"New" same approach that isn't factorization
a * b = N
if N mod 4 = 1 then 2 * N + 2 * a ^ 2 + ((ba) / 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 + ((ba) / 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? 
[QUOTE=Alberico Lepore;556558]2) How would you fix the system?[/QUOTE]You should fix it by doing far more work before spewing out your posts.

[QUOTE=Alberico Lepore;556558]
1) When is this true? 2 * N + 2 * 1 ^ 2 + y ^ 2  ((3 * a + b) / 2) ^ 2 = 0 [/QUOTE] [QUOTE=xilman;556588]You should fix it by doing far more work before spewing out your posts.[/QUOTE] When is this true? 
[QUOTE=Alberico Lepore;556589]When is this true?[/QUOTE]
Only on the 4th Friday of each month. By extension, that's the only day you should post. 
[QUOTE=VBCurtis;556590]Only on the 4th Friday of each month. By extension, that's the only day you should post.[/QUOTE]
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... 
[QUOTE=chalsall;556592]Just wondering why this troll is being allowed to continue to post noise? Even after he pledged to stop doing so...[/QUOTE]Because only Thor can lift that ban hammer and has not chosen to do so.

[QUOTE=Alberico Lepore;556558]a * b = N
if N mod 4 = 1[/QUOTE] [$]ab \equiv 1 \pmod4,[/$] so either [$]a \equiv b \equiv 1 \pmod4[/$] or [$]a \equiv b \equiv 3 \pmod4[/$]. [QUOTE=Alberico Lepore;556558]then 2 * N + 2 * a ^ 2 + ((ba) / 2) ^ 2 = ((3 * a + b) / 2) ^ 2[/QUOTE] [$$]2ab + 2a^2 + (b^2  2ab + a^2)/4 = (9a^2 + 6ab + b^2)/4[/$$] Yep, this checks out, both as an identity and with the relevant quantities as multiples of 4. [QUOTE=Alberico Lepore;556558]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[/QUOTE] Are you asking for which y this equality holds? 
[QUOTE=CRGreathouse;556596]
Are you asking for which y this equality holds?[/QUOTE] 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 
[QUOTE=Alberico Lepore;556558]... N = 65[/QUOTE]What? Back to this uselessness again!
I thought you had promoted yourself to 18 digit numbers. What happened to that? 
[QUOTE=retina;556615]What? Back to this uselessness again!
I thought you had promoted yourself to 18 digit numbers. What happened to that?[/QUOTE] 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+((ba)/2)^2z^2=0 , 2*1171934679702142929*(4*k+1)+2*1^2+y^2z^2=0 
[QUOTE=Alberico Lepore;556617]I have not abandoned CRGreathouse number[/QUOTE]Good. So why all this useless N=65 stuff?[QUOTE=Alberico Lepore;556617]Now I need to understand what k values this system returns integer values[/QUOTE]Well go on then, have at it.

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