Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$.

Question

I did most of it but I stuck here I attached my working tell me if I did correct or not thanks

My working:

EDITED: I wrote the notes as TEX Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.

Statement is true for $n = 1$: $$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$ $$5 \mod{5} = 0$$

Now for $n = p$ and $n = p + 1$: $$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$ $$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$

Please use MathJax to format mathematical expressions. For the rest, please use punctuation and proofread your writing. Where is your work so far? — Théophile, Nov 16 '17 at 17:39
Also remember to show the statement is not true for $n \leq 1$ — SuperSjoerdie, Nov 16 '17 at 17:43
Note that there's a difference between $2^3n-3^n$ and $2^{3n}-3^n$. The former is not always divisible by $5$. — Dustan Levenstein, Nov 16 '17 at 17:46
... because 1) $n$ is presumed to be a natural number so $n\ge 1$ is a assumed and 2) to proof something is true for CONDITION does NOT mean it is false if the condition is not true. — fleablood, Nov 16 '17 at 17:46
@fleablood Slightly, but it's true though. Even though we all know that for $n = 0$ $2^{3n} - 3^n = 0$ and that if $n < 0$ then $0 < r < 5$, it should be written down (I guess. I would've noted it at least). — SuperSjoerdie, Nov 16 '17 at 17:48
@SuperSjoerdie It's true for $n=0$, and for $n<0$ the expression is not even an integer, so it doesn't make sense to ask the question of divisibility by $5$. — Dustan Levenstein, Nov 16 '17 at 17:49
It's not relevent. To prove whales dance when the moon is full, you are under no obigation to prove whales do not dance when the moon is not full and the statement could still be true even if whales do dance when the moon is not full because our statement has nothing to do with non-full moons. — fleablood, Nov 16 '17 at 17:50
@fleablood Ah, now I see :') You're right, he doesn't have to — SuperSjoerdie, Nov 16 '17 at 17:50
In step three, your first line is what you need to prove., which is confusing. Your work after that doesn't make sense. Also, your handwriting is rather inconsistent. — Acccumulation, Nov 16 '17 at 17:59
Possible duplicate of Why $a^n - b^n$ is divisible by $a-b$? — lab bhattacharjee, Nov 16 '17 at 18:16

score 2 · Answer 1 · answered Nov 16 '17 at 17:42

2

Hint: $2^{3(n+1)}-3^{n+1} = 8(2^{3n}-3^{n})+3^n(8-3).$

answered Nov 16 '17 at 17:42

Math Lover

15,153

score 1 · Answer 2 · answered Nov 16 '17 at 17:44

1

we have $$2^{3n}-3^{n}=8^n-3^n\equiv 3^n-3^n\equiv 0\mod 5$$

answered Nov 16 '17 at 17:44

Dr. Sonnhard Graubner

95,283

score 1 · Accepted Answer · answered Nov 16 '17 at 17:53

We first show for the $n=1$ case:

$2^3-3 = 8-3 = 5$

which is clearly divisible by 5. Next is the inductive step where we assume

$2^{3n} - 3^n=5m $ for some integer $m$.

We can rewrite this as

$2^{3n}=5m+3^n$.

Now we want to prove that $2^{3(n+1)}-3^{n+1}$ is divisible by 5. We first can simplify this a bit:

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

We can now make the substitution from the previous statement to write this as

$2^3(5m+3^n)-3\cdot 3^n=5(8m)+8\cdot 3^n-3\cdot 3^n = 5(8m)+5\cdot3^n=5(8m+3^n)$

which is clearly divisible by 5 and proves the statement.

score 0 · Answer 4 · answered Nov 16 '17 at 17:47

0

By I.H. we have $2^{3n}-3^{n}=5k$ for some integer $k$ and thus $2^{3n}=3^{n}+5k$.

Now we have:

$$2^{3(n+1)}-3^{n+1} = 8\cdot 2^{3n}-3^{n+1}= 8(3^{n}+5k)-3^{n+1} = 3^n(8-3)+40k = 5\cdot (3^n+8k)$$

and we are done.

answered Nov 16 '17 at 17:47

nonuser

90,026

Very interesting, I put my solution at least 5 minutes before TheStrangeQuark and his solution is accepted, which is the same as mine, while my is not even useful. – nonuser Nov 16 '17 at 18:03

score 0 · Answer 5 · answered Nov 16 '17 at 18:01

It's hard to read your handwriting but it looks like you have the right idea but were sloppy in your execution and made so distributive error.

Assume if $n=k$ then statement is true and

$2^{3k} - 3^k = 5P$ for some integer $P$. (Always a good idea to specify what a variable is whenever you introduce it.)

$2^{3k+1} -3^{k+1}= 2*2^{3k} - 3$.

Theres two errors here: $2^{3(k+1)} \ne 2^{3k+1}$ and ... why did the $3^{k+1}$ turn into a $3$.

You should have:

$2^{3(k+1)} - 3^{k+1} = 2^{3k + 3} - 3^{k+1} = 8*2^{3k} - 3*3^{k}$

Can you go from there?

$8*2^{3k} - 3*3^{k} = 5*2^{3k} + 3*2^{3k} - 3*3^k = 5*2^{3k} + 3(2^{3k} - 3^k) = 5*2^{3k} + 3(5P) = 5(2^{3k} + 3P)$

Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$.

5 Answers5