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 March 12th, 2017, 09:42 AM #1 Senior Member   Joined: Dec 2013 Posts: 1,101 Thanks: 40 Prove or disprove Prove that for any positive integer k > 2 it always exists at least one number n>0 such as : n^2=a*(a+(2*k)) Example : k=3 a=2 n=4 4^2=2*(2*(2*3))=2*8=16 k=4 a=1 n=3 3^2=1*(1*(2*4))=1*9=9
 March 12th, 2017, 10:49 AM #2 Senior Member   Joined: Dec 2013 Posts: 1,101 Thanks: 40 I Edit to correct my error : Example : k=3 a=2 n=4 4^2=2*(2+(2*3))=2*8=16 k=4 a=1 n=3 3^2=1*(1+(2*4))=1*9=9 Thanks from agentredlum
 March 12th, 2017, 11:34 AM #3 Senior Member   Joined: Dec 2013 Posts: 1,101 Thanks: 40 The equation holds. Someone (not me) proved it. What are the consequences of such equality on factorization of odd semi-primes? Go ahead! There is something very important to derive from this equality. Good luck! I see you in 2018.
 March 12th, 2017, 02:50 PM #4 Senior Member   Joined: Dec 2013 Posts: 1,101 Thanks: 40 Sadly the proof is partially wrong. Sorry for having been in hurry to tell you that it was proven. Last edited by skipjack; March 12th, 2017 at 08:40 PM.
 March 12th, 2017, 05:50 PM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,263 Thanks: 198 $a^2 + 2ak = n^2$ $k$ is given therefore complete the square on $a$ $(a + k)^2 = n^2 + k^2$ $a$ doesn't matter , we just need to show there is a Pythagorean Triple with one leg any integer $k> 2$ because the RHS of the equation represents two legs of a right angled triangle. Web link excerpt follows To prove any number, X, can be the side of some Pythagorean triangle, we use the m,n generating method on two cases: X even and X odd. if X is even: let n=1 and m=X/2 then the side generated by 2 m n is X If X is odd: then we can halve it and let m = (X+1)/2 and n = (X–1)/2 so that the side generated by m2 – n2 = (m – n)(m + n) = 1×X is X End of web link excerpt I found the above proof at the web link below Pythagorean Triangles and Triples Thanks from mobel
 March 13th, 2017, 04:43 AM #6 Senior Member   Joined: Dec 2013 Posts: 1,101 Thanks: 40 Right!
March 15th, 2017, 09:37 AM   #7
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 Originally Posted by agentredlum $a^2 + 2ak = n^2$ $k$ is given therefore complete the square on $a$ $(a + k)^2 = n^2 + k^2$ $a$ doesn't matter , we just need to show there is a Pythagorean Triple with one leg any integer $k> 2$ because the RHS of the equation represents two legs of a right angled triangle. Web link excerpt follows To prove any number, X, can be the side of some Pythagorean triangle, we use the m,n generating method on two cases: X even and X odd. if X is even: let n=1 and m=X/2 then the side generated by 2 m n is X If X is odd: then we can halve it and let m = (X+1)/2 and n = (X–1)/2 so that the side generated by m2 – n2 = (m – n)(m + n) = 1×X is X End of web link excerpt I found the above proof at the web link below Pythagorean Triangles and Triples
It's well known from the old past 300 a.c. approx. I forgot the name of the 2 old greek mathematician, one discover the first the other 100 year later discover the second.

Thanks god mobel is back !

March 16th, 2017, 08:20 AM   #8
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 Originally Posted by mobel The equation holds. Someone (not me) proved it. What are the consequences of such equality on factorization of odd semi-primes? Go ahead! There is something very important to derive from this equality. Good luck! I see you in 2018.
No one got it! No one will anyway.
I gave you a hint. Be aware that there is lot or work to do before building a new algorithm of factorization.

So I see you in 2018.

 March 16th, 2017, 11:04 PM #9 Senior Member   Joined: Dec 2012 Posts: 951 Thanks: 23 Mobel you've to follow Larry Cornell on Academia.edu

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