Final Answer:
Yes, the sum of the squares of two odd numbers always leaves a remainder of 2 when divided by 4.
Step-by-step explanation:
Let's consider two odd numbers, a and b. We can express them as follows:
a = 2m + 1
b = 2n + 1
where m and n are integers.
Now, let's calculate the sum of their squares:
a² + b² = (2m + 1)² + (2n + 1)²
Expanding the squares, we get:
a² + b² = 4m² + 4m + 1 + 4n² + 4n + 1
Combining like terms, we get:
a² + b² = 4(m² + n²) + 4(m + n) + 2
Since m² and n² are integers, 4(m² + n²) is also an integer. Similarly, 4(m + n) is also an integer. Therefore, the sum of the squares of two odd numbers can be expressed as:
a² + b² = 4k + 2
where k is an integer.
This means that the sum of the squares of two odd numbers is always of the form 4k + 2, which when divided by 4 leaves a remainder of 2.