Prove that $7 | (3^{2n} − 2^{n} )$ for every nonnegative integer $n$.

Question

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Prove that $7 | (3^{2n} − 2^{n})$ for every nonnegative integer $n$.

my proof:

Basic step (0): 7 | 1-1 which is true.

Ind. step: if $p(k) \implies p(k+1)$:

$3^{2k} − 2^{k}=7x$

$3^{2(k+1)} − 2^{k+1}=$

$3^{2k+2} − 2^{k+1}=$

$3^{2k}3^{2} - 2^{k}*2=$

$3^{2}(7x)*2= $ is this step correct?

$9*(7x)*2=$

126x which can be written as 7(18x) and therefore is divisible by 7.

Answer 1

You can notice that $3^{2n}=9^n$ and $9=2[7]$

Answer 2

Hint: use for the induction step $3^{2n+2}-2^{n+1}=9*3^{2n}-2*2^n=9(3^{2n}-2^n)+7*2^n$

and then the induction hypothesis!

Edit: long explanation for the 2nd step:$9*3^{2n}-2*2^n=(9*3^{2n}-2*2^n)-7*2^n+7*2^n=9*3^{2n}-9*2^n+7*2^n=9(3^{2n}-2^n)+7*2^n$

Answer 3

$$ \begin{aligned} \because 3^{2 n}-2^n &=9^n-2^n \\ &=(9-2)\left(9^{n-1}+9^{n-2} \cdot 2+9^{n-3} \cdot 3+\cdots+2^{n-1}\right) \\ &=7\left(9^{n-1}+9^{n-2} \cdot 2+9^{n-3} \cdot 3+\cdots+2^{n-1}\right) \\ \therefore 7|3^{2 n} &-2^n \end{aligned} $$

Answer 4

$$ \because 3^{2 n}-2^n=9^n-2^n \equiv 2^n-2^n \equiv 0(\bmod 7) $$ $$\therefore 7|3^{2n}-2^n$$