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 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.

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