Quick version: given natural $n$ and a row of $n$ integers such that the sum of the squares is another square, call it $m^2.$ For $n=5,6,7$ is it always possible to fill in the rest of an $n$ by $n$ matrix of integers, call it $M,$ so that $M M^T = m^2 I? $ If so, $M/m$ is rational orthogonal.


Notes: this is true for $n=1,2,3,4,8.$ 1 is trivial 2 uses complex numbers, 4 uses quaternions, 8 uses octonions. 3 uses quaternion stuff applied to ternary quadratic forms, papers of Jones and Pall mostly, the main one 1939. The naive adaptation of the Jones-Pall formalism to our $n=7$ does not work very well, see [Realizing proper pure octonions as conjugates](https://mathoverflow.net/questions/117212/octonions-and-the-dance-of-the-seven-veils)


This is false for $n = 9,17,25,33,\ldots.$ Indeed, take any odd $n = k^2,$ let the first row have all entries $1,$ no second row is possible that is orthogonal to the given first row, consists of integers, and has the same length. Problem mod 2, insofar as the dot product of the two rows is odd, therefore nonzero. Actually, for any $n >1, \; \; \; n \equiv 1 \pmod 8,$ one may specify any $n-3$ odd numbers, then find the final three (also odd) by Gauss three square theorem to get an odd square sum, no luck.


Anyway, I did some computer checks, entirely successful for small entries for $n=5,6,7,$ and instinct tells me that it only gets easier with larger entries.


So, that is the short version, does this work for any first row of integral length (sum of squares is another square) in dimension $n=5,6,7?$



Filling in a rational orthogonal matrix given one row
Yes, it is possible to fill in. Your problem is a particular case of a completion problem and is treated in the following paper: 


Hsia, J.S.
Two theorems on integral matrices. Linear Multilinear Algebra 5, 257-264 (1978). 


