关于Divisibility Induction的作业归纳法证明难题求助

Hey there, I totally get how frustrating it is to hit a wall with divisibility induction proofs—those final steps can feel like trying to fit a square peg in a round hole, especially when you’ve been tinkering with mods and integer division (a/b where a is divisible by b) and can’t quite cross the finish line. Let’s break this down with actionable tips and a concrete example to help you push past that stuck point.

• Lock in the core induction structure first
  It’s easy to get lost in algebra or mod calculations and forget the induction’s backbone:

  1. Base Case: Prove the statement holds for the smallest integer (usually n=1 or n=0).
  2. Inductive Hypothesis: Assume the statement is true for some integer k (e.g., "Suppose f(k) is divisible by d for some integer k ≥ 1").
  3. Inductive Step: This is where most people get stuck—you need to rewrite the expression for k+1 to explicitly include the inductive hypothesis, then prove the remaining portion is also divisible by d.

• Mods vs. integer division: Pick the right tool for the job

  • Use mod arithmetic if you want to directly check that the remainder of f(k+1) divided by d is 0. For example, if you know f(k) ≡ 0 mod d, you can compute f(k+1) mod d and show it equals 0.
  • Use integer division (writing f(k) = d * m for some integer m) when you need to rearrange algebra to isolate the inductive hypothesis term. This is often more straightforward for expanding polynomial expressions.

• Don’t skip factoring the "leftover" term
  After pulling out the inductive hypothesis from f(k+1), you’ll have a remaining expression. This is where hidden divisibility rules come in:

  • Consecutive integers (k and k+1) always include one even number, so k(k+1) is divisible by 2.
  • Three consecutive integers include a multiple of 2 and a multiple of 3, so k(k+1)(k+2) is divisible by 6.
  • Terms like a^(k+1) - a^k factor to a^k(a-1), which can often be tied back to your divisibility target.

Example Walkthrough: Prove n³ - n is divisible by 6 for all positive integers n #

Let’s apply these steps to a classic problem to see how the final click happens:

1. Base Case (n=1): 1³ - 1 = 0, which is divisible by 6. Done.
2. Inductive Hypothesis: Assume k³ - k is divisible by 6 for some integer k ≥ 1, so k³ - k = 6m where m is an integer.
3. Inductive Step: Prove (k+1)³ - (k+1) is divisible by 6.
   • Expand the expression:

     (k+1)³ - (k+1) = k³ + 3k² + 3k + 1 - k - 1 = k³ - k + 3k² + 3k
     

   • Substitute the inductive hypothesis:

     6m + 3k(k + 1)
     

   • Now look at the leftover term 3k(k+1): since k and k+1 are consecutive integers, one is even—so k(k+1) is divisible by 2. That makes 3*2=6, so 3k(k+1)=6p for some integer p.
   • Combine terms: 6m + 6p = 6(m + p), which is clearly divisible by 6. We’re done!

If you’re stuck on a specific problem, try writing out each of these steps explicitly—chances are you just missed a factoring trick or a way to rearrange the k+1 expression to include your inductive hypothesis.

内容的提问来源于stack exchange，提问作者Jeffrey Chiu