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Can anyone help me? Tried several times, but didn't get a solution.

Bill Dubuque
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gvncore
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 08 '19 at 07:55
  • I recomend learning modular arithmetic. It's not difficult, and it trivializes remainder problems. –  Oct 08 '19 at 07:56
  • It's essentially the same as the linked dupe: $\large\bmod 9!:\ 4^3\equiv 1,\Rightarrow, (4^3)^n\equiv 1^n\equiv 1\ \ $ – Bill Dubuque Oct 08 '19 at 14:27

5 Answers5

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Hint:

(1) Start with the induction base (here $n=0$): $4^{3\cdot 0} + 8 = 1+8 = 9$ which is divisible by $9$.

(2) In the induction step, assume that $4^{3n}+8$ is divisible by $9$ for some $n\geq 0$.

Prove that $4^{3(n+1)}+8$ is divisible by $9$ using that $4^{3n}+8$ is divisible by $9$. Just fill in the details.

Wuestenfux
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The base case should be clear. For the induction step write $4^{3(n+1)}+8$ in the form

$$4^{3(n+1)}+8=(4^{3n}+8) \cdot 4^3+8(1-4^3)$$

and show that both summands on the RHS are divisible by $9$.

Fred
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It is easy to see that equation is true for $P(1)$

Let us assume it is true for $P(n).$

$$ P(n)\implies 9|4^{3n}+8$$ $$ \implies 4^{3n} = 9\lambda - 8 $$ for some value of $\lambda.$

for $P(n+1)$ , we have: $$P(n+1) = 4^{3(n+1)} + 8$$ $$ =4^{3n}.4^3 + 8 $$ $$ = 8(4^{3n}.8 + 1) $$ $$ =8(72\lambda - 64+1) $$ $$=9(64\lambda -56)$$ which is clearly divisible by $9$.

The Demonix _ Hermit
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Hint:

If a term is $m=9n$, the next is $$4^3(9n-8)+8=64\cdot9n-56\cdot9.$$

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The solution here may become clearer if you notice that

$4^{3n} = (4^3)^n = 64^n$

and

$9|m+8 \iff m \equiv 1 \mod 9$

So you are asked to prove by induction that

$64^n \equiv 1 \mod 9 \quad \forall n \in \mathbb{N}$

gandalf61
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