# Prove that if $a \equiv_4 b$ then $3^a \equiv_{10} 3^b$

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?

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

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

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

• Thank you! But i don't see why $$3^b ( 81^k-1)\equiv_{10} 0$$ holds ? Nov 5 '18 at 18:00
• $81^k$ ends up with $1$ therefore $81^k-1$ ends up with $0$ which makes it a multiple of $10$ Nov 5 '18 at 18:06

Apply Euler's theorem. (notice that 3 and 10 are coprimes plus $$\varphi(10)=4$$.)

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

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