Infinity, thou art a heartless bitch - Page 2

Raman · 2013-02-20, 15:11

And then,

I am ready to offer a million dollar prize
to anyone who would be able to find out

a non-zero (positive) integer that

can be simultaneously written into the form a²+6b²
AND into the form 2c²+3d²

LaurV · 2013-02-20, 16:23

[QUOTE=Raman;330191]And then,

I am ready to offer a million dollar prize
to anyone who would be able to find out

a non-zero (positive) integer that

can be simultaneously written into the form a²+6b²
AND into the form 2c²+3d²[/QUOTE]
[TEX]15=\sqrt 3^2+6*\sqrt 2^2=2*\sqrt 6^2+3*1^2[/TEX]

Where's the million?

Raman · 2013-02-20, 16:37

[QUOTE=LaurV;330196]
[TEX]15=\sqrt 3^2+6*\sqrt 2^2=2*\sqrt 6^2+3*1^2[/TEX]

Where's the million?[/QUOTE]

Sorry, that a, b, c, d
all of them needs to be integers!

Thus

[COLOR=White]A number can be written into the form a²+6b² if and only if it has got
- no prime factors congruent to {13, 17, 19, 23} mod 24 raised to an odd power
- sum of exponents of {2, 3, 5, 11} mod 24 prime factors for the number is being [B]even[/B].

A number can be written into the form 2a²+3b² if and only if it has got[/COLOR] [COLOR=White]
- no prime factors congruent to {13, 17, 19, 23} mod 24 raised to an odd power
- sum of exponents of {2, 3, 5, 11} mod 24 prime factors for the number is being [B]odd[/B].[/COLOR]

Why so?

Raman · 2013-02-20, 16:51

[QUOTE=Raman;330198]
Sorry, that a, b, c, d
all of them needs to be integers!
[/QUOTE]

As since,

if in case that this condition is not being imposed,
then I could even be able to solve for the equation
Fermat's Last Theorem,

into the same way...

As such

xⁿ + yⁿ = zⁿ, n ≥ 3

Whether or not
to be being

science_man_88 · 2013-02-20, 20:53

[QUOTE=Raman;330199]As since,

if in case that this condition is not being imposed,
then I could even be able to solve for the equation
Fermat's Last Theorem,

into the same way...

As such

xⁿ + yⁿ = zⁿ, n ≥ 3

Whether or not
to be being[/QUOTE]

if a^2+6*b^2 = 2*c^2+3*d^2

?+even= even+? so to be odd both a and d are odd, if both sides are considered even both a and d are even this cuts the search in half.

2013-02-20, 16:37	#14
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts	[QUOTE=LaurV;330196] [TEX]15=\sqrt 3^2+6\sqrt 2^2=2\sqrt 6^2+31^2[/TEX] Where's the million?[/QUOTE] Sorry, that a, b, c, d all of them needs to be integers! Thus [COLOR=White]A number can be written into the form a²+6b² if and only if it has got - no prime factors congruent to {13, 17, 19, 23} mod 24 raised to an odd power - sum of exponents of {2, 3, 5, 11} mod 24 prime factors for the number is being [B]even[/B]. A number can be written into the form 2a²+3b² if and only if it has got[/COLOR] [COLOR=White] - no prime factors congruent to {13, 17, 19, 23} mod 24 raised to an odd power - sum of exponents of {2, 3, 5, 11} mod 24 prime factors for the number is being [B]odd[/B].[/COLOR] Why so? Last fiddled with by Raman on 2013-02-20 at 16:46*

2013-02-20, 16:51	#15
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts	[QUOTE=Raman;330198] Sorry, that a, b, c, d all of them needs to be integers! [/QUOTE] As since, if in case that this condition is not being imposed, then I could even be able to solve for the equation Fermat's Last Theorem, into the same way... As such xⁿ + yⁿ = zⁿ, n ≥ 3 Whether or not to be being Last fiddled with by Raman on 2013-02-20 at 16:52

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2013-02-20, 16:23	#13
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 41×251 Posts	[QUOTE=Raman;330191]And then, I am ready to offer a million dollar prize to anyone who would be able to find out a non-zero (positive) integer that can be simultaneously written into the form a²+6b² AND into the form 2c²+3d²[/QUOTE] [TEX]15=\sqrt 3^2+6\sqrt 2^2=2\sqrt 6^2+3*1^2[/TEX] Where's the million?

2013-02-20, 20:53	#16
science_man_88 "Forget I exist" Jul 2009 Dartmouth NS 20415₈ Posts	[QUOTE=Raman;330199]As since, if in case that this condition is not being imposed, then I could even be able to solve for the equation Fermat's Last Theorem, into the same way... As such xⁿ + yⁿ = zⁿ, n ≥ 3 Whether or not to be being[/QUOTE] if a^2+6b^2 = 2c^2+3*d^2 ?+even= even+? so to be odd both a and d are odd, if both sides are considered even both a and d are even this cuts the search in half.