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March 12th, 2017, 09:42 AM   #1
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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
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March 12th, 2017, 10:49 AM   #2
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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
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March 12th, 2017, 11:34 AM   #3
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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.
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March 12th, 2017, 02:50 PM   #4
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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.
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March 12th, 2017, 05:50 PM   #5
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$a^2 + 2ak = n^2$

$k$ 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
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March 13th, 2017, 04:43 AM   #6
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Right!
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March 15th, 2017, 09:37 AM   #7
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Quote:
Originally Posted by agentredlum View Post
$a^2 + 2ak = n^2$

$k$ 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 !
Thanks from agentredlum
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March 16th, 2017, 08:20 AM   #8
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Quote:
Originally Posted by mobel View Post
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.
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March 16th, 2017, 11:04 PM   #9
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Mobel you've to follow Larry Cornell on Academia.edu
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