- Hello all:

Let c an even positive integer number fixed.

Consider the quadratic equation in two variables:

-3x^2+y^2-2xy-4cx+4cy+4=0

Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1) of the equation with x and y both positive integers? i.e. (y>x>1 integers)

Sincerely

Sebastián Martín Ruiz

P.S. (It is related to the prime numbers. I'll tell you soon.)

[Non-text portions of this message have been removed] > Hello all:

--Let z=y-x. Then your equation is equivalent to

>

> Let c an even positive integer number fixed.

>

>

> Consider the quadratic equation in two variables:

>

>

> -3x^2+y^2-2xy-4cx+4cy+4=0

>

> Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1) of the equation with x and y both positive integers? i.e. (y>x>1 integers)

>

>

> Sincerely

>

> Sebastián Martín Ruiz

4x^2 - z^2 - 4cz = 4

with x>1 and z>0 integer unknowns.

Now make the further change of variable t=z+2c.

Then the equation is

(2x)^2 - t^2 = 4+4c^2

with x>0 and t>2c integer unknowns,

which forces t to be even. Letting t=2r

we finally transform the equation to the form

x^2 - r^2 = 1+c^2

in integer unknowns x>1 and r>c.

This in turn may be written

(x-r)*(x+r) = 1+c^2.

Does a solution of this always exist if c is large enough?

If c is even, then x=r+1 where 2r=c^2 will work.

If c is odd, then there is no solution because

right hand side is 2 mod 4 and since x and r must have same parity,

the left hand side is divisible by 4.

So, in conclusion, your conjecture is DISPROVEN.

Assuming I did not make any stupid errors.

--Warren D. Smith.> Hello all:

--Let z=y-x. Then your equation is equivalent to

>

> Let c an even positive integer number fixed.

>

>

> Consider the quadratic equation in two variables:

>

>

> -3x^2+y^2-2xy-4cx+4cy+4=0

>

> Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1) of the equation with x and y both positive integers? i.e. (y>x>1 integers)

>

>

> Sincerely

>

> Sebastián Martín Ruiz

4x^2 - z^2 - 4cz = 4

with x>1 and z>0 integer unknowns.

Now make the further change of variable t=z+2c.

Then the equation is

(2x)^2 - t^2 = 4+4c^2

with x>0 and t>2c integer unknowns,

which forces t to be even. Letting t=2r

we finally transform the equation to the form

x^2 - r^2 = 1+c^2

in integer unknowns x>1 and r>c.

This in turn may be written

(x-r)*(x+r) = 1+c^2.

Does a solution of this always exist if c is large enough?

If c is even, then x=r+1 where 2r=c^2 will work.

If c is odd, then there is no solution because

right hand side is 2 mod 4 and since x and r must have same parity,

the left hand side is divisible by 4.

So, in conclusion, your conjecture is DISPROVEN.

Assuming I did not make any stupid errors.

--Warren D. Smith.- you made a sign error (+-) in(*****)

________________________________

De: WarrenS <warren.wds@...>

Para: primenumbers@yahoogroups.com

Enviado: Domingo 26 de Mayo de 2013 6:15

Asunto: [PrimeNumbers] Re: Diophantine equation

> Hello all:

--Let z=y-x. Then your equation is equivalent to

>

> Let c an even positive integer number fixed.

>

>

> Consider the quadratic equation in two variables:

>

>

> -3x^2+y^2-2xy-4cx+4cy+4=0

>

> Can anyone prove that for every c large enough there is at least one other than the trivial solution (1,1) of the equation with x and y both positive integers? i.e. (y>x>1 integers)

>

>

> Sincerely

>

> Sebastián Martín Ruiz

4x^2 - z^2 - 4cz = 4

with x>1 and z>0 integer unknowns.

Now make the further change of variable t=z+2c.

Then the equation is

(2x)^2 - t^2 = 4+4c^2 (*******)

with x>0 and t>2c integer unknowns,

which forces t to be even. Letting t=2r

we finally transform the equation to the form

x^2 - r^2 = 1+c^2

in integer unknowns x>1 and r>c.

This in turn may be written

(x-r)*(x+r) = 1+c^2.

Does a solution of this always exist if c is large enough?

If c is even, then x=r+1 where 2r=c^2 will work.

If c is odd, then there is no solution because

right hand side is 2 mod 4 and since x and r must have same parity,

the left hand side is divisible by 4.

So, in conclusion, your conjecture is DISPROVEN.

Assuming I did not make any stupid errors.

--Warren D. Smith.

[Non-text portions of this message have been removed] - On 5/26/13, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
> you made a sign error (+-) in(*****)

--you are correct... trying again below. Hopefully no stupid errors this time, but

I again am not checking carefully.

>> Hello all:

--Let z=y-x. Then your equation is equivalent to

>>

>> Let c an even positive integer number fixed.

>>

>>

>> Consider the quadratic equation in two variables:

>>

>>

>> -3x^2+y^2-2xy-4cx+4cy+4=0

>>

>> Can anyone prove that for every c large enough there is at least one other

>> than the trivial solution (1,1) of the equation with x and y both

>> positive integers? i.e. (y>x>1 integers)

>>

>>

>> Sincerely

>>

>> Sebastián Martín Ruiz

>

4x^2 - z^2 - 4cz = 4

with x>1 and z>0 integer unknowns.

Now make the further change of variable t=z+2c.

Then the equation is

(2x)^2 - t^2 = 4-4c^2 (****have corrected sign error here***)

with x>0 and t>2c integer unknowns,

which forces t to be even. Letting t=2r

we finally transform the equation to the form

x^2 - r^2 = 1-c^2

in integer unknowns x>1 and r>c.

This in turn may be written

(x-r)*(x+r) = 1-c^2.

or equivalently

(r-x)*(r+x) = c^2 - 1 = (c-1)*(c+1)

Does a solution of this always exist if c is large enough?

If c is even, then r=x+1 where 2r+2=c^2 will work.

If c is odd (then my old erroneous disproof with the sign error

fails to work) then since r>c and x>1 are demanded, the

obvious solution r=c, x=1 is not eligible.

But since the right hand side is divisible by 4 since c-1 and c+1 both are even,

r=x+2, 4x+4=(c-1)*(c+1) which implies x=(c^2-5)/4 works.

So, now your conjecture is PROVEN -- solutions always exist for c large enough.

--

Warren D. Smith

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