25099Re: Diophantine equation
- May 25, 2013
> 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:
> 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)
> 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.
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