Show the number n= `sqrt(7-4sqrt3)+sqrt(7+4sqrt3)`

is natural ?

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You should rewrite the numbers under the radical symbol as a perfect square, such that:

`7 - 4sqrt 3 = (4 + 3) - 4sqrt 3`

Comparing `(4 + 3) - 4sqrt 3` to the following special product `(a - b)^2 = a^2 - 2ab + b^2` yields:

`(4 + 3) - 4sqrt 3 = 2^2 - 2*2*sqrt 3 + (sqrt 3)^2`

`(4 + 3) - 4sqrt 3 = (2 - sqrt 3)^2`

Reasoning by analogy, yields:

`7 + 4sqrt 3 = (2 + sqrt 3)^2`

Taking the square roots yields:

`sqrt (7 - 4sqrt 3) = sqrt((2 - sqrt 3)^2) = |2 - sqrt 3| = 2-sqrt 3`

`sqrt (7 + 4sqrt 3) = sqrt((2 + sqrt 3)^2) = |2 + sqrt 3| = 2+sqrt3`

Replacing `2-sqrt 3` for `sqrt (7 - 4sqrt 3)` and `2+sqrt3` for `sqrt (7 + 4sqrt 3)` yields:

`sqrt (7 - 4sqrt 3) + sqrt (7 + 4sqrt 3) = 2-sqrt 3+2+sqrt 3`

Reducing duplicate members yields:

`sqrt (7 - 4sqrt 3) + sqrt (7 + 4sqrt 3) = 4 in N`

**Hence, performing the indicated operations and using the special product `(a +- b)^2` yields that `sqrt (7 - 4sqrt 3) + sqrt (7 + 4sqrt 3) = 4` , thus`sqrt (7 - 4sqrt 3) + sqrt (7 + 4sqrt 3)` is natural number.**

Given `n=sqrt(7-4sqrt(3))+sqrt(7+4sqrt(3))` :

Square both sides (note that this might introduce extraneous solutions):

`n^2=7-4sqrt(3)+2sqrt((7-4sqrt(3))(7+4sqrt(3)))+7+4sqrt(3)`

`=>n^2=14+2sqrt(49-48)`

`=>n^2=16`

`=>n=+-4`

However the sum of two positives is positive so -4 is an extraneous solution.

So `n=sqrt(7-4sqrt(3))+sqrt(7+4sqrt(3))=4` is a natural number.

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