find all two-digit natural numbers such that the expression x= sqrt((n + 36) / (n - 36)) is also a natural number."

We are tasked with finding all two-digit natural numbers \( n \) such that the expression:

\[
x = \sqrt{\frac{n + 36}{n - 36}}
\]

is also a natural number. Let's work through this step-by-step:

### Step 1: Express \( x^2 \)
We begin by squaring both sides of the equation:

\[
x^2 = \frac{n + 36}{n - 36}
\]

### Step 2: Solve for \( n \)
Rearrange the equation to isolate \( n \):

\[
x^2 = \frac{n + 36}{n - 36} \implies x^2 (n - 36) = n + 36
\]

Expanding both sides:

\[
x^2 n - 36 x^2 = n + 36
\]

Move all terms involving \( n \) to one side and constants to the other:

\[
x^2 n - n = 36 x^2 + 36
\]

Factor out \( n \) on the left-hand side:

\[
n(x^2 - 1) = 36(x^2 + 1)
\]

Now, solve for \( n \):

\[
n = \frac{36(x^2 + 1)}{x^2 - 1}
\]

### Step 3: Simplify the Expression
The expression for \( n \) simplifies to:

\[
n = \frac{36(x^2 + 1)}{(x - 1)(x + 1)}
\]

For \( n \) to be a natural number, the denominator \( (x^2 - 1) = (x - 1)(x + 1) \) must divide evenly into the numerator \( 36(x^2 + 1) \).

### Step 4: Check Integer Solutions
Let’s check small integer values of \( x \) and see if \( n \) becomes a two-digit number.

- For \( x = 2 \):

\[
n = \frac{36(2^2 + 1)}{2^2 - 1} = \frac{36(4 + 1)}{4 - 1} = \frac{36 \times 5}{3} = 60
\]

Thus, \( n = 60 \).

- For \( x = 3 \):

\[
n = \frac{36(3^2 + 1)}{3^2 - 1} = \frac{36(9 + 1)}{9 - 1} = \frac{36 \times 10}{8} = 45
\]

But \( n = 45 \) is not a two-digit number, so we discard this solution.

- For \( x = 4 \):

\[
n = \frac{36(4^2 + 1)}{4^2 - 1} = \frac{36(16 + 1)}{16 - 1} = \frac{36 \times 17}{15} = 40.8
\]

\( n \) is not an integer, so discard this case.

- For \( x = 5 \):

\[
n = \frac{36(5^2 + 1)}{5^2 - 1} = \frac{36(25 + 1)}{25 - 1} = \frac{36 \times 26}{24} = 39
\]

Again, not a two-digit number.

### Conclusion
The only two-digit solution is \( n = 60 \).

### Would you like further details or have any questions?
Here are 5 related questions to deepen your understanding:
1. Can we apply similar steps to find solutions for three-digit numbers?
2. How does changing the constants (e.g., 36) in the original equation affect the solution set?
3. What happens if the square root expression involves a cube instead of a square?
4. How can this problem be solved using inequalities?
5. Can this method extend to irrational numbers under certain conditions?

**Tip**: Checking small integer solutions for \( x \) can often help simplify problems that involve fractions or roots.

Find all two-digit natural numbers such that the expression x = sqrt((n + 36) / (n - 36)) is also a natural number.

Explore how to find all two-digit natural numbers where x = sqrt((n + 36) / (n - 36)) results in a natural number. This problem involves algebraic manipulation, properties of square roots, and factorization. Perfect for students in Grades 9-12.

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