## Diophantine equation

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• 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
Message 1 of 5 , May 24, 2013
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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]
• ... --Let z=y-x. Then your equation is equivalent to 4x^2 - z^2 - 4cz = 4 with x 1 and z 0 integer unknowns. Now make the further change of variable t=z+2c.
Message 2 of 5 , May 25, 2013
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> 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

--Let z=y-x. Then your equation is equivalent to
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.
• ... --Let z=y-x. Then your equation is equivalent to 4x^2 - z^2 - 4cz = 4 with x 1 and z 0 integer unknowns. Now make the further change of variable t=z+2c.
Message 3 of 5 , May 25, 2013
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> 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

--Let z=y-x. Then your equation is equivalent to
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.
Message 4 of 5 , May 26, 2013
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you made a sign error (+-) in(*****)

________________________________
De: WarrenS <warren.wds@...>
Enviado: Domingo 26 de Mayo de 2013 6:15

> 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

--Let z=y-x. Then your equation is equivalent to
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]
• ... --you are correct... trying again below. Hopefully no stupid errors this time, but I again am not checking carefully. ... --Let z=y-x. Then your
Message 5 of 5 , May 26, 2013
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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 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
>
--Let z=y-x. Then your equation is equivalent to
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