<s>[INST] If $\text{lcm}(a,b,c,d)=a+b+c+d$, then $\gcd(abcd,15)>1$ $\text{LCM}(a,b,c,d)=a+b+c+d$, where $a,b,c,d\in\mathbb Z^+$.  
Prove that $abcd$ is divisible by at least one of $3$ and $5$.
I am not sure how to create this proof. Does it involve divisibility tests? How should I set it out?
 [/INST] We suppose that $a\ge b\ge c\ge d\ge 1$ and $s=a+b+c+d$ is the lowest common multiple of $a$, $b$, $c$ and $d$.
Now, by contradiction, we suppose that $s$ is not a multiple of $3$ nor $5$.
Note that if $a=b=c=d$ then $4a=s$ (because $s$ is the sum) and $s=a$ (because $s$ is the lcm). This is not possible.


*

*As $a<s<4a$ and $s$ is a multiple of $a$, then $s=2a$ (can't be $3a$) and $b+c+d=a$, so $b<a$

*$a=b+c+d\le 3b$. Hence $2b<s=2a\le 6b$. So $s=4b$ (can't be $3b$, $5b$ or $6b$). $a=2b$ and $c+d=b$, so $c<b$

*$b=c+d\le 2c$. Hence $4c<s=4b\le 8c$. So either $s=7c$ or $s=8c$ (can't be $5c$ or $6c$).


Now two possibilities :


*

*$s=8c=4b=2a$ Hence $d=c$, and $s=lcm(a,b,c,d)=a$. A contradiction

*$s=7c=4b=2a$ Hence, there is a $k$ such that $a=14k$, $b=7k$, $c=4k$ and $d=3k$ (because $s=a+b+c+d=2a=28k$), so $d$ is a multiple of $3$, contradiction.


Hence, $s$ is a multiple of $3$ or $5$ and as $s$ divides $abcd$, so is $abcd$.
Note that it is obviously always a multiple of $2$ also, because if $a$, $b$, $c$, $d$ are all odd, then $s=a+b+c+d$ would be even, a contradiction.
Some examples :


*

*only multiple of $3$ : 4,  3,  3,  2 

*only multiple of $5$ : 5,  2,  2,  1

*multiple of $15$ : 15, 10,  3,  2



After that, you can prove with a similar method that there only 9 different (with $a\ge b\ge c\ge d$) solutions with $gcd(a,b,c,d)=1$, and that all other solutions are multiples of one of this nine primitive solution.
$$
\begin{array}{cccc|c}
a&b&c&d&s\\\hline
 4&  3&  3&  2 &   12\\
  4&  4&  3&  1 &   12\\
  5&  2&  2&  1 &   10\\
  6&  4&  1&  1 &   12\\
  9&  6&  2&  1 &   18\\
 10&  5&  4&  1 &   20\\
 12&  8&  3&  1 &   24\\
 15& 10&  3&  2 &   30\\
 21& 14&  6&  1 &   42\\
\end{array}
$$
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