March 12th, 2017, 09:42 AM  #1 
Senior Member Joined: Dec 2013 Posts: 1,039 Thanks: 29  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: 1,039 Thanks: 29 
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 
March 12th, 2017, 11:34 AM  #3 
Senior Member Joined: Dec 2013 Posts: 1,039 Thanks: 29 
The equation holds. Someone (not me) proved it. What are the consequences of such equality on factorization of odd semiprimes? 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: 1,039 Thanks: 29 
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: 3,166 Thanks: 173 
$a^2 + 2ak = n^2$ $k$ 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 
March 13th, 2017, 04:43 AM  #6 
Senior Member Joined: Dec 2013 Posts: 1,039 Thanks: 29 
Right!

March 15th, 2017, 09:37 AM  #7  
Senior Member Joined: Dec 2012 Posts: 917 Thanks: 23  Quote:
Thanks god mobel is back !  
March 16th, 2017, 08:20 AM  #8  
Senior Member Joined: Dec 2013 Posts: 1,039 Thanks: 29  Quote:
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: 917 Thanks: 23 
Mobel you've to follow Larry Cornell on Academia.edu


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