Prove $5 \mid 2^{n+1} + 3^{3n+1}$

Question

I tried induction, so I assume the hypothesis and attempt to show $5 \mid 2^{n+2} +3^{3n + 4}$ but this doesn't help. I tried breaking it down into prime factorizations, but I do not see it.

Please reconcile the title & the body – lab bhattacharjee Sep 21 '13 at 15:51 — lab bhattacharjee, Sep 21 '13 at 15:51

lab bhattacharjee · Accepted Answer · 2013-09-21T16:41:08.477

5

$2^{n+1}+3^{3n+1}=2\cdot2^n+3\cdot(3^3)^n$

$\equiv2\cdot2^n+3\cdot(2)^n\pmod5$ as $3^3=27\equiv2\pmod5$

$\equiv2^n(2+3)\pmod5\equiv0$

edited Sep 21 '13 at 16:41

answered Sep 21 '13 at 15:48

lab bhattacharjee

274,582

Great answer, I didn't know I was supposed to use modulo arithmetic. – Don Larynx Sep 21 '13 at 15:55
It's $3^{3n+4}$ instead of $3^{n+4}$ – Stefan4024 Sep 21 '13 at 16:02
@Stefan4024, have you notified the difference between the title & the body of the question? – lab bhattacharjee Sep 21 '13 at 16:10
As far as I understand he tried to use unduction to solve the question, assuming that $5|2^{n+1} + 3^{3n+1}$ holds, then he tried to prove that this holds too: $5|2^{(n+1) + 1} + 3^{3(n+1) + 1} = 2^{n+2} + 3^{3n + 4}$ – Stefan4024 Sep 21 '13 at 16:37
@Stefan4024, good observation. – lab bhattacharjee Sep 21 '13 at 16:40

score 3 · Answer 2 · answered Sep 21 '13 at 15:50

3

Hints:

$$2^{n+1}+3^{3n+1}=2(2^n+3^{3n-2})+25\cdot 3^{2n-1}$$

and use induction on $\;n\;$ ...

answered Sep 21 '13 at 15:50

DonAntonio

211,718
17
136
287

score 2 · Answer 3 · answered Sep 21 '13 at 15:48

2

Hint: 1) $27 = 2 \mod 5$ and

2) $3=-2 \mod 5$

Can u do it now?

answered Sep 21 '13 at 15:48

Gautam Shenoy

10,318

user91500 · Answer 4 · 2014-04-23T09:52:59.173

0

Hint: $$\begin{align}2^{n+2}+3^{3n+4}&=3^3\left(3^{3n+1}+2^{n+1}\right)+\left(2^{n+2}-3^3(2^{n+1})\right)\\&=5k+2^{n+1}(2-3^3)\\&=5k+5\ell\end{align}$$

edited Apr 23 '14 at 09:52

answered Sep 21 '13 at 15:56

user91500

5,606

Prove $5 \mid 2^{n+1} + 3^{3n+1}$

4 Answers4