Question

please solve the above question {ques 4 & 7}

Answer 1

Let x be 2n + 1, and y be 2m + 1

So, x² + y²:

(2n + 1)² + (2m + 1)²

= 4n² + 4n + 1 + 4m² + 4m + 1

= 4(n² + n + m² + m) + 2

Which is of the form 4k + 2, so on dividing by 4, we'll get k + 2/4, implying there is a remainder of 2.

Hence, x² + y² is not divisible by 4.

Yet, the whole thing is of the form 2(f + 1), which is divisible by 2, hence even.

Note: m, n are non-negative integers, k is also a non-negative integer (whole number) and f = 2k.