After commenting on a
question
of Joseph O'Rourke's, I thought it interesting that a number theory result (artihmetic progressions of rational squares cannot be arbitrarily long) had applications to geometry (don't look at mostly regular cones of regular hypercubes for totally rational polytopes). (I hope I got the above right and that it is indeed an application; I proceed on that basis.) Of course I am also impressed by the fact that there is no known geometric proof of the fact that finite geometries satisfy both or neither of the configurations of Pappus and of Desargues.
So of course, a natural question would be to consider applications of number theory to geometry; I'm not going to do that here. Instead, I will ask for assistance with Joseph's program by asking a question about rational squares.
The first question that occurred to me was " (1) Is there a sequence of integer squares whose differences are also integral squares?" For sake of interest I require all squares to be nonzero, though later they may be rational and not just integral.
Before posting this question, I saw the answer was yes, and that indeed there were at least countably many such sequences, although I don't know if there are infinitely many tails. So I nominate question first':
(1') How many infinite sequences of integer squares are there, all of whose first differences are also integer squares?
There is the potential to be uncountably many such, especially if there are (is?) an uncountable infinity of tails. But wait! There's more!
(2) Fix an integer $k$ with $k > 1$. How many infinite sequences of integer squares are there, all of whose first through $k$th differences are also integer squares?
Recall that for a sequence $a_i$, the first difference is the sequence $b_i = a_{i+1} – a_i$, and the $(k+1)$st difference is the first difference of the $k$th difference. I suspect that for $k$ large enough, the answer will be zero. However, those are just warm ups for this question:
(3) How many sequences of rational squares are there such that for every positive integer $k$ all $k$th differences are also rational squares?
Motivation: I think it is a cool set of questions. Also I think that if Joseph is going to get a family of rational polytopes of arbitrarily high dimension, he will find such sequences useful (I am thinking volume of a pyramid being base times height times some rational number in combination with a multidimensional Pythagorean-type expression), and that such a family will imply the existence of such sequences, but I do not see the converse as the polytopes have to satisfy additional relations. As usual, reference requests and related problems are welcome.
Gerhard "Ask Me About System Design" Paseman, 2011.08.03
Best Answer
I was in the process of writing up a "proof" (modulo a conjecture on primes) almost identical to Gjergji Zaimi's when his answer arrived. Thank goodness I didn't have to finish, because examining the slight difference in our approaches made me realize how to get around the conjecture.
To simplify things, let $a_0,\ a_1,\ a_2, \ldots$ be a sequence of odd numbers whose squares have square difference, such as
$$5,\ 13,\ 85,\ 157,\ 12325,\ldots$$ or $$5,\ 13,\ 85,\ 3613,\ 6526885,\ldots$$
Note that the sequence starting at 5 splits into two sequences at 85.
Now factor $12325 = 5b$ with $b=2465$ and continue the first sequence as
$$5,\ 13,\ 85,\ 157,\ 5b,\ 13b,\ 85b,\ 157b, \ 12325b,\ldots $$
or
$$5,\ 13,\ 85,\ 157,\ 5b,\ 13b,\ 85b,\ 3613b,\ 6526885b,\ldots $$
Likewise factor $6526885 = 5B$ with $B= 1305377$ and continue the second sequences as
$$5,\ 13,\ 85,\ 3613,\ 5B,\ 13B,\ 85B,\ 157B,\ 12325B,\ldots$$ or
$$5,\ 13,\ 85,\ 3613,\ 5B,\ 13B,\ 85B,\ 3613B,\ 6526885B,\ldots$$
The point is, starting from any multiple of 5, we reach (in two steps) a multiple of a number (it happens to be 85), from which there are two branches, each of which reaches another multiple of 5. So there's guaranteed to be an infinite binary tree, and hence there are uncountably many sequences of square integers whose first differences are square.
I'm sorry if this is somewhat clumsily explained. Maybe someone can clean it up.