What is #sqrt(7 + sqrt(7 - sqrt (7 + sqrt (7 - sqrt(7 + ...... ∞#?

Let

#x=sqrt(7+sqrt(7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+...oo#

where we constrain our solution to be positive since we are taking only the positive square root i.e. #x>=0#. Squaring both sides we have

#x^2=7+sqrt(7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+...oo#

#=>x^2-7=sqrt(7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+...oo#

Where this time we constrain the left hand side to be positive, since we only want the positive square root i.e.

#x^2-7>=0# #=># #x>=sqrt(7) ~= 2.65#

where we have eliminated the possibility the #x<=-sqrt(7)# using our first constraint.

Again squaring both sides we have

#(x^2-7)^2#=#7-sqrt(7+sqrt(7-sqrt(7-sqrt(7+........oo#

#(x^2-7)^2-7=-sqrt(7+sqrt(7-sqrt(7-sqrt(7+........oo#

The expression in the repeated square roots is the original expression for #x#, therefore

#(x^2-7)^2-7=-x#

or

#(x^2-7)^2-7+x=0#

Trial solutions of this equation are #x=-2# and #x=+3# which results in the following factorization

#(x+2)(x-3)(x^2+x-7)=0#

Using the quadratic formula on the third factor #(x^2+x-7)=0# gives us two more roots:

#(-1+-sqrt(29))/2 ~= 2.19 " and " -3.19#

The four roots of the polynomial are therefore #-3.19..., -2, 2.19..., # and #3#. Only one of these values satisfies our constraint #x>=sqrt(7) ~= 2.65#, therefore

#x=3#

I like to discuss a tricky way to have a solution at a glance on the problem of repeated square roots like the following
# sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo#

where # r # belongs to the following series
#3,7,13,21,31............#, the general term of which is given by
#m^2-m+1# where # m epsilon N # and #m>1#

TRICK
If 1 is subtracted from the given Number #m^2-m+1# the resulting number becomes #m^2-m# which is #m(m-1)# and which is nothing but the product of two consecutive number and larger one of these two will be the unique solution of the problem.

when r = #m^2-m+1# the factor of #m^2-m+1-1# = #(m-1)m# and m is the answer

when r = 3 the factor of (3-1)= 2 = 1.2 and 2 is the answer
when r = 7 the factor of (7-1) =6= 2.3 and 3 is the answer
and so on.......

Explanation
Taking
# x=sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo#
Squaring both sides
#x^2= r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo#

#x^2- r=sqrt(r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo#
Again Squaring both sides
#(x^2- r)^2=r-sqrt(r+sqrt(r-sqrt(r+sqrt(r-sqrt(r+........oo#
#(x^2- r)^2-r=-x#
#(x^2- r)^2-r+x=0#
putting r = #m^2-m+1#

#(x^2- (m^2-m+1))^2-(m^2-m+1)+x=0#

if we put x = m in the LHS of this equation the LHS becomes

LHS =
#(m^2- (m^2-m+1))^2-(m^2-m+1)+m#
#=(cancel(m^2)- cancel(m^2)+m-1))^2-(m^2-m+1-m)#

#=(m-1))^2-(m-1)^2=0#
the equation is satisfied.
Hence m is the answer

let's put

#x=sqrt(7+sqrt(7- sqrt(7+sqrt(7-sqrt....#

We can easily see that

#sqrt(7+sqrt(7-x))=x#

So let's solve the equation:

#7+sqrt(7-x)=x^2#

#sqrt(7-x)=x^2-7#

#7-x=(x^2-7)^2=x^4-14x^2+49#

#x^4-14x^2+x +42=0#

This is not a trivial equation to be solved. One of the other persons that answered the question referred the solution 3. If you try it, you will see it's true.