How to prove that coloring each point with $1$ of $n$ colors for specific $n$ creates a isosceles-right triangle For which natural numbers $n>1$ can you always find $3$ points of the same colour forming an isosceles-right triangle in any colouring of $\mathbb Z^2$ with $n$ colors.
This problem was given to me and as I have no idea where to start, I'd appreciate some help
 A: Not sure if it's overkill but Gallai's extension of van der Waerden's theorem says that for any finite coloring of $\mathbb{Z}^2$, there exists a color which has arbitrarily large square sub-arrays. It follows that for any finite coloring of $\mathbb{Z}^2$, you can always find a mono-colored isoceles right-triangle. You can find an elementary proof in the case of 2 and 3 colors here. You can find stronger results for triangles in $n\times n$ grids in Monochromatic Equilateral Right Triangles on
the Integer Grid - Graham, et. al.. 
A: We shall first give the argument to show a monochromatic right-angled isosceles triangle (MCRIT) exist for two colours & then show that the argument generalises for any finite number of colours.
Consider the first row, there must exist $3$ points that are equally spaced; this is equivalent to finding a monochromatic arithmetic progression in $[n]$. (We leave it to the reader to convince themselves that a two colouring of $[7]$ will contain such a sequence.)

INSERT FIGURE
Consider the figure above; Points $1$ & $2$ are the same colour so point $4$ will need to be the other colour in order to avoid a MCRIT. Similarly, points $2$ & $3$ mean point $5$ is the other colour & points $1$ & $3$ mean point $6$ is the other colour; but now points $4$,$5$ & $6$ will form a MCRIT & thus the result is shown for two colours.
To show the result for $c$ colours we need a corollary of van der Waerden's Theorem, see theorem's $(2.1)$ & $(2.10)$ here http://u.cs.biu.ac.il/~tsaban/RT/Book/Chapter5.pdf
Let $N(n,c)$ denote the minimal length of a $c$-coloured sequence for a monochromatic subsequence of length $n$ to exist.
In the first $N(N( \cdots N(N(3,2),3) \cdots,c-1),c)$ elements of the first row there exists a sequence a monochromatic sequence of equally spaced points of length $N( N( \cdots N(N(3,2),3) \cdots,c-2,c-1)$. Now consider the "triangle" of points above these, they must avoid the colour of the first colour in order to avoid a MCRIT. Repeating this argument inductively will lead us to the case of two colours in $N(3,2)=7$.
