The Square-Sum problem

[LaurV]

Originally Posted by R. Gerbicz

There is a solution for every n>=25 !

Very nice! Man, you are good!

[henryzz]

Very nice. Maybe someone should let numberphile know.
Also I believe papers have been published on less worthwhile subjects.

[MattcAnderson]

I posted to the numberphile youtube page that this problem has been solved.

Regards,
Matt

[science_man_88]

Originally Posted by MattcAnderson

I posted to the numberphile youtube page that this problem has been solved.

Regards,
Matt

I posted a link to this thread on the video page near the start of it as well.

[R. Gerbicz]

Originally Posted by henryzz

Also I believe papers have been published on less worthwhile subjects.

I've thought the same in the recent days, maybe I'll come up with a paper.

[Auto Felix]

How do you know that all sequences start with 1, and end with 8 (for even) or 3 (for odd)? Is there a proof?

[R. Gerbicz]

Originally Posted by Auto Felix

How do you know that all sequences start with 1, and end with 8 (for even) or 3 (for odd)? Is there a proof?

Thanks for your interest!
I"ve constructed all basic sequences that holds this, and then by induction we still maintain this. Note that all integers in [1,24] is free, because T(c,0) and T(c,1) doesn't use them. So we can use these small (1,3,8) integers at the endpoints.

You could ask why we haven't used a shifted and reversed representation, so the sequences starts with 1,3 and 1,8; because in that case we don't know the last term of seq0 and seq1, so we can't glue the sequences. Or why we haven't used the constant 3 at the end for every sequence, because in that case 3=seq0(n)=seq1(n+1), but (n+1-n)=1 is odd so the parity position condition wouldn't be true. There are some traps here.