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Let $a,b \in \mathbb{Z}$. How can I prove that $a \equiv_4 b \rightarrow 3^a \equiv_{10} 3^b$ with basic number theory?

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Note that from $$a \equiv b \mod (4) $$ we get $a=4k+b$

Thus $$3^a-3^b = 3^{4k+b}-3^b $$

$$=3^b ( 81^k-1)\equiv 0 \mod (10)$$

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  • $\begingroup$ Thank you! But i don't see why $$3^b ( 81^k-1)\equiv_{10} 0 $$ holds ? $\endgroup$
    – kerlifa
    Nov 5 '18 at 18:00
  • $\begingroup$ $81^k$ ends up with $1$ therefore $81^k-1$ ends up with $0$ which makes it a multiple of $10$ $\endgroup$ Nov 5 '18 at 18:06
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Apply Euler's theorem. (notice that 3 and 10 are coprimes plus $\varphi(10)=4$.)

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Hint: $3^{4+n}=3^4\cdot3^n\equiv_{10}3^n$.

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$\phi(10)=4$, so $x^4\equiv_{10}1\iff\gcd(x,10)=1$. Now, $a-b\equiv_4 0$, so $$3^{a-b}\equiv_{10}1\iff 3^a\equiv_{10}3^b$$

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