$\operatorname{lcm}(n,m,p)\times \gcd(m,n) \times \gcd(n,p) \times \gcd(n,p)= nmp \times \gcd(n,m,p)$, solve for $n,m,p$?

Question

$\newcommand{\lcm}{\operatorname{lcm}}$

I saw this in the first Moscow Olympiad of Mathematics (1935), the equation was : $$\lcm(n,m,p)\times \gcd(m,n) \times \gcd(n,p)^2 = nmp \times \gcd(n,m,p)$$

My attempt :

I've multiplied both sides of the equation by $\frac{1}{\gcd(n,m,p)}$ to get this ( I don't know why i did ): $$\frac{\lcm(n,m,p)}{\gcd(n,m,p)}\times \gcd(m,n) \times \gcd(n,p)^2 = nmp$$ then I've multiplied both sides by $\gcd(n,m,p)$, I got this but I get stuck here actually: $$\frac{nmp\times \not{nmp}\times \gcd(m,n) \times \gcd(n,p)^2}{\gcd(n,m,p)}=\not{nmp}$$ Finally: $$\gcd(m,n)\times \gcd(n,p)^2\times nmp=\gcd(n,m,p).$$

Answer 1

HINT:$\newcommand{\lcm}{\operatorname{lcm}}$

Use the prime power decomposition. Let $\ell,$ $m$ and $n$ be your integers. We can write $$\begin{eqnarray*} \ell &=& p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k} \\ \\ m &=& p_1^{b_1}p_2^{b_2}\ldots p_k^{b_k} \\ \\ n &=& p_1^{c_1}p_2^{c_2}\ldots p_k^{c_k} \end{eqnarray*}$$

where each of the $p_i$ are distinct primes and the $a_i$, $b_i$ and $c_i$ are non-negative integers. E.g.:

$$\begin{eqnarray*} \ell &=& 2^3 \times 3^0 \times 11^1 \\ \\ m &=& 2^0 \times 3^2 \times 11^3 \\ \\ n &=& 2^9 \times 3^2 \times 11^0 \end{eqnarray*}$$

Two nice peroperties are that

\begin{eqnarray*} \gcd(\ell,m,n) &=& p_1^{\min(a_1,b_1,c_1)}\ldots p_k^{\min(a_k,b_k,c_k)} \\ \\ \lcm(\ell,m,n) &=& p_1^{\max(a_1,b_1,c_1)}\ldots p_k^{\max(a_k,b_k,c_k)} \\ \\ \end{eqnarray*}

Similar statements fold for two numbers (ignore $n$ and all of the $c_i$).

Can you use these to help you solve the problem?

Answer 2

Why not write it as $ \min(a,b,c) + \max(a,b) + 2 \max(b,c) = (a+b+c) + \max(a,b,c)$?

By unique factorization we can look at the the exponents for reach prime $m = q^a\times m_1, n = q^b \times n_1, p = q^c \times p_1$. Then observe$\newcommand{\lcm}{\operatorname{lcm}}$

$$ \gcd(m,n) = p^{\max(a,b)}\times m' \text{ and } \lcm(m,n) = p^{\min(a,b)}\times m'' $$

I don't know if you mean $p$ to be prime in your notation.

Answer 3

Using the standard trick: $$d=\gcd(n, m, p), u=\frac{\gcd(n, m)}{\gcd(n, m, p)}, v=\frac{\gcd(n, p)}{\gcd(n, m, p)}, w=\frac{\gcd(m, p)}{\gcd(n, m, p)}$$

we may write$\newcommand{\lcm}{\operatorname{lcm}}$

$$n=duvn_1, m=duwm_1, p=dvwp_1$$

where

$$\gcd(vn_1, wm_1)=\gcd(un_1, wp_1)=\gcd(um_1, vp_1)=1$$

This gives

$$\gcd(u, v)=\gcd(u, w)=\gcd(v, w)=1$$ $$\gcd(n_1, m_1)=\gcd(n_1, p_1)=\gcd(m_1, p_1)=1$$ $$\gcd(u, p_1)=\gcd(v, m_1)=\gcd(w, n_1)=1$$

Then we have $$\lcm(n, m, p)=duvwn_1m_1p_1, \gcd(m, n)=du, \gcd(n, p)=dv, nmp=d^3u^2v^2w^2n_1m_1p_1$$

Thus the equation becomes

$$(duvwn_1m_1p_1)(du)(dv)^2=(d^3u^2v^2w^2n_1m_1p_1)d$$ $$v=w$$

Since $\gcd(v, w)=1$, $v=w=1$. Conversely if $v=w=1$ and all the $\gcd$ conditions above hold, then the given equation holds.

Thus all solutions are given by

$$n=dun_1, m=dum_1, p=dp_1$$ where $d, u, n_1, m_1, p_1$ are any positive integers satisfying

$$\gcd(u, p_1)=\gcd(n_1, m_1)=\gcd(n_1, p_1)=\gcd(m_1, p_1)=1$$