Re: Diophantine equation

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• ... --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 1 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