Final answer:
The student's question is answered using direct proof for each part. x3 is proved to be odd for any odd integer x, the product of two odd integers x and y is shown to be odd, and it is demonstrated that every odd integer can be written as the difference of two squares.
Step-by-step explanation:
When approaching proofs in mathematics, we rely on established rules and properties to construct logical arguments. Let's tackle each part of the student's question using direct proof.
a) Proof that if x is an odd integer, then x3 is odd:
An odd integer can be represented as 2n + 1, where n is an integer. If we cube this expression, we have (2n + 1)3, which simplifies to 8n3 + 12n2 + 6n + 1. Notice that 8n3, 12n2, and 6n are all multiples of 2, hence even. Adding 1 to an even number yields an odd number, so x3 is odd.
b) Proof that if x and y are odd, then xy is odd:
Using the representation of an odd integer as 2m + 1 and 2n + 1 for x and y respectively, their product xy is (2m + 1)(2n + 1), which expands to 4mn + 2m + 2n + 1. The terms 4mn, 2m, and 2n are even, and adding 1 results in an odd number. Thus, xy is odd.
c) Proof that every odd integer is the difference of two squares:
For an odd integer 2n + 1, we can find two integers a and b, where a = n + 1 and b = n. The difference of their squares a2 - b2 is (n + 1)2 - n2, which equals 2n + 1, verifying the statement.