Consecutive Divisibility Is Impossible

Divisibility is a cornerstone of number theory and appears throughout the HSC Mathematics Extension 2 syllabus, especially in proof and reasoning questions. This problem asks you to show that a number greater than 1 cannot divide two consecutive integers—a result that feels intuitively true but benefits from a rigorous argument. By following the solution, you’ll see how proof by contradiction can turn a simple property of divisibility into an airtight conclusion.

Problem Statement

Prove that for positive integers $a, b$ with $a > 1$, either $b$ is not divisible by $a$ or $b + 1$ is not divisible by $a$ (or both).

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Hints

Use proof by contradiction. Assume both $a|b$ and $a|(b+1)$. What does this tell you about $a|[(b+1) - b]$? What can you conclude about $a$?

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Solutions

Proof by Contradiction:

The statement is equivalent to: "It is not the case that both $a|b$ and $a|(b+1)$."

Assume, for contradiction, that both $a|b$ and $a|(b+1)$.

Then there exist integers $k_1$ and $k_2$ such that:

$$\begin{aligned} b &= ak_1 \\ b + 1 &= ak_2 \end{aligned}$$

A fundamental property of divisibility tells us that if $a$ divides two numbers, it also divides their difference. Applying this to our assumption, $a$ must divide $(b+1) - b = 1$. We can verify this algebraically by subtracting the two equations:

Subtracting the first equation from the second:

$$(b+1) - b = ak_2 - ak_1$$ $$1 = a(k_2 - k_1)$$

Let $K = k_2 - k_1 \in \mathbb{Z}$. Then $1 = aK$, which means $a$ divides $1$.

Since $a$ is a positive integer, the only positive divisor of $1$ is $1$ itself. Therefore, $a = 1$.

This contradicts the given condition that $a > 1$.

Hence, our assumption must be false, and at least one of $b$ or $b+1$ is not divisible by $a$. $\blacksquare$

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Takeaways

• The proof hinges on the divisibility property: if $a \mid x$ and $a \mid y$, then $a \mid (x - y)$. Here, subtracting $b$ from $b+1$ gives $1$, forcing $a$ to divide $1$.
• The only positive divisor of $1$ is $1$ itself, so any $a > 1$ immediately leads to a contradiction—consecutive integers can never share a divisor greater than $1$.
• Proof by contradiction streamlines the argument: assume the opposite, derive an impossibility, and conclude the original statement is true.
• Reconstruct this proof from the hint, justifying each algebraic step, and check how the conditions $a>1$ and $a,b$ positive integers are used.

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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 Differential Equations, HSC Trigonometry