Problem
Max Alekseyev
maxale at gmail.com
Fri Nov 16 20:37:43 CET 2007
On Nov 16, 2007 11:06 AM, Artur <grafix at csl.pl> wrote:
> I'm apologize but but I was do mistake in formula
> Find number m such that 5n^4(4+n^2) is square
> Mathemtica sample Code:
> a={};Do[If[IntegerQ[Sqrt[5n^4 (4+n^2)]],AppendTo[a,n]],{n,1,1364}];a
>
> Out: {1,4,11,29,76,199,521,1364}
If I got you correctly, you're looking for positive integer solutions to
5n^4(4+n^2) = m^2
It implies that n^2 divides m. Let m=n^2*k. Then the equation is equivalent to
5(4+n^2) = k^2
or
k^2 - 5n^2 = 20.
This is a generalized Pell equation and you can plug its coefficients
to Dario Alpern's on-line tool:
http://www.alpertron.com.ar/QUAD.HTM
to get solution in the form of recurrent sequences.
With respect to n, the solution consists of the following 3 sequences,
satisfying the same recurrent relation n(j+1)=18*n(j)-n(j-1) with
different initial conditions:
n(0) = 1, n(1) = 29: { 1, 29, 521, 9349, 167761, 3010349, ... }
n(0) = 4, n(1) = 76: { 4, 76, 1364, 24476, 439204, 7881196, ... }
n(0) = 11, n(1) = 199: { 11, 199, 3571, 64079, 1149851, 20633239, ... }
Your sequence is the union of these three.
Max
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