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• Hi to all, A question was asked recently. Are there infinitely many of n, n+1 both square free? A simple bit of algebra came up with the following, BUT it
Message 1 of 64 , Mar 30, 2001
Hi to all,
A question was asked recently. Are there infinitely many of n,
n+1 both square free? A simple bit of algebra came up with the
following, BUT it may have important consequences. First the algebra,

n is not a square => n = X^2 + a a not equal to 0
n+1 is not a square => n+1 = Y^2 + b b not equal to 0

=> X^2 +a +1 = Y^2 + b. This is a 2nd order Diophantine equation
in 2 variables and `obviously' (i.e. please look up the solution )
has an infinite number of solutions.
So does this spell the beginning of the end of functional number
theory type answers to `infinitely many of' type questions, OR was
this just a one-off. Can anyone else think of other `infinitely
many' conjectures which might be solved in this way? The point being
that `infinitely many' is taken care of by the properties of the
equation.

Regards,
Paul Mills
Kenilworth,
England.
• ... 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
Message 64 of 64 , Oct 20, 2012
On 10/20/2012 10:07 AM, primenumbers@yahoogroups.com wrote:
> 2.2. Re: Question
> Date: Fri Oct 19, 2012 9:02 am ((PDT))
>
>
>
> 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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