# 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:
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

```