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Old 2020-09-09, 15:59   #1
Alberico Lepore
 
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Default "New" same approach that isn't factorization

a * b = N

if N mod 4 = 1

then 2 * N + 2 * a ^ 2 + ((b-a) / 2) ^ 2 = ((3 * a + b) / 2) ^ 2

now i found that in some cases (i don't know which ones)

this is also true

2 * N + 2 * 1 ^ 2 + y ^ 2 - ((3 * a + b) / 2) ^ 2 = 0

So for example for the case N = 65

you would

2 * 65 + 2 * a ^ 2 + ((b-a) / 2) ^ 2 = ((3 * a + b) / 2) ^ 2
,
a * b = 65
,
130 + 2 * 1 ^ 2 + y ^ 2 - ((3 * a + b) / 2) ^ 2 = 0

Could you help me:

1) When is this true?

2 * N + 2 * 1 ^ 2 + y ^ 2 - ((3 * a + b) / 2) ^ 2 = 0

2) How would you fix the system?
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Old 2020-09-09, 19:59   #2
xilman
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Quote:
Originally Posted by Alberico Lepore View Post
2) How would you fix the system?
You should fix it by doing far more work before spewing out your posts.
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Old 2020-09-09, 20:13   #3
Alberico Lepore
 
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Quote:
Originally Posted by Alberico Lepore View Post
1) When is this true?

2 * N + 2 * 1 ^ 2 + y ^ 2 - ((3 * a + b) / 2) ^ 2 = 0
Quote:
Originally Posted by xilman View Post
You should fix it by doing far more work before spewing out your posts.
When is this true?
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Old 2020-09-09, 20:17   #4
VBCurtis
 
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Quote:
Originally Posted by Alberico Lepore View Post
When is this true?
Only on the 4th Friday of each month. By extension, that's the only day you should post.
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Old 2020-09-09, 20:32   #5
chalsall
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Originally Posted by VBCurtis View Post
Only on the 4th Friday of each month. By extension, that's the only day you should post.
I would recommend only on the 5th Friday of a month...

Just wondering why this troll is being allowed to continue to post noise? Even after he pledged to stop doing so...
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Old 2020-09-09, 21:13   #6
Uncwilly
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Originally Posted by chalsall View Post
Just wondering why this troll is being allowed to continue to post noise? Even after he pledged to stop doing so...
Because only Thor can lift that ban hammer and has not chosen to do so.
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Old 2020-09-10, 00:47   #7
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Quote:
Originally Posted by Alberico Lepore View Post
a * b = N

if N mod 4 = 1
\(ab \equiv 1 \pmod4,\) so either \(a \equiv b \equiv 1 \pmod4\) or \(a \equiv b \equiv 3 \pmod4\).

Quote:
Originally Posted by Alberico Lepore View Post
then 2 * N + 2 * a ^ 2 + ((b-a) / 2) ^ 2 = ((3 * a + b) / 2) ^ 2
\[2ab + 2a^2 + (b^2 - 2ab + a^2)/4 = (9a^2 + 6ab + b^2)/4\]

Yep, this checks out, both as an identity and with the relevant quantities as multiples of 4.

Quote:
Originally Posted by Alberico Lepore View Post
now i found that in some cases (i don't know which ones)

this is also true

2 * N + 2 * 1 ^ 2 + y ^ 2 - ((3 * a + b) / 2) ^ 2 = 0
Are you asking for which y this equality holds?
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Old 2020-09-10, 07:19   #8
Alberico Lepore
 
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Quote:
Originally Posted by CRGreathouse View Post



Are you asking for which y this equality holds?
What characteristic must N have for that equality to be true

for example for N = 121

this

2 * 121 + 2 * 1 ^ 2 + y ^ 2- (22) ^ 2 = 0

it's not true

that is, y is not integer

Last fiddled with by Alberico Lepore on 2020-09-10 at 07:22
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Old 2020-09-10, 07:25   #9
retina
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Quote:
Originally Posted by Alberico Lepore View Post
... N = 65
What? Back to this uselessness again!

I thought you had promoted yourself to 18 digit numbers. What happened to that?
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Old 2020-09-10, 08:32   #10
Alberico Lepore
 
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Quote:
Originally Posted by retina View Post
What? Back to this uselessness again!

I thought you had promoted yourself to 18 digit numbers. What happened to that?
I have not abandoned CRGreathouse number

390644893234047643=4*K+3

390644893234047643*3=1171934679702142929=4*H+1


Now I need to understand what k values this system returns integer values

a*b=1171934679702142929*(4*k+1)
,
2*1171934679702142929*(4*k+1)+2*a^2+((b-a)/2)^2-z^2=0
,
2*1171934679702142929*(4*k+1)+2*1^2+y^2-z^2=0

Last fiddled with by Alberico Lepore on 2020-09-10 at 08:34
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Old 2020-09-10, 08:37   #11
retina
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Quote:
Originally Posted by Alberico Lepore View Post
I have not abandoned CRGreathouse number
Good. So why all this useless N=65 stuff?
Quote:
Originally Posted by Alberico Lepore View Post
Now I need to understand what k values this system returns integer values
Well go on then, have at it.
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