Prove that $6^n=6^{n+5} (mod 100)$

Question

How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.

Answer 1

$6^n\equiv6^{n+5}\mod 100$ means $100\mid (6^{n+5}-6^n)$.

Notice that $6^{n+5}-6^n=6^n(6^5-1)=7775\times 6^n$.

Don't forget that $100=2^2\times 5^2$.

Edit: As noticed in the comments, this is true for $n\ge 2$ but not for $n=1$.

Answer 2

$$6^5-1=5(6^4+6^3+6^2+6+1)$$ and since $$6^4+6^3+6^2+6+1\equiv(1+1+1+1+1)(\mod5)=5,$$ we obtain $6^6-1$ is divisible by $25$.

Also, $6^n$ is divisible by $4$ for all $n\geq2.$

Answer 3

$\color{#c00}{a\mid 1\!+\!b,\ n\geq 2}\,\Rightarrow\, (\color{#c00}a\,\color{#0a0}{b})^{\large 2}\!\mid (1\!+\!b)^{\large n+b}\!-{(1\!+\!b)^{\large n}} = \overbrace{(\color{#c00}{1\!+\!b})^{\large\color{#c00} n}}^{\Large \color{#c00}{a^{\Large 2}}\,(\cdots)}\,\underbrace{\overbrace{((1\!+\!b)^{\large b} - 1)}^{\Large \color{#0a0}{b^{\Large 2}}(\cdots)}}_{\rm{ Binomial\ Theorem}}.\ $ Put $\,a=2,\,b=5$.

Answer 4

Well, if you hit a snag that means either... the thing you are trying to prove is false... or it's not actually a snag.

You say $6^{n+5}\equiv 76*6^n \pmod {100}$ does not give the correct result.

How do you know $76*6^n\not \equiv 6^n \pmod {100}$?

...

$76*6^n \equiv 6^n \pmod {100} \iff$

$75*6^n \equiv 0 \pmod {100} \iff$

$100 |75*6^n \iff$

$4|3*6^n = 3^{n+1}2^n \iff$ (as $2$ is prime)

$4|2^n \iff$

$n \ge 2$.

So your snag is only a snag if $n < 2$ in which case the result isn't true.

....

Alternatively we can use Chinese Remainder Theorem.

If $76*6^n \equiv k \pmod{100}$ then there is a unqiue solution to

$k \equiv 76*6^n \equiv 6^n \pmod{25}$ and to

$k \equiv 0 \pmod 4$

If $n = 0$ then $6^n \equiv 1 \pmod 4$ and our solution is $k = 6^n + 25m$ where $25m\equiv m \equiv 3\pmod 4$. (i.e. $m = 3$ and $6^5 = 7776 \equiv 76 = 1 + 75 = 6^0 + 75 \pmod {100}$).

If $n = 1$ then $6^n \equiv 2\pmod 4$ and our solution is $k = 6^n + 25 m $ where $25m \equiv m \equiv 2\pmod 4$ (i.e. $m = 2$ and $6^6= 46656 \equiv 56\equiv 6^1 + 50 \pmod {100}$).

If $n \ge 2$ then $6^n \equiv 0 \pmod 4$ and ore solution is $k =6^n + 25m$ where $25m \equiv m \equiv 0 \pmod 4$ and $6^{n+5}=7776*6^n \equiv 76*6^n = 75*6^n + 6^n \equiv 6^n \pmod {100}$.

...

Third option. $76*6^n = 50*6^n + 25*6^n + 6n^2 = 100(\frac {6^n}2+ \frac {6^n}4) + 6^n$.