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I'm am a little bit stuck on this question, any help is appreciated.

Show that for every $n\in\mathbb{N}$, $3^{4n+2} + 1$ is divisible by $10$.

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@DJC: Do you mean ends with $9$? $81$ is a multiple of $9$... – Aryabhata Feb 28 '11 at 17:16
@Moron. Oops! I meant it is enough to show that $3^{4n+2} \equiv~9~mod~10$. – JavaMan Feb 28 '11 at 17:19
Use Euler's totient function. Eric has given almost the answer. Look at – Sunil Feb 28 '11 at 21:11
up vote 6 down vote accepted

Hint: $$3^{4n+2}=3^{4n}\cdot 9$$ and $$3^{4n}\equiv 1 \pmod{10}$$ since $\phi(10)=4$.

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hint : $3^{4n+2} = (10-1)^{2n+1}$.

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What is the formation law of the remainders of the division by $10$ of the powers $3^{n}$?

(For the notation see modular arithmetic.)

$$\left\{ \begin{array}{c} 3\equiv 3\quad \pmod{10} \\ 3^{2}\equiv 9\quad \pmod{10} \\ 3^{3}\equiv 7\quad \pmod{10} \\ 3^{4}\equiv 1\quad \pmod{10} \end{array}\right. $$

$$\left\{ \begin{array}{c} 3^{5}\equiv 3\quad \pmod{10} \\ 3^{6}\equiv 9\quad \pmod{10} \\ \cdots \\ \cdots \end{array}\right. $$

So, with respect to the divisor $10$ the remainders of $3^{n}$ are periodic ($3,9,7,1,3,9,7,1,\dots$) with period $4$. This together with $4n+2\equiv 2\quad \pmod{4}$ yields $3^{4n+2}\equiv 3^{2}\quad \pmod{10}$. Also $1\equiv 1\quad \pmod{10}$. Hence, for all $n\ge 1$ $$3^{4n+2}+1\equiv 3^{2}+1\equiv 0\quad \pmod{10},$$

which means the remainders of the devision of $3^{4n+2}+1$ by $10$ are $0$.

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$$9+1=10$$ $$81 \times \left( 3^{4n-2} +1 \right) = 3^{4n+2} + 1 +80$$

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Yes, induction works too. – lhf Mar 2 '11 at 14:16

HINT $\rm\ \ \ A^K\: \equiv -1\ \ \Rightarrow\ \ A^{\:J+2\:K}\ \equiv\ A^{\:J}$

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$3^{4n+2} = 3^{2(2n+1)} = 9^{2n+1}$. So your question is really asking about odd powers of 9 and then adding 1. Let's look at powers of 9, and focus on the last digit (ie, take them mod 10): $9^1 = 9,\ 9^2 = 1,\ 9^3=9^1\cdot 9^2=9 \cdot 1 = 9, \cdots$. So the last digit in consecutive powers of 9 goes 9, 1, 9, 1, 9, 1, ...

Now consider adding 1 to only the odd terms above.

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