Express the square numbers lying between 200 and 300 as the product of two consecutive natural numbers increased by 1.

Steps to Solve

1. Identify the square numbers between 200 and 300

First, we need to find the natural numbers whose squares are within the range of 200 to 300. The square root of 200 is approximately $14.14$ and the square root of 300 is approximately $17.32$. Thus, we will check the squares of the integers from 15 to 17:

$$ 15^2 = 225 $$

$$ 16^2 = 256 $$

$$ 17^2 = 289 $$

The square numbers between 200 and 300 are 225, 256, and 289.

2. Express square numbers as products of consecutive integers

Next, we will express each square number found as the product of two consecutive natural numbers plus 1. We can represent this mathematically as:

$$ n(n + 1) + 1 = n^2 + n + 1 $$

Now we'll check each of the square numbers:

• For $225$: We set up the equation $n(n + 1) + 1 = 225$:

$$ n(n + 1) = 224 \quad \Rightarrow \quad n^2 + n - 224 = 0 $$

Now we can use the quadratic formula:

$$ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 896}}{2} = \frac{-1 \pm 30}{2} $$

We get two potential solutions for $n$:

$$ n = \frac{29}{2} \text{ (not an integer)} \quad \text{and} \quad n = -15.5 \text{ (not a natural number)} $$

This means 225 cannot be expressed as the product of two consecutive numbers plus one.

• For $256$: We set up the equation $n(n + 1) + 1 = 256$:

$$ n(n + 1) = 255 \quad \Rightarrow \quad n^2 + n - 255 = 0 $$

Using the quadratic formula:

$$ n = \frac{-1 \pm \sqrt{1 + 1020}}{2} = \frac{-1 \pm 32}{2} $$

This gives us:

$$ n = 15.5 \text{ (not an integer)} \quad \text{and} \quad n = -16.5 \text{ (not a natural number)} $$

Thus, 256 cannot be expressed this way.

• For $289$: We set up the equation $n(n + 1) + 1 = 289$:

$$ n(n + 1) = 288 \quad \Rightarrow \quad n^2 + n - 288 = 0 $$

Using the quadratic formula:

$$ n = \frac{-1 \pm \sqrt{1 + 1152}}{2} = \frac{-1 \pm 34}{2} $$

This gives us:

$$ n = 16.5 \text{ (not an integer)} \quad \text{and} \quad n = -17.5 \text{ (not a natural number)} $$

Therefore, 289 cannot be expressed as a product of consecutive integers plus one.

3. Conclusion about square numbers

After evaluating all three square numbers (225, 256, and 289), we find that none of them can be expressed as the product of two consecutive natural numbers with an increment of 1.