Proof that ${2}^{3n+2}+{5}^{n+1}\text { is divisible by 3}$ using induction

Question

I am having trouble with a proof by induction exercise.

My book shows the typical steps for proving divisibility induction with the number 3 lets say are as following:

Prove true for $n=1$
Assume true for $n=k$
$f(k+1)-f(k)$ getting 3 as a factor
Rearrange $f(k+1)=f(k)$(assumed divisible by 3) + previous result (3 as a factor so divisible)
Conclusion

However I am stuck at this question:

Prove using induction that

${2}^{3n+2}+{5}^{n+1}\text { is divisible by 3}$

Step 1 $${2}^{3n+2}+{5}^{n+1} \text { divisible by 3 when n=1}$$ Step 2 $$\text { assume true for n=k}$$ $$f(k) ={2}^{3k+2}+{5}^{k+1} $$ Step 3 $$f(k+1) -f(k) ={2}^{3k+5}+{5}^{k+2}-{2}^{ 3k+2}-{5}^{k+1}$$ $$=8(2^{3k+2})-{2}^{3k+2}+5(5^{k+1})-{5}^{k+1}$$ $$=7(2^{3k+2})+4(5^{k+1})$$

And being unable to take 3 as a factor I am stuck at the last part. What should I do next?

So what happens if you split your first term to match the second - four times something which matches your $k$ case plus another term? Can you see a factor three in two terms for two different reasons - one obvious and one inductive? — Mark Bennet, May 20 '12 at 17:12
You have something times 4 and something else times 7 - is that suggestive? — Mark Bennet, May 20 '12 at 17:20
Try $f(k+1) - 5f(k)$ instead of $f(k+1) - f(k)$ so you can cancel the powers of $5$. — Zarrax, May 20 '12 at 17:22

score 6 · Accepted Answer · answered May 20 '12 at 17:20

6

$$ 7\cdot2^{3k+2} + 4\cdot5^{k+1} = \Big(7\cdot2^{3k+2} + 7\cdot5^{k+1}\Big) - 3\cdot5^{k+1} = 7\Big(2^{3k+2} + 5^{k+1}\Big) - 3\cdot 5^{k+1} $$ The expression in parentheses is divisible by $3$ because of the induction hypothesis, which you labeled "step 2". (But in "step 2" you forgot to add the words "is divisible by $3$".)

answered May 20 '12 at 17:20

Michael Hardy

1

I understood what you did in the end but I don't see how you got from $7\cdot2^{3k+2} + 4\cdot5^{k+1}$ to $\Big(7\cdot2^{3k+2} + 7\cdot5^{k+1}\Big) - 3\cdot5^{k+1}$ – εν οίδα ότι ουδέν οίδα May 20 '12 at 17:26
@Panayiotis He just rewrote 4 as 7-3 in order to be able to factor out a 7. – chris May 20 '12 at 17:29
@chris Ah now it makes sense, Thanks! That seems to be the books way of answering it aswell – εν οίδα ότι ουδέν οίδα May 20 '12 at 17:31
@Panayiotis This is a typical situation - the point is to learn how to spot it when you are faced by a question you haven't seen before. – Mark Bennet May 20 '12 at 17:34

Proof that ${2}^{3n+2}+{5}^{n+1}\text { is divisible by 3}$ using induction

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