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• Hi all, I would like to work out the conjecture that says that any sufficiently large and non square integer number is the addition of a prime number and a
Message 1 of 64 , Apr 2, 2001
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Hi all,

I would like to work out the conjecture that says that any sufficiently large and non square integer number is the addition of a prime number and a square number: n = p + a^2 with n integer, p prime and a integer.
I have checked with a short program that it is true for n>21679. For n<=21679, there is 16 exceptions for non square even numbers : 2,10,34,58,130,214,226,370,526,706,730,1414,1906,2986,3676,and 9634. And 20 exceptions for non square odd numbers: 5,13,31,37,61,85,91,127,379,439,571,771,829,991,1255,1351,1549,3319,7549, and 21679.
Obviously, any square number k^2 with (2*k-1) composite do not meet the rule since n=p+a^2 can be written as:
k^2=p+a^2 and p=k^2-a^2=(k+a)*(k-a) which cannot be prime. But, the conjecture is true for the case (k-a)=1 that leads to p=2*k-1 prime.

I'm not aware if this question has already been solved.

Lille, France

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• ... 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
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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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