"New" same approach that isn't factorization

[Alberico Lepore]

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?

[xilman]

Originally Posted by Alberico Lepore

2) How would you fix the system?

You should fix it by doing far more work before spewing out your posts.

[Alberico Lepore]

Originally Posted by Alberico Lepore

1) When is this true?

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

Originally Posted by xilman

You should fix it by doing far more work before spewing out your posts.

When is this true?

[VBCurtis]

Originally Posted by Alberico Lepore

When is this true?

Only on the 4th Friday of each month. By extension, that's the only day you should post.

[chalsall]

Originally Posted by VBCurtis

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...

[Uncwilly]

Originally Posted by chalsall

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.

[CRGreathouse]

Originally Posted by Alberico Lepore

a * b = N

if N mod 4 = 1

\(ab \equiv 1 \pmod4,\) so either \(a \equiv b \equiv 1 \pmod4\) or \(a \equiv b \equiv 3 \pmod4\).

Originally Posted by Alberico Lepore

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.

Originally Posted by Alberico Lepore

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?

[Alberico Lepore]

Originally Posted by CRGreathouse

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

[retina]

Originally Posted by Alberico Lepore

... N = 65

What? Back to this uselessness again!

I thought you had promoted yourself to 18 digit numbers. What happened to that?

[Alberico Lepore]

Originally Posted by retina

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

[retina]

Originally Posted by Alberico Lepore

I have not abandoned CRGreathouse number

Good. So why all this useless N=65 stuff?

Originally Posted by Alberico Lepore

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

Well go on then, have at it.