Question

S={х:(x=n²)^(nєN)} Доказать, что єсли хєS то (x+2√x+1) є S

Answer 1

Пусть х∈S, т.е. х можно представить в виде х = р², где р ∈ N

Тогда х + 2√х + 1 = р² + 2р + 1 = (р + 1)² ∈ S

Answer 2

To prove that if x ∈ S, then (x + 2√x + 1) ∈ S, we need to show that if x is a perfect square raised to the power of a natural number, then (x + 2√x + 1) is also a perfect square raised to the power of a natural number.

Let's start by expressing x in terms of n:

x = n^2^n

Now, we want to show that (x + 2√x + 1) is also of the form m^k, where m is a natural number and k is a natural number.

(x + 2√x + 1) = (n^2^n + 2√(n^2^n) + 1)

Now, let's simplify 2√(n^2^n):

2√(n^2^n) = 2 * n^(2^n / 2)

Now, we can rewrite (x + 2√x + 1) as:

(x + 2√x + 1) = (n^2^n + 2 * n^(2^n / 2) + 1)

Now, let's try to factor this expression and see if it matches the form m^k for some natural numbers m and k:

(x + 2√x + 1) = (n^(2^n) + 2 * n^(2^n / 2) + 1)

Let's notice that this expression is in the form of a perfect square:

(x + 2√x + 1) = (n^(2^n / 2) + 1)^2

Now, we need to check if n^(2^n / 2) + 1 can be expressed as m^k for some natural numbers m and k. For this, we need to show that 2^n / 2 is a whole number.

Since n is a natural number, 2^n is also a natural number. And when we divide a natural number by 2, we get either an integer (if n is even) or a number with a fractional part of 0.5 (if n is odd). In our case, n is raised to a power, so n can't be fractional; it is always an integer.

Thus, 2^n is always even, and 2^n / 2 is always an integer. Let's say 2^n / 2 = k, where k is a natural number.

Now, we can rewrite (x + 2√x + 1) as:

(x + 2√x + 1) = (n^(2^n / 2) + 1)^2 = (n^k + 1)^2

Now, (x + 2√x + 1) is indeed in the form of m^k, where m = n^k + 1 and k = 2. Both m and k are natural numbers because n and k are natural numbers.

Therefore, we have shown that if x ∈ S (x is a perfect square raised to the power of a natural number), then (x + 2√x + 1) ∈ S (it is also a perfect square raised to the power of a natural number).