How many pairs $n$, $m$ exist such that $n \wedge m=50$ and $n \vee m=75$?

And such that $n \wedge m=50$ and $n \vee m=600$?
For the first one, the answer is 0! Indeed, a bit of reflection shows that we must always have $n \wedge m \mid n \vee m$, but $50 \nmid 75$.

For the second, we start by decomposing into prime factors: $50=2 \times 5^{2}$ and $600=2^{3} \times 3 \times 5^{2}$. We also know that if $n$, $m$ are solutions, then $n m=50 \times 600=2^{4} \times 3 \times 5^{4}$. Finally, $50 \mid n$ and $50 \mid m$.

In conclusion, the solutions will be exactly of the form $m=50 \times 2^{a} \times 3^{b}$ and $\mathfrak{n}=50 \times 2^{2-a} \times 3^{1-b}$, with $0 \leqslant a \leqslant 2$ and $0 \leqslant b \leqslant 1$. In conclusion, we have 3 choices for $a$ and 2 for $b$, making a total of 6 pairs of solutions.

We conclude this section with a fundamental result due to Gauss, hence its name.

Gauss's Lemma: Let three integers $a$, $b$, and $c$ be such that $a \mid b \times c$. If $a \wedge b=1$, then $a \mid c$.

With what we have seen previously, this result is quite intuitive. The hypothesis $a \mid b \times c$ can be translated as the fact that we find all the elements of the prime factorization of $a$ in the product $b \times c$. But since $a \wedge b=1$, none of these elements appear in the decomposition of $b$. They must therefore all come from $c$, hence the conclusion.