So... I came up with kind of a problem.

Is there any two dimensional polygon that has a multiple of four as the number of its sides(let us call the number of sides n) where all sides are equal and all angles are equal that is tileable only with itself where n is bigger then 1*4(i.E. NOT a square.), and if not, where and what is the mathematical proof for this?

-☭

>>1
also I mention that this is probably a well known mathematical question/problem and has been solved/proofen/falsified already, but couldn't find any informations. 8/

>>1
-tileable only with itself +tileable with itself only
i mean, it can of course be tileable with anything else too as long as you can tile it with only itself too

assuming that by "all sides are equal and all angles are equal" you mean a regular polygon:
no regular polygon with >4 sides is tileable with itself only (only the regular triangle and the square are)

as for the proof, I don't have a clue :)

wait, cant you tile a regular triangle with two right triangles

A regular *hexagon* tiles in the honeycomb pattern.

>>4
you ever heard of the hexagon? that doesn't have a multiple of 4 as number of its sides tough.
someone showed me the proof already btw.. It works with the angles, but I forgot the exact wording. X)

The hexagon is the largest regular polygon that tiles with itself.  You don't need a mathmatical proof to see that this is true, just logic.  The more sides a regular polygon has, the wider its angles are.  The wider the angles, the less polygons there will be at an intersection of the tiled pattern (triangles have 6 at each intersection, squares have 4, hexagons have 3).  There is no way to have an intersection of 2 polygons, short of a straight line.

call the number of sides in the polygon 4n, where n=1,2,3..... (i like this notation better)

recall that if a regular polygon can be tiled with itself, then an integer (call it x) multiplied by the angle measurement (which is equal to [180*(4n-2)] / 4n) must equal 360.

hence 180*(4n-2) / 4n = 90*(2n-1)/n has to be a factor of 360. this is the same thing as saying (2n-1)/n is a factor of 4.

for n=1, (2n-1)/n = 1, which is a factor of 4
for n=2, (2n-1)/n = 3/2, which is not a factor of 4.
for n=3, (2n-1)/n = 5/3, which is not a factor of 4.
...
for n=n+1, (2[n+1]-1)/[n+1] = 2n+1/n+1 = 2 - (1/n+1), which is not a factor of 4.

so by induction (i'm sure there's a better way but this is how i think, sorry) (2n-1)/n is not a factor of 4 for 1<n<infinity.

therefore i basically proved what you wanted

Yeah he's right.It leaves you he entire right to blame the short life of alcalin batteries upon the packets wich themselves are problably recyclable.Fight for a free world!=)......