Divisibility of a Shifted Power Expression

Proof by algebraic manipulation is a core skill in HSC Mathematics Extension 1. In this problem, we’ll prove a divisibility statement involving powers of 2 by expanding a perfect square and cleverly factoring the result. Working through this example will strengthen your ability to recognise common factors and apply index laws in a proof.

Problem Statement

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

---

Hints

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

---

Solutions

Let $E = (2^m - 1)^2 - 1$.

We start by expanding the square to simplify the expression. Since we need to show divisibility by a power of two, getting everything in terms of powers of two will be helpful.

Step 1: Expand the square:

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

Now we want to factor out $2^{m+1}$. Observe that the first term $2^{2m}$ can be rewritten using index laws as $2^{m+1} \cdot 2^{m-1}$.

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

Note that $2^{2m} = 2^{m+1} \cdot 2^{m-1}$ (using exponent laws).

Therefore:

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

The factored form $2^{m+1}(2^{m-1} - 1)$ shows that $E$ is $2^{m+1}$ times an integer, provided the factor in parentheses is indeed an integer.

Step 3: Conclude divisibility:

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

Therefore, $E = 2^{m+1} \cdot k$ where $k = 2^{m-1} - 1 \in \mathbb{Z}$.

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

---

Takeaways

• Learn to expand and simplify expressions to expose a common factor.
• Recognise how index laws help rewrite a power to match the desired divisor.
• Always check that the coefficient after factoring is an integer; here $m-1 \ge 0$ ensures the factor $2^{m-1}$ is an integer.

---

Further Readings

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