On 10/20/2012 10:07 AM,

primenumbers@yahoogroups.com wrote:

> 2.2. Re: Question

> Posted by: "djbroadhurst"d.broadhurst@... djbroadhurst

> Date: Fri Oct 19, 2012 9:02 am ((PDT))

>

>

>

> --- Inprimenumbers@yahoogroups.com,

> Kermit Rose<kermit@...> wrote:

>

>> >Methods to solve Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

>> >

>> >for what values of x would

>> >x^2 + 5 x + 6 be a perfect square.

> Set A = 1, B = 0, C = -1, D = 5, E = 0, F = 6.

>

> and you will get the obvious answer from Dario:

>

> x = -2

> y = 0

> and also:

> x = -3

> y = 0

> Calculation time: 0h 0m 0s

>

> David

Thank you David.

I had been confused by consideration of the following:

if x y = z and y > x,

then if we set t = (y+x)/2 and s = (y-x)/2

then z = t^2 - s^2

and x = y - 2 t.

So z = x y = (y - 2t) y = y^2 - 2 t y

y^2 - 2 t y - z = 0

Transforming w = y - k yields

(w + k)^2 - 2 t (w + k) - z = 0

w^2 + (2 k - 2 t) w + (k^2 - 2 t k - z) = 0

which has integral solution if and only if

(2k - 2 t)^2 + (z + 2 t k - k^2) is an integral square

In all that derivation, I forgot that I still did not know what value t was.

I confused myself too easily.

Kermit

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