## Growing list of nested radicals

There is an empty list initially. First, 2 is inserted into the list. Then, at every step, every number in the list is replaced with (2 + √n) and (2 − √n) where n is the number being replaced.

What is the product of all numbers in the list when this step has been performed 1000 times?

[SOLVED]

#### Prunthaban Kanthakumar solved this puzzle:

Let P(n) denote the set of numbers generated after n steps. We are interested in finding the product of all numbers in P(1000). Let S(n) denote the set of all square root of numbers in P(n).

We start with P(1000) and shrink it using multiplications (I am mixing union, multiplication and addition in the set theory notation to keep it compact. :-)).

Π P(1000)
= Π [{2 + S(999)} ∪ {2 − S(999)}]
= Π {4 − P(999)}
= Π [{4 − {2 + S(998)}} ∪ {4 − {2 − S(998)}}]
= Π [{2 + S(998)} ∪ {2 − S(998)}]
= …
= Π [{2 + S(1)} ∪ {2 − S(1)}]
= 22 − 2
= 2

#### Saurav Prakash solved this puzzle:

Let the product of all the numbers after n steps performed be P(n).

For n = 1, P(n) = (2 + √2)(2 − √2) = 2

For n = k, if we multiply 2 + √x with 2 − √x present in the kth step, we obtain conjugate of x present in the (k − 1)th step. Therefore, for n = k, after multiplying every conjugate pair we get all the numbers of the (k − 1)th step. This implies that product of all numbers in kth step is equal to the product of all the numbers in (k − 1)th step.

So, P(1000) = P(999) = … = P(1) = 2.

#### Ilan Mayer solved this puzzle:

The list after one step:

(2 + √2, 2 − √2)

The list after two steps:

(2 + √(2 + √2), 2 − √(2 + √2), 2 + √(2 − √2), 2 − √(2 − √2))

The list consists of pairs of expressions of the form 2 + x and 2 − x.

In each step 2 + x is replaced with 2 + √(2 + x) and 2 − √(2 + x) and 2 − x is replaced with 2 + √(2 − x) and 2 − √(2 − x).

The product of these 4 new expressions is (2 + x)(2 - x) which is same as the original product.

Therefore the product of all the numbers in the list after any number of steps is 2.

#### quasi solved this puzzle:

Proposition:

Let A be a finite, nonempty subset of (0, 4) such that A is symmetric about x = 2, and let B be defined by

B = {2 - √(a) | a ∈ A} ∪ {2 + √(a) | a ∈ A}

B is also a finite, nonempty subset of (0, 4) such that B is symmetric about x = 2. Moreover, the product of the elements of B equals the product of the elements of A.

Proof:

Let a ∈ A. Since a ∈ (0, 4), 0 < √a < 2, hence both 2 − √a and 2 + √a are both in (0, 4). It follows that B is a finite, nonempty subset of (0, 4).

Since the reflection of 2 − √a about x = 2 is 2 + √a, it follows that B is symmetric about x = 2.

The map a → 2 − √a gives a one-to-one correspondence between A and B ∩ (0, 2). Similarly, the map a → 2 + √a gives a one-to-one correspondence between A and B ∩ (2, 4). Also, note that the number 2 cannot be an element of B (although 2 might be an element of A).

Thus, B has exactly twice as many elements as A and we can group the elements of B in pairs such that for each a ∈ A, we pair 2 − √a with 2 + √a.

To find the product of the elements of B, it suffices to find the product of the products of the matched pairs. Each matched pair has product 4 − a, hence the product of the elements of B is the same as the product of the elements in the set {4 − a | a ∈ A}. But since A is symmetric about x = 2, the set {4 − a | a ∈ A} equals A. Hence, the product of the elements of B is the same as the product of the elements of A.

As a corollary, if we start with the set {2} and apply the specified transformations a finite number of times, the product of the elements in each of the new sets stays at 2.

So the answer is 2.

### Credit

This puzzle is taken from folklore.