squared is going to be even, and any odd number squared is going to be odd.  Furthermore, since two times any number is going to be even, and finally, any even number plus one is going to be odd.  For example: 

IF:              (n +1)^2 - n^2                   =  (2n +1) 

According to Pythagorean Theorem 

THEN:       n^2 + (2n +1)                    =  (n +1)^2 

For some number n 

SO:            (2n +1)^(1/2)  =  integer (odd) 

THEN:      

For some even number n and some odd integer. 

HENCE:   (4n +1)^(1/2)  =  integer (odd) 

In the last example above every odd integer, greater than one, may be represented for some root square composed of 4n +1 only when n is some even number.  Since all odd integers when treated in this way, may be yielded in series, and every other odd integer is of the 4n +1 type, and the remaining alternate odd integers are of the 4n +3 type.  When any odd integer is squared, then it becomes a 4n +1 type.  Since at least one of the primes in a twin prime set must be either of the two types, and a 4n +1 solution is equivalent to a 2n +1 solution, and 2n +1 is the difference between n^2 and (n +1)^2 , furthermore, since (2n +1)^(1/2) = integer, for some even number n, and this may represent the central term in a Pythagorean Theorem.  

Since the addition of two odd integers will always yield an even number, and the addition of two even integers will always yield an even number.  For example: 

WHERE:  P = prime 

IF:           P + P  =  2P 

THEN:     2P  =  integer (even) 

FOR:       2P +1 = Prime

OR:         2P -1  = Prime

IF:           4P -1 = Prime

THEN:     Prime = 4n +3

IF:           4 * 137 -1 = 547 Prime

THEN:   ((4 * 137 -1) -3) /4 /4 = 2 * 17, or 2P

IF:          (19 -1)(2 * 19) -1 = 4n +3 = 683  Prime (No Square Summands)

THEN:    (19 -1)(2 * 19) -3 = 4n +1 = 681  Not Prime (No Square Summands)

A proof attributed to Euclid states that:  When 1 is added or subtracted from the product of all prime numbers thought to exist, whether in series or not, then the remainder will always be divisible by a prime not contained within the set.  According to the Fundamental Theorem of Arithmetic, all numbers are reducible to one, by primes, and of course a prime is only factorable by itself and one.  Now, here it is...

When all prime numbers contained within a set, are represented in series, for example: 

For

Wilson-Opperman Bridge:  To Be Continued...

___________________________________________________

IF:            (o+1)^2 /4 + ((e+1)^2 -1) /4            =  e (e+1) /2 

AND:         n^2 + (n+1)^2 - 1                         =  2n (n+1) 

THEN:       n^2 + (n+1)^2                              =  2n (n+1) +1 

HENCE:     x  =  (x+1)^2 /x^2                        =  2.147899035705 

The last of which is the Opperman-Polignac configuration as well as the Euler-Goldbach series. 

THEN:       (n-1)^2 + (n^2 -1)      &nbssp;                  =  2n (n-1) 

___________________________________________________

IF:            ((2P +1) * 2P /2 - P^2)                   =  P (P + 1) 

OR:  a value that falls between P^2 and (P +1)^2 

THEN:      ((2P +1) * 2P /2 - P^2) /P^2           =  fraction greater than 1 

SO:          ((2P +1) * 2P /2 - P^2)/(P +1)^2     =  fraction less than 1 

IF:            ((2P +1) * 2P /2 - P^2) /2              =  P (P +1) /2 

AND: 

According to Karl Freidric Gauss, the sum of all integers in series beginning with 1, may be expressed by the following induction equation: 

For all integers n 

IF:            1 + 2 + 3... + n                            =  n (n +1) /2 

THEN:      1 + 2 + 3... + P                            =  P (P +1) /2 

IF:            n + ((n-1)/2)^2  =  ((n+1)/2)^2

THEN:     (n + ((n-1)/2)^2 + ((n+1)/2)^2 = 2 ((n+1)/2)^2

For some Integer n, and Prime P

HENCE:   2^n -1 + ((2^n -2) /2)^2 + ((2^n) /2)^2

                                                                  = 2^P

OR:         2^n -2 + ((2^n -2) /2)^2 + ((2^n) /2)^2

                                                                  = 2^P -1

THEN:

2^((P+1)/2) -2 + ((2^((P+1)/2) -2) /2)^2 + ((2^((P+1)/2) /2)^2

                                                                  = 2^P -1

AND:  2 (2^n -1 + ((2^n -2) /2)^2 + ((2^n) /2)^2)

                                                                  = 2^(2n)

THEN:  2 (2^n -1 + ((2^n -2) /2)^2 + ((2^n) /2)^2) +1

                                                                  =  2^(2n) +1

HENCE:

2 (2^((P+1)/2) -1 + ((2^((P+1)/2) -2) /2)^2 + ((2^((P+1)/2) /2)^2) +1

                                                                 =  2^(P+1) +1

___________________________________________________

For all twin primes. 

IF:           (P1 + 2)  =  P2 

THEN:     (P (P +2) +1)^(1/2)                        =  integer 

IF:            P + P  =  2P 

THEN:     (2P +1)  = 

/(2 * 31) /2^4 

___________________________________________________

PHYSICS 

IF:                x  =  (1/x + 1)^x                        =  2.293166287408 

AND:            e  =  (1/k + 1)^k                        =  2.718281828 

THEN:          e^2.2931425038 * 6 /pi^2           =  6.02213713  NA 

OR:              e^2.293166287408 * 6 /pi^2       =  6.02228036  Infinite Volume 

___________________________________________________

IF:                2.147899035705^(1/x)               =  1.608477493173 

THEN:          e^1.608477493173                    =  4.995200209 

IF:                x  =  (1/x + 1)^x                        =  2.293166287408 

AND:            e  =  (1/k + 1)^k                        =  2.718281828 

IF:               1 /e^(1/2.279944390) + 1            =  pi^2 /6 

AND:            (pi^2/6 - 1) * e^(1/x)                   =  1.351574830515 

___________________________________________________

Unfinished: 

This is also unfinished, so I will include it like this. 

According to "The Mathematical Universe" by William Dunham, pp. 65-74; Pierrie de Fermat lived from (1601-1665).  What we will concern ourselves with here is Fermat's assertion that prime numbers of the form (4n+1) are only once the sum of two perfect squares.  Just one quick note before we continue. 

There is an entire series of numbers (all of which I have not listed in sequence) that end in 5... beginning with the number 5, next 25, 45, 225, 245... 1225, then 2025, 3025 etc. that are only once the sum of two perfect squares.  Furthermore, the number 117, for example is not prime and is only once the sum of two perfect squares.  So while this method of reasoning may be utilized to prove a negative, it may never be used to prove a positive.  There is of course another interesting pattern embedded within this line of reasoning, but we will get to that at a future date.  With that said, I will now continue. 

Fermat proposed a scheme with which to determine the form of all real numbers. 

They must consist of four forms. 

>> 4n      = integer (even) 

>> 4n+1 = integer (odd)

>> 4n+2 = integer (even)

>> 4n+3 = integer (odd)

IF:           (2^(P-1) -1)/P = integer   Fermat, circa 1650 

WHERE:   P = Prime or Pseudo-Prime 

AND:        (2^((P-1)/2) + or -1)/P = N integer  Fermat 

WHERE:   P = Prime or Pseudo-Prime 

HENCE:   (2^(P-1) - 1) /P = N (PN + or - 2)   McMillan circa 2003

If we further divide the exponent by 2 in the technique illustrated above.  There appear to be certain prime numbers of the form (4n+1) which more than once obey this rule.  By this I mean as a continuation to the prime number theorem illustrated above as the reverse form of (4n+1) is applied to the position of the exponent.  We will call these prime numbers; Fermat Primes, in honor of the man which discovered this form.  For example: 

>> (2^((17-1)/4) +1)/17 = 1

>> (2^((41-1)/4) +1)/41 = 25

>> (2^((73-1)/4) -1)/73 = 3591

>> (2^((89-1)/4) -1)/89 = 47127

>> (2^((97-1)/4) +1)/97 = 172961

THEN:     (2^((FP-1)/4) ±1)/FP = integer 

>> (2^((113-1)/4) -1)/113 = 2375535

>> (2^((137-1)/4) +1)/137 = 125400505

So for the Fermat Primes which appear before 100, there are only 5 of them.  Now these primes have an unusual property which I will discuss later.  If we continue expansion of the sequence to the next series. 

>>(2^((113-1)/8) +1) /113 = 145

>>(2^((233-1)/8) -1) /233 = 2304167>

>>(2^((337-1)/8) -1) /337 = 13050583119