Let's work through these two proofs step by step, leaning into proof by contradiction as requested—this is a classic approach for these number theory problems.

1. Proving (a^3 + b^3 = c^3) has no non-zero integer solutions #

We start with a contradiction assumption, then show it leads to an impossible scenario:

• Core Assumption: There exists a smallest set of positive integers ((a, b, c)) where (a^3 + b^3 = c^3), and (a, b, c) are coprime (if they shared a common divisor, we could divide the entire equation by that divisor to get a smaller valid triple—breaking our "smallest" claim).
• (a) and (b) can't both be even: If they were, (c^3) would be even (sum of two evens), so (c) is even too. That means all three are divisible by 2, which contradicts our coprimality condition.
• They also can't both be odd: Odd cubed is odd, so odd + odd = even—so (c) would have to be even. But wait, odd + even = odd, so exactly one of (a) or (b) must be even, and (c) is odd (since the sum of an odd and even cube is odd).
• Rewrite the equation as (c^3 - a^3 = b^3), which factors to ((c - a)(c^2 + ac + a^2) = b^3). Now, (c - a) and (c^2 + ac + a^2) are coprime: suppose a prime (p) divides both. Then (p) divides (c^2 + ac + a^2 - a(c - a) = 2a^2), and also divides (c^2 + ac + a^2 - c(c - a) = a(2c + a)). Since (a) and (c) are coprime, (p) must divide 2—but if (p=2), (c -a) is even, meaning (c) and (a) have the same parity, which contradicts one being odd and one even.
• Since their product is a cube and they're coprime, each factor must be a cube on its own: let (c - a = m^3) and (c^2 + ac + a^2 = n^3), so (b = m \cdot n).
• Substitute (c = a + m^3) into the second equation: expanding gives (3a^2 + 3a m^3 + m^6 = n^3). This can be rearranged to form another cube-sum equation with smaller integers, which directly contradicts our "smallest triple" assumption.
• Since our initial assumption leads to a contradiction, no such positive integer triples exist. Negative solutions are just sign-flipped positive ones (e.g., ((-a)^3 + (-b)^3 = (-c)^3) is equivalent to (a^3 + b^3 = c^3)), so there are no non-zero solutions at all.

2. Proving no three consecutive positive integers satisfy (a^3 + b^3 = c^3) #

Again, contradiction is straightforward here:

• Core Assumption: There exist three consecutive positive integers (a, b, c) such that (a^3 + b^3 = c^3). Since cubes grow quickly, the largest number has to be (c) (smaller numbers can't sum to a smaller cube). So we'll denote the three numbers as (n-1), (n), (n+1) where (n ≥ 2) (since (n-1) has to be positive).
• Plug these into the equation: ((n-1)^3 + n^3 = (n+1)^3)
• Expand all the cubes:
  • Left side: ((n^3 - 3n^2 + 3n - 1) + n^3 = 2n^3 - 3n^2 + 3n - 1)
  • Right side: (n^3 + 3n^2 + 3n + 1)
• Subtract the right side from both sides to simplify:
  (2n^3 - 3n^2 + 3n - 1 - n^3 - 3n^2 - 3n - 1 = 0)
  This reduces to: (n^3 - 6n^2 - 2 = 0)
• Now check for positive integer solutions:
  • (n=1): (1 - 6 - 2 = -7 ≠ 0)
  • (n=2): (8 - 24 - 2 = -18 ≠ 0)
  • (n=3): (27 - 54 - 2 = -29 ≠ 0)
  • (n=4): (64 - 96 - 2 = -34 ≠ 0)
  • (n=5): (125 - 150 - 2 = -27 ≠ 0)
  • (n=6): (216 - 216 - 2 = -2 ≠ 0)
  • (n=7): (343 - 294 - 2 = 47 > 0)
  • For (n ≥7), (n^3) grows way faster than (6n^2): (n^3 -6n^2 -2 = n^2(n-6) -2). When (n≥7), (n-6≥1), so (n^2(n-6) ≥49*1=49), making the whole expression at least (49-2=47>0) (it only gets bigger as (n) increases).
• Since there are no positive integers (n) that satisfy the cubic equation, our initial assumption is false. No three consecutive positive integers can satisfy (a^3 + b^3 = c^3).

内容的提问来源于stack exchange，提问作者Tina Hsieh