Exercise 1: Direct Proof That n Odd Implies n^2 Odd — Possible Solution
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THE PROOF
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Claim: If n is odd, then n^2 is odd.

Proof. Assume n is odd. By definition, an odd integer can be written
as n = 2k + 1 for some integer k.

Then:
  n^2 = (2k + 1)^2
      = 4k^2 + 4k + 1
      = 2(2k^2 + 2k) + 1

Since k is an integer, 2k^2 + 2k is also an integer (integers are
closed under multiplication and addition). So n^2 has been written as
2 times an integer, plus 1 - which is exactly the definition of an
odd number.

Therefore n^2 is odd. QED.

WHY THIS FOLLOWS THIS CHAPTER'S OWN DIRECT-PROOF STRUCTURE
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Per this chapter, a direct proof of "if P then Q" means "assume P is
true, and show through a valid chain of reasoning that Q must then
also be true." Here P is "n is odd" and Q is "n^2 is odd" - the proof
assumes P (n = 2k+1), then performs a valid algebraic chain (expanding
the square, factoring out 2) to arrive at Q (n^2 fits the exact
definition of odd), with no gaps in the reasoning at any step.

WHY THE ALGEBRA SPECIFICALLY PROVES ODDNESS, NOT JUST "SOME FORMULA"
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The definition of an odd integer is precisely "2 times some integer,
plus 1." The proof doesn't just simplify n^2 to some expression - it
specifically arranges the expression into exactly that form,
2(2k^2 + 2k) + 1, with the quantity in parentheses confirmed to be an
integer. That exact match to the definition is what makes the proof
complete, rather than merely suggestive.

WHY THIS WORKS AS AN ANSWER
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It follows the direct-proof pattern precisely - assuming the premise,
performing a valid algebraic chain, and arriving at an expression that
matches the formal definition of the conclusion exactly, with each
step justified.