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$n \in N$ is positive integer, and $64^n-7^n$ can be divisible by 57. Prove that $8^{2n+1}+7^{n+2}$ is also divisible by 57.

jiadong
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  • it is gg if you have not tried – meonstackexchange Mar 17 '15 at 13:47
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    I have tried to change the form of $8^{2n}+7^{n+2}$ to have relation to $64^n-7^n$ and I also have tried with Mathematical induction, but I failed. – jiadong Mar 17 '15 at 13:47
  • @jiadong If you know congruences (modular aritmetic) then it is best to view it simply as the multiplication of two congruences - see my answer. It is essential to learn congruences in order to master elementary number theory. – Bill Dubuque Mar 17 '15 at 14:40

2 Answers2

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HINT

$$8^{2n+1}+7^{n+2} = 8\cdot 64^n + 49\cdot 7^n =8\cdot 64^n + (57-8)\cdot 7^n =\color{blue}{ 8(64^n-7^n) + 57\cdot 7^n} $$

AgentS
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Conceptually it's just congruence $\rm\color{#c00}{multiplication}$ using $\,\color{#0a0}{\rm CPR} =\,$ Congruence Product Rule

$$\begin{array}{rrl}{\rm mod}\,\ 57\!:\!\!\!\! & -8\!\!\!\!&\equiv\, 49\\ &64^n\!\!\!\!&\equiv\, 7^n\\ \color{#c00}\Rightarrow &-8\cdot 64^n\!\!\!\!&\equiv\, 49\cdot 7^n\ \ {\rm by}\ \ \color{#0a0}{\rm CPR}\\ {\rm i.e.} &\,-8^{2n+1}\!\!\!\!&\equiv\, 7^{n+2}\end{array}\qquad\qquad$$

The proof is ganeshie8's answer is precisely the proof of CPR specialized to these numbers.

Remark $\ $ Similarly $\ 64\equiv 7\,\Rightarrow\, 64^n\equiv 7^n\,$ is an instance of the Congruence Power Rule, which is simply an inductive extension of the Product Rule.

Community
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Bill Dubuque
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  • Thanks!! Having the CPR in mind, I feel that it is much more natural to find a way to prove this problem! – jiadong Mar 17 '15 at 15:16
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    @jiadong Yes, congruence arithmetic allows us to convert many divisibility problems into trivial congruence arithmetic (which, being so similar to integer arithmetic, allows us to reuse our grade-school intuition, which is lacking for arithmetic of divisibility relations). – Bill Dubuque Mar 17 '15 at 15:19