a^4+b^3=c^2
in [Math]
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From: Vincenzo Librandi on 30 Jul 2007 10:28 a,b,c, integers, with gcd(a,b,c)=1 to solve a^4+b^3=c^2 Regards, Vincenzo Librandi
From: hagman on 30 Jul 2007 16:26 On 30 Jul., 20:28, Vincenzo Librandi <vincenzo.librandw...(a)alice.it> wrote: > a,b,c, integers, with gcd(a,b,c)=1 > > to solve > > a^4+b^3=c^2 > > Regards, > Vincenzo Librandi E.g. a=1, b=2, c=3. Hint: Investigate b*(b-1) = 2*a*a first.
From: Vincenzo Librandi on 30 Jul 2007 23:06 On 30 Jul., 20:28, Vincenzo Librandi <vincenzo.librandw...(a)alice.it> wrote: > a,b,c, integers, with gcd(a,b,c)=1 > > to solve > > a^4+b^3=c^2 > > Regards, > Vincenzo Librandi >>hagman wrote: >>E.g. a=1, b=2, c=3. >> >>Hint: Investigate b*(b-1) = 2*a*a first. The solutions are finite or infinite ? Regards, Vincenzo Librandi
From: Vincenzo Librandi on 31 Jul 2007 11:54 On 30 Jul., 20:28, Vincenzo Librandi <vincenzo.librandw...(a)alice.it> wrote: > a,b,c, integers, with gcd(a,b,c)=1 > > to solve > > a^4+b^3=c^2 > > Regards, > Vincenzo Librandi >>hagman wrote: >>E.g. a=1, b=2, c=3. >> >>Hint: Investigate b*(b-1) = 2*a*a first. >>>Vincenzo Librandi wrote: >>> >>>The solutions are finite or infinite ? The solutions are infinite. Regards, Vincenzo Librandi
From: Vincenzo Librandi on 1 Aug 2007 22:28 >Vincenzo Librandi wrote: > >a,b,c, integers, with gcd(a,b,c)=1 > >to solve > >a^4+b^3=c^2 b^3=c^2-a^4 b^3=(c+a^2)(c-a^2) integers m,n, such that m^3=c+a^2; n^3=c-a^2 c=(m^3+n^3)/2 a^2=(m^3-n^3)/2 let m=k+1 and n=k-1 a^2=3k^2+1 Pell equation a^2-3k^2=1 by k=0,1,4,15,56,209,780,... 26^4+224^3=3420^2 with k=4: m=5; n=3 which the solution 7^4+15^3=76^2 with k=56, m=57, n=55 97^4+3135^3=175784^2 and so on Regards, Vincenzo Librandi
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