Induction for a Divisibility Pattern Modulo Ten

Mathematical induction is a standard proof technique in HSC Extension 1, often tested through divisibility arguments. This problem addresses a subtle twist: the statement is only claimed for even positive integers. By rewriting $n=2k$, we convert the statement into a regular induction over all positive integers $k$, then prove divisibility by 10 using algebraic manipulation and the inductive hypothesis. Working through it will reinforce how to adapt induction when a domain restriction is present.

Problem Statement

Prove by induction that $4^{n+1} + 6^n$ is divisible by $10$ when $n$ is an even positive integer.

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Hints

Since $n$ is even, let $n = 2k$ for $k \geq 1$. Prove by induction on $k$ that $4^{2k+1} + 6^{2k}$ is divisible by $10$.

For the inductive step:

• Express $4^{2(k+1)+1} + 6^{2(k+1)}$ as $16 \cdot 4^{2k+1} + 36 \cdot 6^{2k}$
• From IH: $6^{2k} = 10M - 4^{2k+1}$ for some integer $M$
• Substitute and factor out $10$

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Solutions

Since $n$ is an even positive integer, let $n = 2k$ where $k \in \mathbb{Z}^+$.

We prove by induction on $k$ that $P(k)$: "$4^{2k+1} + 6^{2k}$ is divisible by $10$" holds for all $k \geq 1$.

To set up the base case, we check the smallest value $k=1$ (which gives the smallest even $n=2$).

Base case ($k=1$, i.e., $n=2$):

$$4^{2(1)+1} + 6^{2(1)} = 4^3 + 6^2 = 64 + 36 = 100 = 10 \times 10$$

Divisible by $10$. \checkmark

Inductive hypothesis:

Assume $P(k)$ holds: $4^{2k+1} + 6^{2k} = 10M$ for some integer $M$.

From this: $6^{2k} = 10M - 4^{2k+1}$

Inductive step:

Prove $P(k+1)$: $4^{2(k+1)+1} + 6^{2(k+1)}$ is divisible by $10$.

We first expand the exponents so the expression involves $4^{2k+1}$ and $6^{2k}$ — exactly the terms from $P(k)$ — which will let us use the inductive hypothesis.

$$\begin{aligned} 4^{2(k+1)+1} + 6^{2(k+1)} &= 4^{2k+3} + 6^{2k+2} \\ &= 4^2 \cdot 4^{2k+1} + 6^2 \cdot 6^{2k} \\ &= 16 \cdot 4^{2k+1} + 36 \cdot 6^{2k} \end{aligned}$$

Now we replace $6^{2k}$ using the relation from the inductive hypothesis.

Substitute $6^{2k} = 10M - 4^{2k+1}$ from IH:

$$\begin{aligned} &= 16 \cdot 4^{2k+1} + 36(10M - 4^{2k+1}) \\ &= 16 \cdot 4^{2k+1} + 360M - 36 \cdot 4^{2k+1} \\ &= 360M + (16 - 36) \cdot 4^{2k+1} \\ &= 360M - 20 \cdot 4^{2k+1} \\ &= 10(36M - 2 \cdot 4^{2k+1}) \end{aligned}$$

Since $M$ and $4^{2k+1}$ are integers, $36M - 2 \cdot 4^{2k+1}$ is an integer.

Therefore, $4^{2(k+1)+1} + 6^{2(k+1)}$ is divisible by $10$. \checkmark

By the principle of mathematical induction, $P(k)$ holds for all $k \geq 1$.

Therefore, $4^{n+1} + 6^n$ is divisible by $10$ for all even positive integers $n$. $\blacksquare$

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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, HSC Polynomials, HSC Complex Numbers