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Statement

Setting as in Theorem 1: $q$ is an odd prime power and $Q(A,B)$ is the quadrance $(x_2-x_1)^2+(y_2-y_1)^2$ of points of $\mathbb F_q^2$.

Lemma 1 (p. 2). Let $a\in\mathbb F_q$ be such that $a^2+1$ is not a square in $\mathbb F_q$. Then for any two distinct points $A\ne B$ on the line $y=ax+i$ (for any $i\in\mathbb F_q$), $Q(A,B)\ne1$.

Equivalently, every line of such a slope $a$ is an independent set of the unit-quadrance graph $D_q$.

Source. Le Anh Vinh, On chromatic number of unit-quadrance graphs (finite Euclidean graphs), arXiv:math/0510092v1 (2005), Lemma 1 and its proof on p. 2; the edition read is identified on the source card.

Read depth. Proof verified: the one-line proof below was checked here. Nothing here is independently reviewed.

Proof pointer

Page 2. For $A=(x_1,ax_1+i)$ and $B=(x_2,ax_2+i)$ with $x_1\ne x_2$, the quadrance is $(a^2+1)(x_1-x_2)^2$, a non-square times a nonzero square, hence a non-square and in particular not $1$.

Dependencies

None.

Bears on

• Problem 188: the paper does not treat Problem 188. The lemma gives, in the finite-field analogue of the plane, whole lines with no unit pair, so such a line could be colored red in that analogue without a red unit pair. In the real plane every line contains unit pairs, so the lemma has no real counterpart and gives no bound on the problem's $K_*$.