- I am interested in knowing proven results regarding the possibility of
generating perfect squares with expressions like
a + bt
1) it is obvious that not always is possible to get squares. E.g.
a = 281941 = 11*19*19*71
b = 510510 = 2*3*5*7*11*13*17
because a + b*t = 11*(25631 + 46410*t) and therefore if the expression
between parentheses does not give an odd power of 11..
2) when a +bt generates squares, t can be written as a number of quadratic
polynomials. How many ? The number depends on
the factorization of b ?
Hope somebody can provide answers
- On 10/20/2012 10:07 AM, firstname.lastname@example.org wrote:
> 2.2. Re: QuestionThank you David.
> 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
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.
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