Numero pari come somma di due numeri primi - Congettura forte di Goldbach
(Even number as sum of two prime numbers - Goldbach's strong conjecture)
Poiché l’insieme di tutti i numeri pari equivale all’insieme di tutti i numeri?
4n - 4? e? 4n - 6 ?con? n > 1 ?,
la congettura di Eulero (o congettura forte di Goldbach, ogni numero pari maggiore di 2 può essere scritto come somma di due numeri primi), trova conferma nelle seguenti affermazioni:
?? (n >1)?? ?? (p, p′, m)? |? n = √(p ? p′+ m^2)
?? n? ?? z? |? z = 4n - 4
?? n? ?? x? |? x = 6n - 9
?? x? ?? y |? y = x/3
?z = 4n - 4 = p + p′
x - y = 4n - 6 = p + p′
Statement
Geometric description
Semi-prime table
Conclusions
Corollary
With p and p' prime numbers being the same or different
If
n^2 - m^2 = (n - m) ? (n + m) = p ? p′
then
2n = (n - m) + (n + m) = p + p′
We will demonstrate that for every n > 1 there are always two equal or different prime numbers
(n - m), (n + m) = (p, p′)
such that
n = √(p ? p′+ m^2)
therefore 2n can always also be the sum of two equal or different primes.
In the example there are 13 repeated pairs 2n = p + p', in fact the sum of all possible 2n = p + p' up to 30 is greater than the sum of the consecutive numbers (2n > 2) up to 30.
Semi-primes table
The table returns all numbers n ≠ 4n - 6.
The quantity q_d of non-repetitive semi-primes p ? p′ (in light blue) for each n^2 is equal to the quantity q of p + p = 2n (in pink).
Quantities aligned horizontally and then vertically, from n to 0 to m
Diagonal (1, 39) contains all odd numbers and therefore repeats some p ? p′ (in purple).
The diagonal (A) orders consecutively for each p all its infinite odd semi-primes.
The edges and intersections of the diagonals (A) of each p identify all the odd semi-primes:
the edges identify the semi-primes p ? p = p^2
the intersections identify all the semi-primes p ? p′ ≠ p^2
With (n > 0) = m + 1 and (p, p) always different, the diagonal (p,p) contains all contains all the prime numbers and all odd numbers
n^2 ? m^2 = y = 2n ? 1
With (n > 1) = m + 2 the diagonal (4, 4n ? 4) contains all even numbers greater than 2, except the numbers 4n ? 6
n^2 ? m^2 = z = 4n ? 4
With (n > 2) = m + 3 the diagonal (9, 6n ? 9) contains
n^2 ? m^2 = 6n ? 9 and therefore also all odd semi-primes
n^2 ? m^2 = x = 6n ? 9 = 3p
Since the set of all even numbers is equivalent to the set of all numbers (4n - 4) and (4n - 6) with n > 1, Euler's conjecture or Goldbach's strong conjecture,
"every even number greater than 2 can be written as the sum of two prime numbers",
is confirmed by the following statements:
?? (n >1)?? ?? (p, p′, m)? |? n = √(p ? p′+ m^2)
?? n? ?? z? |? z = 4n - 4
?? n? ?? x? |? x = 6n - 9
?? x? ?? y |? y = x/3
?z = 4n - 4 = p + p′
x - y = 4n - 6 = p + p′
Total quantity q of pairs p + p' existing in the interval [4, 2p_i]
Total quantity q_d of odd semi-primes p ? p′ existing in the interval [9, (p_i)^2]
Quantity q_p of odd semi-primes p ? p′ generated by (p_i) >2 in the interval [p_i, (p_i)^2]
With n ≥ 5, the sum ∑ of all possible pairs of equal or different primes p + p′ = 2n existing in the interval [4, 2n] is always greater than the sum of only the even numbers existing in the same interval.
Table (p ? p′) & (p + p′)
There is a semi-prime (p ? p′) for every pair (p + p′)
100% stacked line chart with table indicators (p ? p′) and (p + p′)
Scatterplot with straight table lines (p ? p′) and (p + p′)
Twin prime numbers
Cousin prime numbers
Sexy prime numbers
Other pairs
Summations that return p^2
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