The sides $a>b>c$ of a triangle are integers such that $3^a, 3^b, 3^c$ leave the same remainder when divided by $10000$. The sides $a>b>c$ of a triangle are integers such that $3^a, 3^b, 3^c$ leave the same remainder when divided by $10000$. Find the minimum value of the perimeter of such a triangle.

Since $3^a, 3^b,3^c$ leave the same remainder $$3^a \equiv 3^b\equiv 3^c\pmod{10000}$$
I feel stuck after this... Any help ?
 A: $10000\mid 3^b-3^c=3^c(3^{b-c}-1)$ So $10000\mid 3^{b-c}-1$.
And similarly $10000\mid 3^{a-b}-1$.
So you need those two conditions and $b+c>a$.
A: Hint: If $r, n$ are coprime, then the sequence of remainders of powers $r^m$ modulo some number $n$ is a sequence of powers of an element $[r]$ in the multiplicative group $\mathbb{Z}_n^{\times}$. Since $10000 = 2^4 5^4$, the Chinese Remainder Theorem gives that the ring $\mathbb{Z} / 10000 \mathbb{Z}$ is isomorphic to $(\mathbb{Z} / 2^4 \mathbb{Z}) \times (\mathbb{Z} / 5^4 \mathbb{Z})$, and so $$(\mathbb{Z} / 10000 \mathbb{Z})^{\times} \cong (\mathbb{Z} / 2^4 \mathbb{Z})^{\times} \times (\mathbb{Z} / 5^4 \mathbb{Z})^{\times} \cong \mathbb{Z} / (2^3 \cdot 1) \mathbb{Z} \times \mathbb{Z} / (5^3 \cdot 4) = \mathbb{Z} / 8 \mathbb{Z} \times \mathbb{Z} / 500 \mathbb{Z}.$$ So, every element of $\mathbb{Z} / \mathbb{Z}_{10000}^{\times}$ has order dividing $500$ (this is the Carmichael Number for $n = 10000$). On the other hand, $3^k \not\equiv 
1 \bmod 10000$ for $k = 100, 250$, so $3$ has order $500$. Thus, $3^a \equiv 3^b \equiv 3^c \bmod 10000$ iff $a \equiv b \equiv c \bmod 500$, that is, iff $a = c + 500 k, b = c + 500 l$, for some integers $k, l$ such that $k > l > 0$.
Now, the positive numbers $a, b, c$, $a \geq b \geq c$, form the sides of a triangle iff $a < b + c$, or in our case, $500 (k - l) < c$.
