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	Bill Allombert on Sat, 27 Jan 2024 13:42:03 +0100
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        Re: Any chance to compute system of Diophantine exquations in 26 variables in GP?
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- To: pari-users@pari.math.u-bordeaux.fr
 
- Subject: Re: Any chance to compute system of Diophantine exquations in 26 variables in GP?
 
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
 
- Date: Fri, 26 Jan 2024 20:28:48 +0100
 
- Delivery-date: Sat, 27 Jan 2024 13:42:03 +0100
 
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- In-reply-to: <ZbOI46rQIx62D8xd@seventeen>
 
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- References: <9f454ef82d96ebd7401532329ff7bef9@stamm-wilbrandt.de> <ZbOI46rQIx62D8xd@seventeen>
 
On Fri, Jan 26, 2024 at 11:26:43AM +0100, Bill Allombert wrote:
> On Fri, Jan 26, 2024 at 09:53:47AM +0100, hermann@stamm-wilbrandt.de wrote:
> > I used this system of Diophantine equations:
> > https://en.wikipedia.org/wiki/Formula_for_primes#Formula_based_on_a_system_of_Diophantine_equations
> I assume this is this one:
> <https://maa.org/sites/default/files/pdf/upload_library/22/Ford/JonesSatoWadaWiens.pdf>
> 
> If you follow carefully the proof, you should be able to write a program to solve
> it for any small k ! See middle of page 455. 
The issue is that the smallest solution is doubly exponential in k^4,
so you will probably not be able to compute it.
(for us k=10)
we solve 
(3) (2*k)^3*(2*k+2) * (n+1)^2 + 1 = f^2
? my(Q=quadunit((2*k)^3*(2*k+2)*4)); n=imag(Q)-1; f=real(Q);
? n
%41 = 343772642385433639988435123780
? f
%42 = 144220715637070429940775452568001
(at least!)
but then
p=(n+1)^k
q=(p+1)^n
is going to be too large.
Cheers,
Bill.