Find the least value of $m+n$, where $1\le mI am no so sure how to do this without a calculator...
My method is like this.
So, $k=1978^n-1978^m$ is divisible by $1000$ which means $k$ is divisible by $8$ and $125$.
Since $8$ does not divide into $1978=2\times7\times 127$, $8$ divides $1978^m$ implies that $m\ge 3$.
Then obviously $125$ does not divide into $1978$ as well, so I need to figure out what $r$ is such that $1978^r\equiv 0\mod 125$. I have no idea how to find $r$ without using a calculator...
Could anyone help please?
 A: You are on the right track but you've made a small error.  Let $a = 1978$.  Then as you noted, $1000 \mid a^n - a^m$, which leads to $a^m (a^{n-m} - 1) = 1000k$ for some integer $k$.  Since $a^r - 1$ is odd for all positive integer $r$, we must have $8 \mid a^m$ so $m \ge 3$, and $125 \mid a^{n-m} - 1$.  So instead of having a power of $a$ being congruent to $0$ modulo $125$, it leaves a remainder of $1$.  From Euler's theorem, for $\gcd(a, b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(b)$ is the Euler totient function that counts the number of positive integers between $1$ and $b$ that are relatively prime to $b$.  One property of $\varphi$ is that if $b$ is the power of a prime, say $b = p^k$, then $$\varphi(p^k) = p^k - p^{k-1}.$$
Since $\gcd(1978,125) = 1$ and $125 = 5^3$, it follows that $\varphi(125) = 5^3 - 5^2 = 100$ and $$1978^{100} \equiv 1 \pmod {125}.$$  Thus $(m,n) = (3, 103)$ satisfies the given conditions.  All that is left is to show that such an $n$ is minimal; i.e., there is no smaller exponent $r^* < 100$ for which $a^{r^*} \equiv 1$.  This is not difficult to show and I leave as an exercise for the reader.
A: If $n>m$ are such that $1978^n - 1978^m$ is a multiple of $1000$, then we factorize as $1978^{m}(1978^{n-m} - 1)$ is a multiple of $1000$.
This is equivalent to $1978^{m}(1978^{n-m}-1)$ being a multiple of $8$ and $125$.

In the first case, note that $1978^{n-m} - 1$ is odd because $n \neq m$, hence coprime to $8$. $1978$ as a multiple of $2$ has multiplicity $1$ . So it follows that $m \geq 3$ is necessary and sufficient for divisibility by $8$ to hold.

In the second case, $1978^m$ is coprime to $125$ (by prime factorization) so we need the smallest value of $T = n-m$ such that $1978^T \equiv 1 \pmod{125}$.
Note that $1978 \equiv 103 \pmod{125}$. So we need to find $T$ such that $103^T \equiv 1 \pmod{125}$. Now, since $\phi(125) = 100$ it follows that $T$ is a divisor of $100$.
Start with the basic : if $103^T \equiv 1 \pmod{125}$ then in particular $103^T \equiv 1 \pmod{5}$, which comes to $3^T \equiv 1 \pmod{5}$, which is true only if $T$ is even. Thus, all odd values of $T$ are ruled out.
Next, we go modulo $25$ to see that $3^T \equiv 1 \pmod{25}$ must be true. We see that $\phi(25) =20$, so $T$ must be a factor of $20$. It is easy (because $3$ is small) to see that no number smaller than $20$ works. Thus, $T=20$ or $T = 100$.
Unfortunately, checking that $103^{20} \not \equiv 1 \pmod{125}$ is to be checked by hand. One can use repeated squaring to make this easier. (Also refer to the answer linked by J.W.Tanner below).
Thus, $T=100$, and $n = 3+100 = 103$ and $m=3,n=103$ form the smallest set of values with $m+n = 106$.
A: As mentioned, it boils down to computing the order $\,K\,$ of $\,a = 1978\equiv -22\pmod{\!5^3}$.
Note $\!\bmod 5^3\!:\ 1\equiv a^K\Rightarrow\bmod 5\!:\  1\equiv a^K\equiv 3^K\Rightarrow 4\mid K,\,$ so $\, K = 4k$
and $\bmod 5^2\!:\ a^4 \equiv 3^4\not\equiv 1,\,$ therefore we infer that $\,\color{#c00}{\nu_5(a^4\!-1) = 1}$
so applying LTE = Lifting The Exponent as here
$$\begin{align}
&5^{\large 3}\!\mid a^{\large 4k}-1\\[.2em]
\iff\ &3 \le \nu_5(a^{\large 4k}-1) = \color{#c00}{\nu_5(a^{\large 4}-1)}+\nu_5(k)\\[.2em]
\iff\ & 2\le \nu_5(k)\\[.2em]
\iff\ &5^{\large 2}\!\mid k\end{align}\qquad$$
so the least such $\,k\,$ is $\,5^2,\,$ so $\,a\,$ has order $\,K = 4k = 4\cdot 5^2=100\pmod{\!5^3}$.
