Can somebody fill me in on a visual explanation for the following:
If there exist integers $x, y$ such that $x^2 + y^2 = c$, then there also exist integers $w, z$ such that $w^2 + z^2 = 2c$
I know why it is true (ex. take $w = x-y, z = x+y$), but I would think there is a visual explanation hiding somewhere because of squared terms (we can make squares!!)
On
Hint $\ $ The transformation is simply multiplication by $\, 1+i\,$ in the complex plane
$$(1+i)(x+yi)\, =\, x\!-\!y + (x\!+\!y)i $$
Thus if $\ |x+yi| = x^2+y^2 = c \ $ then
$$ (x\!-\!y)^2+(x\!+\!y)^2 =\, |x\!-\!y+(x\!+\!y)i| = |(1+i)(x+yi)|\, =\, 2\,|x+yi|\,=\,2c$$
Geometrically, multiplication by $\ 1+i = \sqrt{2}\, e^{\large i\pi/4}\,$ may be visualized as an expansion by the factor $\,\sqrt{2}\,$ composed with a rotation by $\,\pi/4.\,$ Let's visualize this in the diagram from Oleg567's answer, excerpted below. Let $\,\rm\color{#c00}{v = x+y\, i}\,$ be the lower red vector. Multiplying it by $\,1+i\,$ yields $\, (1+i){\rm v = v} + i\rm v,\,$ where $\,\rm\color{#c00}{i\,v = -y + x\, i}\,$ is the rotation of $\,\rm v\,$ by $\,\pi/4.\,$ Adding these red vectors yields the result $\,\rm\color{blue}{v + i\, v = x-y + (x+y)\,i}.$
$\qquad\quad$ 
On
If you imagine an integer-sized coordinate grid (as in Oleg567's answer) overlaid with another coordinate grid that has been shrunk by a factor of $\sqrt2$ and rotated by $45^{\circ}$ about the origin, the original unit vectors still have integer coordinates.
So the point $(x,y)$ also has integer coordinates in the new system, being a linear combination of unit vectors. Its distance to the origin is now $\sqrt2$ times the old distance, because we did shrink the unit vectors.
On
In terms of the areas of squares:
Edit: Thanks to Michael T for pointing out a simpler rearrangement.
It will take me forever to post the diagram so here is a description.
Draw the circle with centre $(0,0)$ and radius $\sqrt c\,$. Locate the point $(x,y)$ on this circle: by assumption, $x$ and $y$ are integers. Draw the tangent to this circle starting at $(x,y)$ and having length $\sqrt c\,$. This will give a point distant $\sqrt{2c}$ from the origin (because we have a right angled isosceles triangle), and the point will have integer coordinates because it is obtained from $(x,y)$ by a displacement of $(y,-x)$ or $(-y,x)$, depending which way we drew the tangent.
Update: see another answer for the picture. Thanks Oleg!