Pythagorean Theorem intuition I actually have a degree in Pure Mathematics, and this has always bugged me. Like a tiny, hardly-noticeable stone in your shoe you just can't get out.
Despite being at the core of geometry, arithmetic, and infinity, despite being able to prove it a dozen different ways, and despite it being the first "formula" anyone learns and one of the oldest known mathematical ideas - I can only see shadows and silhouettes of its truth.
Looked at from the point of view of geometry, I can see the statement plainly as a statement regarding the intrinsic notions of angle and area (using the proof of similar right-angle triangles) - however, the last step requires a correspondence between the notions of length and area to make complete (namely, the area of a shape increases quadratically as its scale increases linearly). This has always bothered me, since, the Theorem is an idea solely between angle and length.
There are, of course, simple proofs that don't depend on the notion of area (ex: Pythagorean Theorem Proof Without Words 6), however, sadly, they don't further my intuition, despite being clever/cute.
Staring at the statement blank in the face, I understand it far less intuitively then I do half of the ideas in algebra, analysis, number theory, etc.
 A: I wonder if some way of looking at Einstein's proof of the Pythagorean theorem might help.
First recall that the areas of figures of the same shape are proportional to the squares of the distances involved. For example, if you have two regular pentagons, and the length of the side of one of them is $6$ times the length of the side of the other, then its area is $36$ times the area of the other.
Now look at the right triangle. Suppose the angles other than the right angle are $\theta^\circ$ and $(90-\theta)^\circ.$ Drop a perpendicular from the vertex of the right angle to the hypotenuse, splitting the triangle into two parts. The area of the original right triangle is the sum of the areas of those two parts. One of those two parts shares the angle of $\theta^\circ$ with the original triangle and has a right angle at the foot of the perpendicular that you dropped. It thus has two angles in common with the original triangle; therefore all three in common; hence it has the same shape as the  original triangle, and its hypotenuse is one of the legs of the original triangle. Similarly the other part has the same shape as the original triangle and its hypotenuse is the other one of the legs of the original triangle. You've split the triangle into two triangles each having the same shape as the original triangle. And areas are proportional to the squares of distances.
A: The Pythagorean Theorem does seem to follow nicely from the similarity of right triangles formed within a right triangle by dropping a perpendicular from the right angle to the hypotenuse, and the fact that similar polygons are as the squares on corresponding sides (@Michael Hardy). These are Props. 8 and 20 in Book VI of Euclid's Elements. However he proves the PT in I,47 relying mainly on congruency of triangles (I,4) and the fact that a parallelogram is double a triangle on the same base and under the same height (I,41). The latter rests on I,37 (triangles on same base under same height are equal), which Newton said was the first thing in Euclid not obvious to him. Euclid I,47 makes no appeal to area increasing as the square of length, since his theory of proportion doesn't come until Book V, followed by its geometric application in VI. The "proof without words" (@Zach) seems to rest on Euclid VI,8, which does rest on words. Do you find even Euclid I,47 less than satisfying as proof of the Pythagorean Theorem?
