You can represent ANY positive integer

*n* using only three twos by writing

*n*= -log_{2}(log_{2}(√√...√2))

where log_{2} is the logarithm base 2, and there are *n* square roots.

Can you manage to do the same (that is, represent all positive integers) using three *THREEs* instead? And by using only *TWO* twos?

Using only two 2's:

sec (arctan (sqrt x)) = sqrt (x + 1) is true for all positive values of x, so starting at x=2:

(sec (arctan (sqrt 2)))^2 = 3

(sec (arctan (sec (arctan (sqrt 2)))))^2 = 4

(sec (arctan (sec (arctan (sec (arctan (sqrt 2)))))))^2 = 5

(sec (arctan (sec (arctan (sec (arctan (sec (arctan (sqrt 2)))))))))^2 = 6

Using only two 3's:

The exponent 2 can be written as (sec (arctan (sqrt 3))) = 2, so starting at x=3:

(sec (arctan (sqrt 3)))^(sec (arctan (sqrt 3))) = 4

(sec (arctan (sec (arctan (sqrt 3)))))^(sec (arctan (sqrt 3))) = 5

(sec (arctan (sec (arctan (sec (arctan (sqrt 3)))))))^(sec (arctan (sqrt 3))) = 6

(sec (arctan (sec (arctan (sec (arctan (sec (arctan (sqrt 3)))))))))^(sec (arctan (sqrt 3))) = 7

*Edited on ***December 10, 2003, 11:19 am**