Reverse process sum of odd numbers from x to y. What is the fastest method?
 

If I know the sum S of odd numbers from x to y

[(y+1)/2]^2-[(x-1)/2]^2=S

What is the computationally fastest way to know x and y by excluding x = y?

Example

[(y+1)/2]^2-[(x-1)/2]^2=249

thank you

[QUOTE=Alberico Lepore;466517]If I know the sum S of odd numbers from x to y

[(y+1)/2]^2-[(x-1)/2]^2=S

What is the computationally fastest way to know x and y by excluding x = y?

Example

[(y+1)/2]^2-[(x-1)/2]^2=249

thank you[/QUOTE]

depends, do you assume only one set of values works ? 9 for example, is the difference of 16 and 25, it's also the difference between 0 and 9 . 25, is the difference of 0 and 25, and it's also the difference between 144 and 169.

Please see the attached screenshot from Wolfram Alpha.

[QUOTE=Alberico Lepore;466517]If I know the sum S of odd numbers from x to y

[(y+1)/2]^2-[(x-1)/2]^2=S

What is the computationally fastest way to know x and y by excluding x = y?

Example

[(y+1)/2]^2-[(x-1)/2]^2=249

thank you[/QUOTE]

Let a=(y+1)/2 and b=(x-1)/2.
Then you have a^2 - b^2 = S, or, apparently, (a-b)(a+b)=S.

So: you factor your S and for each divisor pair, you can determine a and b.

For example for S=249, you have
a-b=1, a+b=249
or
a-b=3, a+b=83.

[b]Batalov[/b] has already given the essentials: x and y are determined from integer factorizations of S. I mention a few nit-picking details.

Formulaically,

[(y+1)/2]^2-[(x-1)/2]^2

is invariant under the substitutions

y <-- -y - 2 and

x <-- -x + 2.

The former does not fit the original "sum of odd numbers from x to y" description in which we are apparently assuming x and y are odd, 0 < x < y.

However, the x-substitution does work; it merely prepends canceling negative and positive terms to the original sum when x > 1.

If x = y, the substitution gives the pair (-y + 2, y); if y > 1, then x = -y + 2 is not equal to y.

So if S > 1 is odd, the formula (-S + 2, S) is probably the computationally fastest solution with x <> y, consistent with the "sum of consecutive odd numbers" description. If S is an odd prime, this is the [i]only[/i] such solution.

For even values of S, the problem can only have solutions when S is divisible by 4. In this case, the solutions are again as [b]Batalov[/b] described, except one uses factorizations of S in which both factors are even.

[QUOTE=Dr Sardonicus;466578][b]Batalov[/b] has already given the essentials: x and y are determined from integer factorizations of S.[/QUOTE]Did no-one else spot the likelihood that the OP was playing around with factorization algorithms?

The example, showing a difference of squares (the basis of many factoring algorithms since Fermat) was the giveaway for me.

Thanks for the answers.

Adding

620677=p^2+6*(y+1)*p

629641=[p+6*(y-x+2)/2]^2+6*(x-1)*[p+6*(y-x+2)/2]

There are other ways ?

For fairness I tell you what this is for

[url]https://www.academia.edu/34339494/Fattorizzazione_RSA_di_Lepore_Complessit%C3%A0_random[/url]

I found a solution
But I do not have the means to test it
Wolfram does not allow me to do that
Kindly you could test

(6*(103446-6*a^2-2*a)/(6*a+1)+((6*a+1)^2-(6*(a-1)+1)^2)/6)=(6*(104940-6*b^2-2*b)/(6*b+1)+((6*b+1)^2-(6*(b-1)+1)^2)/6)
,
a+(y-x+2)/2+2*sqrt[(9*a^4+6*a^3-319301*a^2-106434*a+2675268480)/(6*a+1)^2]=b
,
620677=(6*a+1)^2+6*(y+1)*(6*a+1)
,
629641=(6*b+1)^2-6*(x-1)*(6*b+1)

thank you

If you solve the problem (sum of odd numbers of odd number items) you solve the problem of factorization and therefore of primality
look at this

I write in Italian as I can not explain it in English

Considerazioni di Lepore su Primalità e Fattorizzazione

Definizione 1

Ogni numero dispari, non primo, non divisibile per due e per tre si scrive come somma di numeri dispari consecutivi positivi.

Dimostrazione

Sia N un numero non divisibile per due e per tre allora
N=p*q=q*(p-1+1)=(2*p-2)*q/2+q=(y+x)*(y-x)/4+(y+x)/2=((y+1)/2)^2-((x-1)/2)^2


Definizione 2

Ogni numero fattorizzabile N=p*q si scrive come somma di numeri dispari consecutivi positivi da p+q-1 a (q-p)+1

Dimostrazione

N=((y+1)/2)^2-((x-1)/2)^2=((p+q-1+1)/2)^2-(((q-p)+1 -1)/2)^2=p*q


Definizione 3

Ogni numero primo P non si scrive come somma di numeri dispari consecutivi positivi.

Dimostrazione

Dalla definizione 2 e dalla definizione di numero primo si ha che p=1 e q=P quindi si ha p+q-1=P e (q-p)+1=P quindi si scrive soltanto come se stesso.



Alberico Lepore 4 Settembre 2017

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[QUOTE=Batalov;467079][COLOR=LemonChiffon].[/COLOR][/QUOTE]

You can calculate the distribution of the prime numbers and consequently the primality of a number