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Old 2016-07-11, 17:07   #1
Godzilla
 
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Default Formubla-bla-bla to calculate the sum of two Prime numbers just by knowing the product

Italiano:

Ho trovato una formula per calcolare la Somma di due numeri Primi conoscendo solamente il loro Prodotto , ovviamente dato il Risultato N c'è un margine di errore di confidenza nel senso nel senso che la formula produce due numeri N1 N2 e in questo intervallo che è appunto di confidenza c'è S ovvero la somma dei due numeri Primi , l'ho fatto esaminare , e ho avuto una risposta positiva nel senso che , it work. Qualcuno di Voi conosceva già qualcosa di simile oppure no ? perché a tutti quelli che ho chiesto non risulta , ho anche auto-pubblicato un libro .

Google translate:

I found a formula to calculate the sum of two Prime numbers only knowing their product obviously given the result N is a margin of error in the sense of confidence in the sense that the formula produces two numbers N1 and N2 in this range that is S is precisely the confidence which is the sum of the two Prime numbers , I did look, and I've had a positive response in the sense that it work. Some of you already knew something like this or not? because everyone I've asked is not, I also self-published a book.
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Old 2016-07-11, 17:23   #2
CRGreathouse
 
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Great, would you demonstrate this with N =
Code:
25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357
?
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Old 2016-07-11, 17:51   #3
Godzilla
 
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Great , wonderful



Allora :

la formula è :

sqrt(2178*(P1*P2)) / 22 = N1 = (P1+P2) = (N1 Limite minimo --->P1 + P2<-----N1 limite massimo)
809 * 811 = 656099 sqrt(2178*(656099)) / 22 = 1718

la formula
da ( 1718 = limite massimo )e la somma
( 1620 = P1 + P2 )
allora prendendo il numero 656099 - 99999 = 556100 (
prendo cinque 9 perché 656099 è formato da sei cifre ,
quindi i 9 aumentano in proporzione al numero )


mettiamo nella stessa formula 556100 quindi sqrt(2178
* 556100 ) / 22 e otteniamo ( 1581 = limite minimo
adesso 1718 - 1581 = 137 quindi sappiamo che c'è un
margine di 137 numeri su cui c'è la somma esatta
poi una cosa
molto strana è che se io prendo il numero

più grande 1718 e prendo dal 137 solo il numero 37
(quindi non conto solo la prima cifra di qualsiasi
numero che ottengo ) e lo sottraggo a 1718 quindi sarà
1718 - 37 = ( 1681 = limite massimo ) e restringo
ancora di più il campo.

Autore
Alessandro Boatto

Google translate:

then:

the formula is:

sqrt (2178 * (P1 * P2)) / 22 = N1 = (P1 + P2) = (N1 Minimum limit ---> P1 + P2 <----- N1 Maximum limit)

809 * 811 =656099 = sqrt (2178 * (656099)) / 22 = 1718
the formula
from (1718 = Maximum limit) and the sum
 (1620 = P1 + P2)
allora taking the number 656099-99999 = 556100 (
prendo five 9s because 656099 is made up of six digits,
so the increase in proportion to the number 9)


we put in the same formula 556100 so sqrt (2178
 * 556100) / 22 and we get (1581 = minimum limit
adesso 1718 - 1581 = 137 so we know that there is a
margine of 137 numbers on which there is the exact sum
poi something
very strange is that if I take the number

largest from 137 in 1718 and take only the number 37
 (so do not account only the first digit of any
I get that number) and then subtract it to 1718 will
1718-37 = (1681 = upper limit) and restrict myself
even more the field.
Author
Alessandro Boatto

Last fiddled with by Godzilla on 2016-07-11 at 18:48
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Old 2016-07-11, 18:47   #4
CRGreathouse
 
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It seems that you're giving a range for the sum (or the individual prime factors?) rather than the sum. But it could still be interesting.

Would you make it more explicit, though? Are you saying that the larger of the prime factors of a semiprime N is at most sqrt (2178 * N) / 22?
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Old 2016-07-11, 18:57   #5
Godzilla
 
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Quote:
Originally Posted by CRGreathouse View Post
It seems that you're giving a range for the sum (or the individual prime factors?) rather than the sum. But it could still be interesting.

Would you make it more explicit, though? Are you saying that the larger of the prime factors of a semiprime N is at most sqrt (2178 * N) / 22?

exactly one interval of the sum between 1718 becomes 1681 and 1581, then in 1681 -1581 = 100 before it was 137; then in 2178 and 22 are special numbers are not chosen at random
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Old 2016-07-11, 19:53   #6
CRGreathouse
 
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I'm trying to understand how to apply this to an arbitrary semiprime. Could you explain? I'm still not sure how to find p + q when p*q =
Code:
243024012172767921272070572273889784413128637316016546174345576716037252140832962072921774230631752471236112161067930718597652764457070824619552435605715745864114863281389313085139156559416423279705566979749854667657283120412462687328553074798354718362816682384351829113366441102477473982521947260421
, say.
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Old 2016-07-11, 20:25   #7
a1call
 
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I would prefer a much smaller numeric example.
How about 15 as the product.
How would you go about arriving at 8 as the sum?
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Old 2016-07-11, 20:37   #8
Godzilla
 
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prendiamo i numeri 1681 e 1581 , adesso noi sappiamo che la somma corretta è 1620 quindi qualsiasi somma è un numero pari , adesso basta partiamo da 1680 + N (è sempre un numero dispari , qundi 1681 + 3 = 1684 , ma 1684 > 1681 quindi no , poi 1679 + 3 = 1682 e 1682 > 1681 quindi no , poi 1677 + 3 = 1680 ma 1677 non è un numero Primo quindi procediamo con il successivo 1677 + 5 = 1682 e 1682 >1681 , poi 1673 + 3 = 1676 ma 1673 non è un numero primo , avanti così…. perché solo esclusivamente 809 + 811 = 1620 , basta fare un programma con il pc. E deve essere compreso tra 1681 e 1581 .


Google translate

we take the 1681 and 1581 numbers, now we know that the correct sum is 1620 so any sum is an even number, that's enough we start from 1680 + N (always an odd number, qundi 1681 + 3 = 1684 but 1684> 1681 so no, then 1679 + 3 = 1682 and 1682 > 1681 then no, then 1677 + 3 = 1680, but 1677 is not a prime number, then we proceed with the next 1677 + 5 = 1682 but 1682 > 1681, then 1673 + 3 = 1676, but 1673 is not a prime number, so far ahead .... because only only 809 + 811 = 1620, just do a program with the PC. It must be between 1681 and 1581.

Last fiddled with by Godzilla on 2016-07-11 at 20:52
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Old 2016-07-11, 20:53   #9
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Quote:
Originally Posted by a1call View Post
I would prefer a much smaller numeric example.
How about 15 as the product.
How would you go about arriving at 8 as the sum?
I'm trying to avoid an example where the number is readily factored. Finding the sum of the prime factors in such a case would demonstrate that the method is powerful.
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Old 2016-07-11, 21:00   #10
a1call
 
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Quote:
Originally Posted by CRGreathouse View Post
I'm trying to avoid an example where the number is readily factored. Finding the sum of the prime factors in such a case would demonstrate that the method is powerful.
In light of the language barrier, I would still prefer a simple numeric example. Nothing prevents him from giving bigger numeric examples afterwards.
Perhaps because I have little confidence in understanding larger numeric examples.
Once the method is understood if correct it could be scaled up indefinitely.

If the concept is valid/legit then it is valid for smaller examples as well as the larger ones.

Last fiddled with by a1call on 2016-07-11 at 21:02
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Old 2016-07-11, 21:16   #11
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if you got more than 2178=99*22 out of this you're more knowledgeable than me.
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