I do a bit of origami in my spare time and i need a way to divide a basic square into three via folds is there any way i can do this?
Sure; fold it like you would a letter. Or am I missing something here?
“Basic Square”? Does this have a specific meaning in origami that we’d need to know to better answer this question? For now, I was thinking the same thing as the previous posted reply. - Jinx
You can also fold the square across diagonally to make a 90-45-45 triangle and then again to make another, smaller 90-45-45.
What you can’t do is evenly divide it into 3 smaller squares.
I’m not sure if this is exactly one third, but at least it’s close to a third.
Fold the paper in half horizontally, then fold one of those halves diagonally. Then fold the entire square in half diagonally so that the two diagonal folds intersect. The intersection is 1/3 from the edge.
Now, would someone like to prove/disprove this?
Huh, Gregory?
The way I have followed your instructions the intersection is at a quarter of the length.
I do not believe you will find an answer here BigDaiv. I asked my Maths teacher once if it were still possible to make a new discovery in Mathematics. He gave me this problem: Cut a piece of paper into 3 equal lengths without measuring each length. It is possible to do 2 lengths because the paper can be folded in half and cut, but 3 he says has yet to be proven.
p.s. If anyone finds the answer, tell me and I’ll keep it safe 
One would think that dividing a piece of paper into equal thirds is easy. Either fold it like a standard mailed letter, or into ‘Z’ sections. Adjust it until the folded sections are equal, then cut.
Unless I’m missing something?
Have you ever tried to fold a piece of paper letter style and have all segments be equal. It’s near impossible without experience and even then it would not be quite right. If you think about it the lengths of each segment will be out by the thickness of the paper. When you fold over the first piece the second piece is folded over that extra piece of paper. In effect it is folded twice to raise itself above this first fold.
Orient the square so that one edge is toward you. Take the bottom right corner, and bring it up so that it meets the middle of the top edge of the square. Flatten and fold. The intersection of the (former) bottom edge and left edge of the square is now exactly one third up the left edge. (This is known as Haga’s theorem).
I’m not sure what the set of allowable origami operations is, but here’s another solution which generalizes to other 1/n folds:
Fold the square in half and then in half again parallel to one edge, so that you have a 1/4 x 1 strip. Unfold into a square, then fold again along the main diagonal (bringing A to C). Unfold again. Call the intersection of the main diagonal with one of the 1/4 folds point P. Make a fold along the line AP (this is the part which I’m not sure is allowed by origami rules). This intersects AB at a point Q, 1/3 of the way from B to C. (Proof: AR:AB::RP:BQ. AR=3/4, AB=1, RP=1/4, so BQ=1/3.)
Warning–bad ASCII art follows:
B---------Q---------------------C
| . : |
| . : |
| . : |
R.......P.......................|
| : . |
| : . |
| : . |
|....:..........................|
| : . |
| : . |
| : . |
|..:............................|
| : . |
| : . |
|: . |
A-------------------------------D
Note that if you initially folded the square not into fourths but into a 1/n x 1 strip, then the same construction gives AR=(n-1)/n and RP=1/n, so BQ=1/(n-1). This allows you to fold by n-1 whenever you can fold by n; since you can clearly make any fold by 2[sup]n[/sup] this lets you make any desired division, in theory.