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Apart from changing $p$ to $q$, this is from an unanswered question that the OP deleted about a week ago. It asks to prove

$$\operatorname{lcm}(m, q)=\operatorname{lcm}(n, q) \;\;\to\;\; \operatorname{lcm}(m + n, q) \ge \operatorname{lcm}(n, q) \tag{1}\label{eq1A}$$

Although the original question doesn't specify what the variables are, this question will have them all be positive integers. Using the $p$-adic valuation function, a sufficient condition to prove the RHS of \eqref{eq1A} is that, for all primes $p$,

$$\nu_p(\operatorname{lcm}(m + n, q)) \ge \nu_p(\operatorname{lcm}(n, q)) \tag{2}\label{eq2A}$$

Using the prime factorizations, we get for the LHS of \eqref{eq1A} for some non-integer $r$ that

$$\nu_p(\operatorname{lcm}(m, q)) = \max(\nu_p(m), \nu_p(q)) = \nu_p(\operatorname{lcm}(n, q)) = \max(\nu_p(n), \nu_p(q)) = r \tag{3}\label{eq3A}$$

If $\nu_p(q) = r$, then $\nu_p(\operatorname{lcm}(m + n, q)) \ge r$. Otherwise, we have $\nu_p(m) = \nu_p(n) = r$, so $p^r \mid m \;\land\; p^r \mid n \;\to\; p^r \mid m + n$. Thus, we once again have $\nu_p(\operatorname{lcm}(m + n, q)) \ge r$. This means \eqref{eq2A} holds in both cases, proving the RHS of \eqref{eq1A} is true.

Are there other methods to prove this?

John Omielan
  • 52,653
  • Basic lcm & gcd properties also quickly prove the inequality by proving $,{\rm lcm}(n,q)\mid {\rm lcm}(m+n,q)$. This yields a "global" proof (vs. your above "local" proof) of this divisibility - see my answer. $\ \ $ – Bill Dubuque May 05 '24 at 02:24

2 Answers2

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Using notation $\begin{align}[x,y] &:= {\rm lcm}(x,y),\\ (x,y)&:= \gcd(x,y)\end{align}\,$ and $\,\color{#c00}{\rm D}:=$ lcm distributes over gcd

$\qquad\underbrace{[q,n]=(\color{#0af}{[q,m]},[q,n])}_{\textstyle \color{#0af}{[q,m]}=[q,n]} \overset{\color{#c00}{\rm D}}= [q,\color{#0a0}{(m,n)}]\ {\large \mid}\ [q,\color{#0a0}{m\!+\!n}]\ $ since $\ \color{#0a0}{(m,n)\mid m\!+\!n}$

Bill Dubuque
  • 282,220
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With the LHS of the question's \eqref{eq1A}, using that $\operatorname{lcm}$ is an integral multiple of $q$, and the result of posts such as Prove $\,\gcd(m,n)\operatorname{lcm}(m,n) = mn$, then for some positive integer $k_1$, we can rewrite it to be

$$\frac{mq}{\gcd(m, q)} = \frac{nq}{\gcd(n, q)} = k_{1}q \;\;\to\;\; m = \gcd(m, q)k_1, \; n = \gcd(n, q)k_1 \tag{1}\label{eq1B}$$

Next, we also have

$$m_1 = \gcd(m, q), \; n_1 = \gcd(n, q), \; e = \gcd(m_1, n_1) \;\to\; m_1 = ef, \, n_1 = eg, \, \gcd(f, g) = 1 \tag{2}\label{eq2B}$$

Substituting this into the RHS of \eqref{eq1B} gives

$$m = efk_1, \; n = egk_1 \tag{3}\label{eq3B}$$

From $m_1 \mid q$ and $n_1 \mid q$, we also get

$$ef \mid q \;\to\; q = efk_2, \;\; eg \mid efk_2 \;\to\; g \mid fk_2 \;\to\; g \mid k_2 \;\to\; k_2 = gk_3 \;\to\; q = efgk_3 \tag{4}\label{eq4B}$$

Due to $\gcd\left(\frac{m}{\gcd(m,q)}, \frac{q}{\gcd(m,q)}\right) = 1$, and the similar thing for $n$, along with the RHS of \eqref{eq1B}, \eqref{eq2B} and \eqref{eq4B}, we have

$$\gcd(k_1, gk_3) = 1 \;\land\; \gcd(k_1, fk_3) = 1 \;\to\; \gcd(k_1, fgk_3) = 1 \tag{5}\label{eq5B}$$

Using the above, and the result of posts like Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $, we get

$$\gcd(m + n, q) = \gcd(e(f + g)k_1, efgk_3) = e\gcd((f + g)k_1, fgk_3) = e\gcd(f + g, fgk_3) \tag{6}\label{eq6B}$$

For positive integers $a$ and $b$, $a \ge \gcd(a, b)$ and $b \ge \gcd(a, b)$, so we finally have

$$\begin{equation}\begin{aligned} e(f + g) & \ge e\gcd(f + g, fgk_3) \\ (e(f + g)k_{1})q & \ge \gcd(m + n, q)k_{1}q \\ \frac{(m + n)q}{\gcd(m + n, q)} & \ge k_{1}q \\ \operatorname{lcm}(m + n, q) & \ge \operatorname{lcm}(n, q) \end{aligned}\end{equation}\tag{7}\label{eq7B}$$

John Omielan
  • 52,653