Question on Alf van Poorten's proof of the Four Square Theorem

Question

I am reading the proof of the Four Square Theorem by Alf van Poorten. So far I think to get it, but I do not see were the $x-6n, x-2n, x+2n, x+6n$ comes from (bottom of the first page). Why are the squares chosen in this way, should we not consider a general arithmetic progression?

Answer 1

The author says that "I may suppose without loss of generality that the four squares in arithmetic progression all are odd". Now the difference of two odd squares is always divisible by $4$: $$(2a+1)^2-(2b+1)^2=4(a^2+a-b^2-b)$$ and it follows that the common difference is $d=4n$ for some integer $n$. Hence you may consider the arithmetic progression $$y, y+4n, y+8n, y+12n$$ that is, if you let $y=x-6n$ in order to have symmetry, $$x-6n, x-2n, x+2n, x+6n.$$

Answer 2

First, the $4$ squares are chosen this way so the author can multiply them together to get, as stated in the paper, an odd integer $y$ such that

$$y^2 = \left(x^2 - 4n^2\right)\left(x^2 - 36n^2\right) = \left(x^2 - 20n^2\right)^2 - 256n^4 \tag{1}\label{eq1A}$$

The paper then uses this Pythagorean triple, i.e., $y^2 + \left(16n^2\right)^2 = \left(x^2 - 20n^2\right)^2$, in its analysis to finish the proof.

As for not considering "a general arithmetic progression", note as far as having all values be perfect squares, this is actually as general as you can get. This is because all perfect squares are congruent to either $0$ or $1$ , modulo $4$. Thus, the common difference between perfect squares in an arithmetic sequence of $2$ or more elements must be congruent to $-1$, $0$ or $1$ modulo $4$. However, for any sequence of $3$ or more, if the common difference is not $0$, then not all of the values will be congruent to $0$ or $1$ modulo $4$ and, thus, be perfect squares. Thus, for $4$ such values, their common difference must be a multiple of $4$, i.e., $4n$ for some integer $n$. If you have $x$ be the first value among the $4$ plus $6n$, you will thus get those $4$ stated values of $x - 6n$, $x - 2n$, $x + 2n$ and $x + 6n$.