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 January 12th, 2018, 12:08 AM #1 Member   Joined: Jan 2015 From: italy Posts: 39 Thanks: 1 Primality Test and Factorization with Pythagorean triples and quadratic diophantine A Pythagorean triple (A,b,c) with a minor cateto A odd always admits a solution (A,b,b+1) where 2*b+1=A^2 For example, let N be a semiprimo N=p*q then N/1, N/N, N/p, N/q will be our four paths Suppose we follow the N/q road then (N/q+1)/2 or (N/q-1)/2 will be odd then [(N/q+1)/2]^2+b^2=(b+1)^2 will admit integer solutions (quadratic diophantine parabolic case) or [(N/q-1)/2]^2++b^2=(b+1)^2 will admit integer solutions (quadratic diophantine parabolic case) to keep in mind that the roads that can be traveled as mentioned above are four N/1, N/N, N/p, N/q so there will be four right paths reiterating we will arrive at b=0 and find the solution q example N=67586227 solve ((((((((((((((67586227/q+1)/2-1)/2-1)/2-1)/2-1)/2)-1)/2+1)/2+1)/2-1)/2+1)/2-1)/2-1)/2)^2+b^2=(b+1)^2 , b=0 the quadratic diophantine parabolic case are obtained thus solve ((67586227/q+1)/2)^2+b^2=(b+1)^2 , b b=(4567898080095529 + 135172454*q - 3*q^2)/(8*q^2) this number 4567898080095529=N^2 this number 135172454/N= integer =2 as N=p*q then 8*b=(N^2)/(q^2)+(2*N)/(q)-3 then 8*b=p^2+2*p-3 if it admits solutions we are on a right path What do you think of this?
January 12th, 2018, 10:10 AM   #2
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Joined: Jan 2015
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Quote:
 Originally Posted by gerva then N/1, N/N, N/p, N/q will be our four paths
thanks to Lutz Donnerhacke
I make this change

N = 187
for N / 1 (-1, + 1, -1, -1, -1, + 1, -1)
for -N / 1 (+ 1, -1, + 1, + 1, + 1, -1, + 1)
for N / N (+ 1, + 1, + 1, + 1, + 1, + 1, + 1)
for -N / N (-1, -1, -1, -1, -1, -1, -1)

As soon as a path is different from this one must follow only that

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