TITLE: Proof that ${2}^{3n+2}+{5}^{n+1}\text { is divisible by 3}$ using induction
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: $$
7\cdot2^{3k+2} + 4\cdot5^{k+1} = \Big(7\cdot2^{3k+2} + 7\cdot5^{k+1}\Big) - 3\cdot5^{k+1} = 7\Big(2^{3k+2} + 5^{k+1}\Big) - 3\cdot 5^{k+1}
$$
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 $3$".)