Show that for all integers

Question

Show that for all integers n ≥ 1 applies that $5|2^{3n} - 3^{n}$

can anyone help?

$2^{3n}-3^n=8^n-3^n=(8-3)(\sum_{i=0}^{n-1} 8^{n-1-i}3^i)$ – Sep 19 '19 at 13:27 — , Sep 19 '19 at 13:27

score 0 · Answer 1 · answered Sep 19 '19 at 13:27

0

It is a one line proof: $$2^{3n}-3^n=8^n-3^n\equiv 3^n-3^n\equiv 0 \mod 5$$

answered Sep 19 '19 at 13:27

score 0 · Answer 2 · answered Sep 19 '19 at 13:27

0

Tips:

$2^{3n}=8^n$

$a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+b^{n-1}\right)$

answered Sep 19 '19 at 13:27

MafPrivate

J. W. Tanner · Answer 3 · 2019-09-19T15:32:24.310

0

Hint:

You could prove it by mathematical induction using

$2^{3(n+1)}-3^{n+1}=8\times2^{3n}-3\times3^n=5\times2^{3n}+3(2^{3n}-3^n)$

edited Sep 19 '19 at 15:32

answered Sep 19 '19 at 13:36

J. W. Tanner

3 Answers3