[seqfan] Re: Minimal k > n such that (4k+3n)(4n+3k) is square

[Kevin Ryde]

charles.greathouse at case.edu (Charles Greathouse) writes:
>
> If there is no n < k < 109n/4 with (4k+3n)(4n+3k) square, then a(n) = 393n.

I tried continuing past 393 where there's further multipliers

    f=393, 76441, 14829361, 2876819793

which is solutions to 12*f^2+25*f+12=square, excluding multiples of
earlier solutions, and which therefore k=f*n gives (4k+3n)(4n+3k)=square
for any n.

It seems when a particular n allows f=109/4=27.25 that there's a single
extra solution between the integer ones.  Eg. n=4

    f = 109/4 = 27.2500
        393
        21949/4 = 5487.2500
        76441
        4258797/4 = 1064699.2500

But when there's more than one solution they're in pairs.  Eg. n=13

    f = 10.1538 = 132/13         \ new pair
        70.4615 = 916/13         /
        393
        2170.4615 = 28216/13        \ new pair
        13870.1538 = 180312/13      /
        76441
        421259.3846 = 5476372/13      \ new pair
        2690939.3846 = 34982212/13    /
        14829361

It's possible to have a mixture of 27.25 and pairs.  Eg. n=24

    f = 157/24 = 6.5417                <----+
        654/24 = 27.2500    <- single       |- pair
        2509/24 = 104.5417             <----+
        393
        35269/24 = 1469.5417
        131694/24 = 5487.2500
        491557/24 = 20481.5417
        76441

Dunno if this is always so or if it has any significance.