Prove by mathematical induction, that $81\times 3^{2n} - 2^{2n}$ is divisible by $5$ where $n \in \mathbb{Z}^+$

Question

For all $k$, the equation I came up with is $3^{4+2k} = 5m + 2^{2k}$ where $m$ is a positive integer.

For all $k+1$, the expression is $3^{6+2k} - 2^{2k+2}$.

I tried to plug in the first equation to reach an expression that can be expressed with the integer $5$ taken common, but I am unable to figure out the required manipulation of the expressions to reach the result.

Answer 1

Modulo $5$, you have

$$ 81\times 3^{2n} - 2^{2n}= 81\times 9^n - 4^n = 81\times (-1)^n - (-1)^n= 80\times (-1)^n = 0. $$

Answer 2

Let us prove this using induction.

Let $\lambda_n=81.(3^{2n}-2^{2n})$ where $n$ is an integer

Base step: $n=1\Rightarrow \lambda_1=81.(9-4)=81\times 5$ which is clearly divisible by five.

$\therefore $ Base step verified.

Induction hypothesis: For every $k\leq n,\space 5|\lambda_n$ where $k$ is an integer

Now for $k=n+1$ we have $\lambda_{n+1}=81.(3^{2n+2}-2^{2n+2})$ $$\lambda_{n+1}=81.(3^{2n}.3^2-2^{2n}.2^2)$$ $$\lambda_{n+1}=81.(3^{2n}.3^2-2^{2n}.2^2-2^{2n}.3^2+2^{2n}.3^2+3^{2n}.2^2-3^{2n}.2^2)$$ On rearranging the terms, we get: $$\lambda_{n+1}=81.((3^{2n}.3^2-2^{2n}.3^2)+(3^{2n}.2^2-2^{2n}.2^2)-(3^{2n}.2^2-2^{2n}.3^2))$$

$$\lambda_{n+1}=81.(9\lambda_n+4\lambda_n-36\lambda_{n-1})$$ now $5|\lambda_n$ and through our hypothesis, $5|\lambda_{n-1}$

$\therefore 5|\lambda_{n+1}$

Now this implies that our hypothesis also holds for integral values higher than $n$.

$\therefore $ our hypothesis holds for all positive integers.

Edit: I'm sorry that I didn't read the question properly. But this still can be solved from this result.

This result says that $5|81(3^{2n}-2^{2n})\Rightarrow 5|(81.3^{2n}-2^{2n})-80.2^{2n}$

Now as $5|80.2^{2n}\space \space (5|80)$, it is evident that $5|(81\times 3^{2n}-2^{2n})$ hence your result.

Answer 3

$$81\cdot3^{2n}-2^{2n}=5k\implies \\\begin{align}81\cdot3^{2(n+1)}-2^{2(n+1)}=81\cdot9\cdot3^{2n}-4\cdot2^{2n}&=9\cdot(81\cdot3^{2n}-2^{2n})+5\cdot2^{2n} \\&=9\cdot5k+5\cdot2^{2n} \\&=5k'.\end{align}$$

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Shorter:

$$81\cdot3^{2n}-2^{2n}=9^n-4^n\mod 5\implies 9^{n+1}-4^{n+1}=9\cdot9^n-4\cdot4^n=4(9^n-4^n)\mod5.$$

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Yet shorter:

$$81\cdot3^{2n}-2^{2n}=9^n-4^n=4^n-4^n\mod 5.$$

Answer 4

$3^{6+2k}-2^{2+2k}=4(3^{4+2k}-2^{2k})+5\times3^{4+2k}=4(5m)+5(l)=5(4m+l)$

Answer 5

Let $a_n=81\times 3^{2n} - 2^{2n}$, then notice $a_{n+1}-9a_n=5\cdot 4^n$ (verify). Now apply the induction hypothesis $5 \mid a_n$.