Which one of the following is a Pythagorean triple in which one side differs from the hypotenuse by two units?

Options

A. `(2n + 1, 4n,2n^(2) +2n)`
B. `(2n,4n, n^(2) + 1)`
C. `(2n^(2),2n, 2n+ 1)`
D. `{2n, n^(2) -1 ,\n^(2) + 1)`

Answer

To determine which one of the given options is a Pythagorean triple where one side differs from the hypotenuse by two units, we can follow these steps:

### Step 1: Understand the Pythagorean Theorem
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b). This can be expressed as:
\[ c^2 = a^2 + b^2 \]

### Step 2: Set up the relationship
According to the problem, we need to find a Pythagorean triple (a, b, c) such that:
\[ c = a + 2 \]
or
\[ c = b + 2 \]
This means that one side (either a or b) differs from the hypotenuse (c) by 2 units.

### Step 3: Substitute and rearrange
If we assume \( c = a + 2 \), we can substitute this into the Pythagorean theorem:
\[ (a + 2)^2 = a^2 + b^2 \]
Expanding the left side:
\[ a^2 + 4a + 4 = a^2 + b^2 \]
Now, simplify the equation:
\[ 4a + 4 = b^2 \]
This gives us:
\[ b^2 = 4a + 4 \]

### Step 4: Check for integer values
To find integer values for a, b, and c, we can start testing small integer values for a:
- If \( a = 1 \):
\[ b^2 = 4(1) + 4 = 8 \] (not a perfect square)
- If \( a = 2 \):
\[ b^2 = 4(2) + 4 = 12 \] (not a perfect square)
- If \( a = 3 \):
\[ b^2 = 4(3) + 4 = 16 \] (perfect square, \( b = 4 \))

Now, we can find \( c \):
\[ c = a + 2 = 3 + 2 = 5 \]

### Step 5: Verify the triple
Now we have:
- \( a = 3 \)
- \( b = 4 \)
- \( c = 5 \)

We can verify:
\[ 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \]
Thus, \( (3, 4, 5) \) is a Pythagorean triple.

### Conclusion
The Pythagorean triple where one side differs from the hypotenuse by 2 units is \( (3, 4, 5) \).

---