I decided to draw the lines y(x) = x² + b * x  on the integer number spiral, because it seems like the most basic form of a quadratic equation, with c=0.

So that then means drawing the equations y(x) = x²,  y(x) = x² + x, y(x) = x² + 2x, y(x) = x² + 3x... etc. 

I also decided that I would start at the value x = 2, because I can't say for sure that the result for x = 1 is a compostite (as per the Quadratic Expanation pages).

Below are the first 12 at 200 x 200 pixels, to try and keep page size reasonable: 

y(x) = x2	y(x) = x2 + x	y(x) = x2 + 2x
y(x) = x2 + 3x	y(x) = x2 + 4x	y(x) = x2 + 5x
y(x) = x2 + 6x	y(x) = x2 + 7x	y(x) = x2 + 8x
y(x) = x2 + 9x	y(x) = x2 + 10x	y(x) = x2 + 11x

 
Notice how the lines start by spiralling and then straighten out after. This is why it was necassary to go through the process I went through the find the 'base quadratic' because I was picking out the series of integers that start at the point when the line is straight.

Lets see what happens when we layer these on top of one another, I have zoomed in so we are looking at 40 x 40 integers here.

y(x) = x2	After layering up to y(x) = x2 + x	After layering up to y(x) = x2 + 2x
After layering up to y(x) = x2 + 3x	After layering up to y(x) = x2 + 4x	After layering up to y(x) = x2 + 10x
After layering up to y(x) = x2 + 20x	After layering up to y(x) = x2 + 50x	After layering up to y(x) = x2 + 100x
After layering up to y(x) = x2 + 200x	After layering up to y(x) = x2 + 500x	After layering up to y(x) = x2 + 795x

 
The last one, after layering 795 interations of  y(x) = x² + b * x, is an Ulam spiral with 1 considered as prime.... and the centre (number 0) considered prime.

I could, perhaps should have, started the Ulam sprial at 1... which is generally the case for an Ulam spiral and the final result at y(x) = x² + 795x would be exactly the same as an Ulam spiral with 1 considered prime.

So really, a prime can be considered any integer that does not fall on the integer solutions of y(x) = x² + b * x, for all values of b and x > 1.