find all positive integers n whose only prime divisors are 2 and 5 and n+25 is a perfect square

We are tasked with finding all positive integers \( n \) such that:

1. The only prime divisors of \( n \) are 2 and 5.
2. \( n + 25 \) is a perfect square.

### Step 1: Express \( n \) in terms of its prime divisors
Since \( n \)'s only prime divisors are 2 and 5, we can express \( n \) as:
\[
n = 2^a \cdot 5^b
\]
where \( a \) and \( b \) are non-negative integers.

### Step 2: Set up the equation
We know that \( n + 25 \) must be a perfect square. Thus, we can write:
\[
n + 25 = k^2
\]
for some integer \( k \), so:
\[
n = k^2 - 25
\]
Thus, the equation becomes:
\[
k^2 - 25 = 2^a \cdot 5^b
\]
or equivalently:
\[
k^2 = 2^a \cdot 5^b + 25
\]

### Step 3: Factorization
The left-hand side of the equation can be factored:
\[
k^2 - 25 = (k - 5)(k + 5)
\]
Therefore, we have:
\[
(k - 5)(k + 5) = 2^a \cdot 5^b
\]
Since the prime divisors of the right-hand side are only 2 and 5, both \( k - 5 \) and \( k + 5 \) must also be products of powers of 2 and 5.

### Step 4: Analyze the factors
Let's consider the possibilities for \( k - 5 \) and \( k + 5 \):

- \( k - 5 \) is divisible by \( 2^a \)
- \( k + 5 \) is divisible by \( 5^b \)

However, the sum and difference \( k+5 \)

Find all positive integers n whose only prime divisors are 2 and 5, and n + 25 is a perfect square.

Explore how to find all positive integers n whose only prime divisors are 2 and 5, where n + 25 forms a perfect square. The solution covers prime factorization, perfect square properties, and step-by-step reasoning. Suitable for Grades 9-12 students.

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