Induction for a Divisibility Pattern Modulo Ten

Problem Statement

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

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 $10$

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 $k \geq 1$ .

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 $10$ .

\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}

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 $n$ . $\blacksquare$

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:

HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
HSC Polynomials: https://vumaths.com/booklets/hsc-polynomials/
HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/

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