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Divisibility of a Shifted Power Expression

Authors
  • avatar
    Name
    Vu Hung
    Twitter

Problem Statement

Prove that (2m−1)2−1(2^m - 1)^2 - 1 is divisible by 2m+12^{m+1} for all integers m≥1m \geq 1.

Hints

Expand the perfect square (2m−1)2(2^m - 1)^2 algebraically, then simplify. Try to factor out 2m+12^{m+1} from the resulting expression.

Solutions

Let E=(2m−1)2−1E = (2^m - 1)^2 - 1.

Step 1: Expand the square:

E=(2m)2−2(2m)+1−1=22m−2m+1+1−1=22m−2m+1\begin{aligned} E &= (2^m)^2 - 2(2^m) + 1 - 1 \\ &= 2^{2m} - 2^{m+1} + 1 - 1 \\ &= 2^{2m} - 2^{m+1} \end{aligned}

Step 2: Factor out 2m+12^{m+1}:

Note that 22m=2m+1⋅2m−12^{2m} = 2^{m+1} \cdot 2^{m-1} (using exponent laws).

Therefore:

E=2m+1⋅2m−1−2m+1⋅1=2m+1(2m−1−1)\begin{aligned} E &= 2^{m+1} \cdot 2^{m-1} - 2^{m+1} \cdot 1 \\ &= 2^{m+1}(2^{m-1} - 1) \end{aligned}

Step 3: Conclude divisibility:

Since m≥1m \geq 1, we have m−1≥0m - 1 \geq 0, so 2m−12^{m-1} is an integer. Thus (2m−1−1)(2^{m-1} - 1) is an integer.

Therefore, E=2m+1⋅kE = 2^{m+1} \cdot k where k=2m−1−1∈Zk = 2^{m-1} - 1 \in \mathbb{Z}.

By definition, (2m−1)2−1(2^m - 1)^2 - 1 is divisible by 2m+12^{m+1}. ■\blacksquare

Takeaways

  • Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
  • Check edge cases and verify where each assumption is used in the argument.

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

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