[seqfan] Re: quintic diophantine q^2=2*t^5-1

Andrew N W Hone A.N.W.Hone at kent.ac.uk
Tue Aug 20 12:36:12 CEST 2019


This equation defines a curve of genus two in the (q,t) plane, so by Faltings' theorem there are only 
finitely many rational points on the curve, so in particular only finitely many integer solutions. However, 
Faltings' result doesn't give an effective upper bound on the number of solutions, so other methods 
are necessary. 

The article by Crescenzo mentioned below apparently only considers the case where q,t are both 
required to be prime, in which case there are no solutions.

All the best,
Andy


________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Walter Kehowski [wkehowski at cox.net]
Sent: 20 August 2019 10:04
To: Sequence Fanatics Discussion list; Richard J. Mathar
Subject: [seqfan] Re: quintic diophantine q^2=2*t^5-1

Peter Crescenzo, A Diophantine Equation Which Arises in the Theory of Groups, Advances in Mathematics 17, 25-29 (1975).

> On August 20, 2019 at 1:51 AM "Richard J. Mathar" <mathar at mpia-hd.mpg.de mailto:mathar at mpia-hd.mpg.de > wrote:
>
>
>     In conjunction with A161460 (pairs of tau-values that repeat
>     in the list of the number-of-divisors function) one may ask
>     whether 50175 is in the sequence, and one of the subtasks
>     seems to be to find pairs of (q,t) such that q^2=2*t^5-1.
>     Apparently this inhomogeneous quintic diophantine equation has only the
>     solution q=t=1. Is there any proof to that?
>
>     Searching for odd q and t with a substitution q=2*q'+1, t=2*t'+1
>     does not lead far,
>     q'*(q'+1)= t'*(16*t'^5 + 40 t'^4+ 40 t'^3 + 20 t'^2 + 5'*t')
>     because the usual unique factorization methods are not applicable:
>     q' and t' are not prime.
>
>     For comparison: the quadratic diophantine Pell equation q^2=2*t^2-1
>     has q=1,7,41,239,1393,8119..
>     t=1,5,29,169,985,..
>     as in A002315 and A001653.
>
>     The cubic q^2=2*t^3-1 seems to have only the "trivial"
>     solution q=t=1.
>
>     The quartic q^2=2*t^4-1 has (obviously) a subset of the
>     quadratic solutions if we sieve for perfect squares in A001653.
>     This yields at least two solutions at
>     q=1,239,..
>     t^2=1,169,..
>
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>

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