1. Adjust your compass so it's circle fits on the paper, and so it almost fill the paper.

l2. Then mark the center point where your circle will be with a pencil (or better yet, poke a hole with the compass).

3. Draw the circle.

4. Fold the circle any way that the line goes through the center point. With all folds: accuracy counts.

5. Put the center of the compass where the fold meets the circle. Mark the two intersection points of the compass with the first circle. You don't need to draw a full circle, just the two intercepts.

6. Now do step 5 again, but on the other intercept of the fold and the circle.

7. Fold the paper two more times through the center, connecting the center point with the circumference intercept marks.

8. Crease the folds with your finger nails and tear.

9. Take each of the 6 pieces you should now have, and fold, crease with fingernails, and tear from one side of the circles remaining arc to the other.

10. Now you should have 6 symmetrical triangles. On each bend, connecting two corners. Gently pinch near the center. Do this all three ways. Where the creases cross is the center. Poke a hole there with the compass.

11. We're going to complete one fibget at a time from here. Write "G" "o" "d" with the G being a spiral and the D being the opposite spiral. The "o" is a circle! On one triangle put the letters clockwise and on a second do it counter-clockwise. Point the top of the letters towards the points.

12. Take the clockwise one, point G up and turn over with G still pointing up. On the top point draw an arrow pointing counter-clockwise. On the bottom right: an equals sign ( = ). Under the equals sign write %3. This means every third time you spin it, it should complete the operations on that side first. On the bottom left: a plus sign ( + )

13. Take the counter clockwise one. Point G up. Turn over, keeping G up. On the top is an equals sign ( = ). On the bottom right put an arrow pointing clockwise. And on the bottom left, a plus sign ( + )

14. Put the point = with the %3 under it facing up underneath and between the "o" and "d" of the other triangle. Put one of the pipe cleaners through the middles of the triangles where you poked holes.

15. Squeeze the triangles together without twisting and put them in the middle of the pipe cleaner by putting the ends together to find the midpoint.

16. Pinch the pipe cleaner and evenly coil the pipe cleaner in a circle around itself. Do this on both sides, one after the other. Make sure to coil it clock-wise so it will have a G on side facing up.

17. Repeat steps 11 through 16, two more times, making two more fibgets, for a total of three. They have seven symbols on each side, with the pipecleaner being the seventh! 777 777

18. On all three write on the face where the equals isn't over "%3" the characters 木,+=,1. Under the spin arrow goes the 木 (or an alpha(A) under and in an omega(Ω)), under plus goes +=, and under equals goes 1. According to Yuri Matiyasevich: In order for v to be the 2uth fibonacci number (v=F_2u), it is necessary and sufficient that there exist numbers g, h, l, m, x, y, z such that:
u<=z<l
l^2-lz-z^2=1
g^2-gh-h^2=1
l^2|g
l|m-2
(2h+g)|(m-3)
x^2-mxy+y^2=1
l|(x-u)
(2h+g)|(x-v)

*** p|d means d is evenly divisible by p

Together with the the Binet formula (ϕ^u-ϕ^-u)/5^.5=F_u=v' , where ϕ is the golden ratio) equations with variables as exponents can be represented as variables without exponents! This shows something like: there is no single diophantine equation to solve all diophantine equations, so we can't show definitively 0=/=1! The Fibjet calculates (G|O|D)=F_2木! And in so doing shows that Fibgets may be God! This is as anything is possible with God and fibjets prove anything possible!

Below is a diagram of how to label the points of a fibje. The triangles under are the backsides of the triangles above!