Consecutive Divisibility Is Impossible

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:

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

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

• 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.

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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: https://vumaths.com/booklets/hsc-proofs/
• HSC Differential Equations: https://vumaths.com/booklets/hsc-differential-equations/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/

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