$1^n-3^n-6^n+8^n$ is divisible by $10$

Question

Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$

It is divisible by $2$ and $5$ if we rearrange it will it be enough

$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$. And $(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.

Hence $\gcd(2,5)$ is $1$ and it is divisible by $2\cdot5=10$.

Is it correct?

Answer 1

Try a proof by induction.

Base Case: $n=1 \to 1-3-6+8=0$ which is divisible by $10$. So true for $n=1$

Then assume it true for some $n=k$; so: $$1^k-3^k-6^k+8^k=10t, t\in \Bbb Z$$ Then use that fact to show it true for $n=k+1$: i.e, prove that: $$1^{k+1}-3^{k+1}-6^{k+1}+8^{k+1}=10u, u \in \Bbb Z$$ For this, I would use that $n^{k+1}=n(n^k)$

Answer 2

Note: $p,q,r,s,t, a ,b, n \in \mathbb{N},$

$f(n):= (8^n-3^n) - (6^n-1^n)= (8-3)p-(6-1)q=5(p-q).$

On the other hand:

$f(n)= (8^n-6^n) -(3^n-1^n)=(8-6)r -(3-1)s=2(r-s)$

Hence:

$f(n)=5(p-q)=2(r-s).$

Euclid's Lemma:

$2$ divides $(p-q)$, i.e.

$p-q=2t$.

Combining:

$f(n)= 5\cdot 2 t.$

Used:

$(a^n-b^n)=$

$(a-b)(a^{n-1}+a^{n-2}b+.....+b^{n-1}).$

$p,q,r$ and $s$ were used for the above second factor.

Euclid's Lemma:

If a prime $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.

Answer 3

Your proof is OK, however, for completeness, you must further show why the binomials are divisible by $2$ and $5$.

If you are familiar with modular arithmetic, it is: $$1^n-3^n-6^n+8^n\equiv 1-3-6+8\equiv 0 \pmod {10}.$$

Answer 4

Another way to see the solution is by using $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})\tag{1}$$ leading to $$1^n-3^n-6^n+8^n=(8^n-3^n)-(6^n-1^n)=\\ (8-3)(8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}+3^{n-1})-\\ (6-1)(6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}+1^{n-1})=...$$ we can see that $8^{n-1}+8^{n-2}\cdot3+...+8\cdot3^{n-2}$ is always even and so it $6^{n-1}+6^{n-2}\cdot1+...+6\cdot1^{n-2}$ then $$...=5(2m+3^{n-1})-5(2n+1)=5\left(2(m-n)+3^{n-1}-1\right)$$ and $3^{n-1}-1$ is even from the same application of $(1)$, thus we can "extract" a 2 from $\left(2(m-n)+3^{n-1}-1\right)$ and the result follows.

Answer 5

Perhaps another method to notice is that, by Euler's Generalization of Fermat's Little Theorem, we have that: $$1^n-3^n-6^n+8^n \equiv 1^{n \mod \phi(10)} - 3^{n \mod \varphi(10)} - 6^{n \mod \varphi(10)} + 8^{n \mod \varphi(10)} \mod 10$$ Since $\varphi(10) = 4$, you really only need to check the cases of $n = 0, 1, 2$ and $3$. For $n = 0$, you have: $$1^0-3^0-6^0+8^0 \equiv 1 - 1 - 1 + 1 \equiv 0 \mod 10$$ For $n = 1$: $$1^1-3^1-6^1+8^1 \equiv 1 - 3 - 6 + 8 \equiv 0 \mod 10$$ For $n = 2$: $$1^2-3^2-6^2+8^2 \equiv 1 - 9 - 36 + 64 \equiv 20 \equiv 0 \mod 10$$ For $n = 3$: $$1^3-3^3-6^3+8^3 \equiv 1 - 27 - 216 + 512 \equiv 270 \equiv 0 \mod 10$$ Since all the other exponents leading forward will be one of these cases, you are done. It is not the most elegant way to prove this, however, it is fairly direct.