...
if you're returning, then please press the "ctrl" and "F5" keys together to update your cache.
just click on the headings above to navigate to the several different pages of this website.
...
first, a short proof that "there exists a prime between x^2 and (x +1)^2.", or Legendre's con-
jecture is actually a theorem! look below...
...
assume that... (x^2)/ln(x^2) +1 < ((x +1)^2)/ln((x +1)^2) using the Prime Number Theorem,
(x^2 +ln(x^2))/ln(x^2) < (x^2 +2x +1)/ln((x +1)^2) combining the formula w/+1 on the LHS,
x^2 +2*ln(x) < (x^2 +2x +1)*[(ln(x^2))/(ln((x +1)^2)] multiplying both sides by ln(x^2),
...
but the limit [(ln(x^2))/(ln((x +1)^2)] -> 1 almost immediately, and we have ln(x^2) < 2x +1,
and taking the derivative wrt 'x', we have... (d/dx)[ln(x^2)] < (d/dx)[2x +1], or... 2/x <2, 1<
x; so our beginning statement has to be true for x > 1, and it is...
...
x= 2; LHS= 3.885 < 4.096 = RHS, and x= 3; LHS= 5.096 < 5.770 = RHS...and so on, and
so on. we simply check between 1 and 4 to see that the primes 2 and 3 exist. I'm using
approximations of pi(x) to secure his conjecture--- It's like I'm teeing off from "being in the
rough" and "still making a superb hole-in-one!"
*QED
...
when considering Legendre's stronger conjecture, please substitute the 'sqrt(x)' in for '1' in the
previous beginning statement. then subtract x^2 to get... sqrt(x)*ln(x^2) < 2x, and w/out loss
of generality... subtract '1' from only the RHS; divide by '2' and square both sides to arrive at...
x*[ln(x)]^2 < [x]^2; take the derivative of both sides twice (using the chain rule), multiply by 'x',
and divide by '2'. we have... 1 +ln(x) < 2x, which is true for x > 0. [[corrected on 04/05/2012]]
*QED
09/17/2011 (both)
...
...
for those of you that have visited my website, I've spent the last 10 years (almost 100 drafts)
searching for a reasonable explanation for why odd-perfect numbers don't exist, and here it is...
...
Euler **famously** proved that 'even-perfect' numbers have the form N= (2^(n-1))*(2^n -1), where
(2^n -1) is a Mersenne prime which is the sum of all integers from 1 to a Mersenne prime. Notice
that the middle term of that sum must be a square if the N > 6, i.e. 1 +2 +3 +(4) +5 +6 +7 = 28.
This fact leads us to believe that (x +1)/2 = y^2 for x= Mp > 3. therefore, x= 2*y^2 -1, and if we
substitute M= p^q*Q^2 in for an odd-perfect 'x', then we have the Diophantine equation p^q*Q^2
+1= 2*y^2, or... a*Q^2 +1 = 2*y^2; no 2 "odd" squares can differ by 1, unless a= 1; Q= 1 & y= 1
or Q= 7 & y= 5; hence, regardless of whether an OPN may be deficient or abundant, we couldn't
properly choose the middle term for a particular "odd-perfect" sum. if you're in doubt, then please
check my work... 2*(n-2)^2 = n^2 +1 implies (n -1)*(n -7) = 0 where 'n' is 'Q'. {corrected 5/4/2012}
...
the question of whether an odd-perfect number will behave like an even-perfect number is com-
parable to asking... will the two lines ever be parallel? the answer is NO. the OPN-line always has
a sag or bow in it as I have shown you with the Diophantine measure: a*Q^2 +1 = 2*y^2, and
the amount of deformity cannot even be predicted except for two erroneous solutions!
...
thus, odd-perfect numbers don't exist!
...
obviously, when someone as prolific as Euler stated that this problem was too complicated to be
easily solved, every mathematician was easily persuaded to believe it; and if you (OR) I decided
to retain Euler's already-proven formula for an odd-perfect number, it's surprising to me that we
could have been easily mislead by his prior conclusion!
...
*QED
Bill Bouris
01/04/2012 {corrected 04/21/2012}
...
what's this stuff ??? just cereal,... supposed to be good for you,... you try it,... I'm not gonna try
it,... let's get Mikey,... yeah, he won't eat it,... he hates everything!... ... he likes it... hey, Mikey!
...
(OR) you can expect to comprehend the heuristic and believe that it's unlikely that they exist!
...
...
here's a better puzzle:
let 'm' be any integer of the form m= 2^k*(2n+1). if sigma(m)= 2m -1, then 'm' can only be of the
form2^k, or... an almost-perfect number.
...
proof:
either sigma(2n +1)= 4n +1, or sigma(2^k*(2n+1)) = sigma(2^k)*sigma(2n +1)= n*2^(k+2) +2^(k+1)
-1implies that sigma(2^k)= [n*2^(k+2) +2^(k+1) -1]/(4n +1) is defined for all 'k'.
...
if n <> 0 in the latter statement, then sigma(2^k) will be fractional; also, if k= 0, n= 0, then sigma(1)
=(0 +2 -1)/(4(0)+1)= 1/1 = 1; if k= k when n= 0, then sigma(2^k)= (2^(k+1))-1 = 2*(2^k) -1. if you
tried to argue w/the former statement that sigma(2n +1)= 4n +1, then you'd have to answer a much
bigger question about the sigma function, since sigma(4n +3)= 8n +5; it's certainly false, since sig-
ma(4n +3)is ALWAYS EVEN; that's a more difficult problem to "prove or disprove".
...
if sigma(4n +3) is ALWAYS EVEN, then an APN must only be of the form 2^k.
03/26/2012
...
...
there are infinitely many Mersenne prime numbers! I didn't create the theorem (strong portion)...
...
according to Dirichlet's theorem, there are infinitely many primes of the form g(n)= 4n +3 where
a= 3, d= 4, (a,d)= 1, and g(n)= a +n*d, and f(n)= n; it implies that 1/f(0) == 1/0 << [series 1/g(n)]
== [1/3 +1/7 + 1/11 + 1/19... + 1/n] which is divergent; 'n' is an element the set of Natural (con-
secutive) Numbers.
...
similarly, there will be infinitely many Mersenne primes of the form M(x)= 2*(2^x) -1 where a= -1,
d= 2, (a,d)= 1, and M(x)= a +N(x)*d, such that N(x)= 2^x is 1-to-1 and onto with respect to the set
of Whole (consecutive) Numbers. Thus, if and only if, 1/N(1) = 1/2 << [series 1/M(x)] == [1/3 +1/7
+1/31 + 1/127 +1/8191 + ... +1/x]. This fact can be confirmed after computing only the first few
terms.
...
therefore, there are infinitely many Mersenne primes of the form M(p) = 2^p -1.
*QE(Dirichlet's Interpretation)
01/23/2011
...
...
there are finitely many Fermat prime numbers! I didn't create the theorem (strong portion)...
...
unless you discount Dirichlet's theorem, again, there will be finitely many Fermat primes of the
form F(x)= 2*(2^(2^x -1)) +1 where a= +1, d= 2, (a,d)= 1, and F(x)= a +G(x)*d, such that G(x)=
2^(2^x -1) is 1-to-1 and onto with respect to the set of Whole (consecutive) Numbers. Thus, iff
I'm correctly understanding his argument, then 1/G(0) = 1/1 must be >> [series 1/F(x)] == [1/3
+1/5 +1/17 + 1/257 +1/65537 + ... +1/x]. This simple fact can be confirmed after computing
just the first few terms.
...
therefore, there are finitely many Fermat primes of the form F(x)= 2^(2^x) +1.
*QE(Dirichlet's Interpretation)
01/25/2011
...
if you're returning, then please press the "ctrl" and "F5" keys together to update your cache.
just click on the headings above to navigate to the several different pages of this website.
...
first, a short proof that "there exists a prime between x^2 and (x +1)^2.", or Legendre's con-
jecture is actually a theorem! look below...
...
assume that... (x^2)/ln(x^2) +1 < ((x +1)^2)/ln((x +1)^2) using the Prime Number Theorem,
(x^2 +ln(x^2))/ln(x^2) < (x^2 +2x +1)/ln((x +1)^2) combining the formula w/+1 on the LHS,
x^2 +2*ln(x) < (x^2 +2x +1)*[(ln(x^2))/(ln((x +1)^2)] multiplying both sides by ln(x^2),
...
but the limit [(ln(x^2))/(ln((x +1)^2)] -> 1 almost immediately, and we have ln(x^2) < 2x +1,
and taking the derivative wrt 'x', we have... (d/dx)[ln(x^2)] < (d/dx)[2x +1], or... 2/x <2, 1<
x; so our beginning statement has to be true for x > 1, and it is...
...
x= 2; LHS= 3.885 < 4.096 = RHS, and x= 3; LHS= 5.096 < 5.770 = RHS...and so on, and
so on. we simply check between 1 and 4 to see that the primes 2 and 3 exist. I'm using
approximations of pi(x) to secure his conjecture--- It's like I'm teeing off from "being in the
rough" and "still making a superb hole-in-one!"
*QED
...
when considering Legendre's stronger conjecture, please substitute the 'sqrt(x)' in for '1' in the
previous beginning statement. then subtract x^2 to get... sqrt(x)*ln(x^2) < 2x, and w/out loss
of generality... subtract '1' from only the RHS; divide by '2' and square both sides to arrive at...
x*[ln(x)]^2 < [x]^2; take the derivative of both sides twice (using the chain rule), multiply by 'x',
and divide by '2'. we have... 1 +ln(x) < 2x, which is true for x > 0. [[corrected on 04/05/2012]]
*QED
09/17/2011 (both)
...
...
for those of you that have visited my website, I've spent the last 10 years (almost 100 drafts)
searching for a reasonable explanation for why odd-perfect numbers don't exist, and here it is...
...
Euler **famously** proved that 'even-perfect' numbers have the form N= (2^(n-1))*(2^n -1), where
(2^n -1) is a Mersenne prime which is the sum of all integers from 1 to a Mersenne prime. Notice
that the middle term of that sum must be a square if the N > 6, i.e. 1 +2 +3 +(4) +5 +6 +7 = 28.
This fact leads us to believe that (x +1)/2 = y^2 for x= Mp > 3. therefore, x= 2*y^2 -1, and if we
substitute M= p^q*Q^2 in for an odd-perfect 'x', then we have the Diophantine equation p^q*Q^2
+1= 2*y^2, or... a*Q^2 +1 = 2*y^2; no 2 "odd" squares can differ by 1, unless a= 1; Q= 1 & y= 1
or Q= 7 & y= 5; hence, regardless of whether an OPN may be deficient or abundant, we couldn't
properly choose the middle term for a particular "odd-perfect" sum. if you're in doubt, then please
check my work... 2*(n-2)^2 = n^2 +1 implies (n -1)*(n -7) = 0 where 'n' is 'Q'. {corrected 5/4/2012}
...
the question of whether an odd-perfect number will behave like an even-perfect number is com-
parable to asking... will the two lines ever be parallel? the answer is NO. the OPN-line always has
a sag or bow in it as I have shown you with the Diophantine measure: a*Q^2 +1 = 2*y^2, and
the amount of deformity cannot even be predicted except for two erroneous solutions!
...
thus, odd-perfect numbers don't exist!
...
obviously, when someone as prolific as Euler stated that this problem was too complicated to be
easily solved, every mathematician was easily persuaded to believe it; and if you (OR) I decided
to retain Euler's already-proven formula for an odd-perfect number, it's surprising to me that we
could have been easily mislead by his prior conclusion!
...
*QED
Bill Bouris
01/04/2012 {corrected 04/21/2012}
...
what's this stuff ??? just cereal,... supposed to be good for you,... you try it,... I'm not gonna try
it,... let's get Mikey,... yeah, he won't eat it,... he hates everything!... ... he likes it... hey, Mikey!
...
(OR) you can expect to comprehend the heuristic and believe that it's unlikely that they exist!
...
...
here's a better puzzle:
let 'm' be any integer of the form m= 2^k*(2n+1). if sigma(m)= 2m -1, then 'm' can only be of the
form2^k, or... an almost-perfect number.
...
proof:
either sigma(2n +1)= 4n +1, or sigma(2^k*(2n+1)) = sigma(2^k)*sigma(2n +1)= n*2^(k+2) +2^(k+1)
-1implies that sigma(2^k)= [n*2^(k+2) +2^(k+1) -1]/(4n +1) is defined for all 'k'.
...
if n <> 0 in the latter statement, then sigma(2^k) will be fractional; also, if k= 0, n= 0, then sigma(1)
=(0 +2 -1)/(4(0)+1)= 1/1 = 1; if k= k when n= 0, then sigma(2^k)= (2^(k+1))-1 = 2*(2^k) -1. if you
tried to argue w/the former statement that sigma(2n +1)= 4n +1, then you'd have to answer a much
bigger question about the sigma function, since sigma(4n +3)= 8n +5; it's certainly false, since sig-
ma(4n +3)is ALWAYS EVEN; that's a more difficult problem to "prove or disprove".
...
if sigma(4n +3) is ALWAYS EVEN, then an APN must only be of the form 2^k.
03/26/2012
...
...
there are infinitely many Mersenne prime numbers! I didn't create the theorem (strong portion)...
...
according to Dirichlet's theorem, there are infinitely many primes of the form g(n)= 4n +3 where
a= 3, d= 4, (a,d)= 1, and g(n)= a +n*d, and f(n)= n; it implies that 1/f(0) == 1/0 << [series 1/g(n)]
== [1/3 +1/7 + 1/11 + 1/19... + 1/n] which is divergent; 'n' is an element the set of Natural (con-
secutive) Numbers.
...
similarly, there will be infinitely many Mersenne primes of the form M(x)= 2*(2^x) -1 where a= -1,
d= 2, (a,d)= 1, and M(x)= a +N(x)*d, such that N(x)= 2^x is 1-to-1 and onto with respect to the set
of Whole (consecutive) Numbers. Thus, if and only if, 1/N(1) = 1/2 << [series 1/M(x)] == [1/3 +1/7
+1/31 + 1/127 +1/8191 + ... +1/x]. This fact can be confirmed after computing only the first few
terms.
...
therefore, there are infinitely many Mersenne primes of the form M(p) = 2^p -1.
*QE(Dirichlet's Interpretation)
01/23/2011
...
...
there are finitely many Fermat prime numbers! I didn't create the theorem (strong portion)...
...
unless you discount Dirichlet's theorem, again, there will be finitely many Fermat primes of the
form F(x)= 2*(2^(2^x -1)) +1 where a= +1, d= 2, (a,d)= 1, and F(x)= a +G(x)*d, such that G(x)=
2^(2^x -1) is 1-to-1 and onto with respect to the set of Whole (consecutive) Numbers. Thus, iff
I'm correctly understanding his argument, then 1/G(0) = 1/1 must be >> [series 1/F(x)] == [1/3
+1/5 +1/17 + 1/257 +1/65537 + ... +1/x]. This simple fact can be confirmed after computing
just the first few terms.
...
therefore, there are finitely many Fermat primes of the form F(x)= 2^(2^x) +1.
*QE(Dirichlet's Interpretation)
01/25/2011
...