The following problem is taken from "All Soviet Union Math Competition 1978":
Prove that there is no $m$ for that $(1000^m - 1) \mid (1978^m - 1)$
A direct solution is using LTE lemma: $$\nu_3(1000^m-1) = \nu_3(1000-1) + \nu_3(m) = 3 + \nu_3(m)$$but $$\nu_3(1978^m-1) = \nu_3(1978-1) + \nu_3(m) = 1 + \nu_3(m)$$so $\nu_3(1978^m-1) < \nu_3(1000^m-1)$.
I still wonder if there is a simpler solution for this problem (since this is a problem for 8th grade pupils so LTE seems overkill, IMHO).
Suppose towards contradiction that there exists such an $m$, and write $$1978^m - 1 = 1978^m - 1000^m + 1000^m - 1.$$ Clearly $1000^m - 1$ is divisible by itself, so we only need to show that $$1978^m - 1000^m = 2^m \cdot (989^m - 500^m)$$is divisible by $1000^m - 1$. Now since $\mathrm{gcd}(2^m, 1000^m - 1) = 1$, we must have that $$(1000^m - 1) \mid (989^m - 500^m).$$ But this doesn't hold, since for $m > 0$, $989^m - 500^m < 1000^m - 1$ (and this inequality is true because clearly $989^m + 1 < 1000^m + 500^m$). Contradiction.
Given said level, likely it was intended to use the (binary, subtractive) Euclidean gcd algorithm. This computes gcds using the following reduction step $\,(A,B) = (A,B\!-\!A)\,$ if $\,B > A,\,$ then factor out any common factors of $\,2,\,$ leaving at most one $\rm\color{#0af}{even}$ argument, from which we cancel all factors of $\,2\,$ (valid by $\,2\nmid m\Rightarrow (m,\color{#0af}{2^k} n) = (m,n)\,$ by Euclid). For example
$$\begin{align} (48,176) &= (48,128)\ \ \text{by subtracting least from greatest}\\[.1em] &= 16\:\!(3,\color{#0af}8)\ \ \ \ \text{by factoring out as many $2$'s as possible}\\[.1em] &= 16(3,\color{#c00}1)\!=\!16,\ \text{by cancelling all $2$'s from $\,\color{#0af}8,\,$ leaving $\,\color{#c00}1$} \end{align}\qquad$$
More complicated gcds require more than this single step, e.g. above instead of $(3,1)$ we may reach $(m,n)\,$ for odd $\,m,n>1\,$ so we need to inductively (recursively) apply the algorithm to compute that smaller gcd. The same idea works for any radix, e.g. for decimal we extract/cancel factors of $\,2\,$ and $\,5\,$ (easily recognizable from the units digit).
Like the above example, the OP requires only a single step - which we perform in the proof below. To aid future use we do a slightly generalized problem, for base $\,e.\,$ Specializing the Lemma below yields essentially the same arithmetic as in the accepted answer (but we show how to discover it).
Lemma $ $ If $\,A =\color{#c00}a\:\!e^m\!+c,\,\ B = \color{#0a0}b\:\!e^m\!+c,\,\ (e,c)\!=\!1\,$ then $\,A\mid B\iff A\mid \color{#0a0}b\!-\!\color{#c00}a$
Proof $\,\ A\mid B\!\iff\! A \!=\! (A,B) \!=\! (A,B\!-\!A) \!=\! (A,(b\!-\!a)\color{#0af}{e^m}) = (A,b\!-\!a)$ $\!\iff\! A\mid\color{#0a0}b\!-\!\color{#c00}a\,$ where we cancelled $\,\color{#0af}{e^m}$ by Euclid, using $\:\!\color{0a0}e\:\!$ is coprime to $A$ by $\,(A,e) = (c,e) = 1$. $\ \ \bf\small QED$
OP has $\:\!A = \color{#c00}{500^m}2^m-1,\ B = \color{#0a0}{989^m} 2^m-1\,$ so $\:\!A\mid \color{#0a0}b\!-\!\color{#c00}a\,$ fails here since $\,A > b\!-\!a>0\,$ (i.e. we have: $\,A\!=\!1000^m\!-\!1 > 989^m > \color{#0a0}{989^m}\!-\!\color{#c00}{500^m}\! = \color{#0a0}b\!-\!\color{#c00}a),\,$ thus $\,A\nmid B\,$ by the Lemma.
Remark $ $ The point of explicitly presenting it as a special case of one form of the Euclidean algorithm is that this ubiquitous method shows us how to discover the solution using general methods. We could instead eliminate all use of the algorithm and present the proof using only divisibility arithmetic - but that would obfuscate the idea that led to the discovery of the proof.
It deserves to be better known that there is an analogous algorithm for polynomials, e.g. see here where I explain how it can be used to understand a note by Joe Silverman on almost inverses.
$$1000^\color{red}{n}-1=999\times(n \pmod 3)$$ $$1978^m-1=1977\times(m\pmod 3)$$ And $999$ does not divide $1977$.
Thinking modulo $3$, I think a desperate Russian student like Nu can conclude...
