If $\text{lcm}(a,b,c,d)=a+b+c+d$, then $\gcd(abcd,15)>1$

Question

$\text{LCM}(a,b,c,d)=a+b+c+d$, 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?

Answer 1

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.

1. As $a<s<4a$ and $s$ is a multiple of $a$, then $s=2a$ (can't be $3a$) and $b+c+d=a$, so $b<a$
2. $a=b+c+d\le 3b$. Hence $2b<s=2a\le 6b$. So $s=4b$ (can't be $3b$, $5b$ or $6b$). $a=2b$ and $c+d=b$, so $c<b$
3. $b=c+d\le 2c$. Hence $4c<s=4b\le 8c$. So either $s=7c$ or $s=8c$ (can't be $5c$ or $6c$).

Now two possibilities :

1. $s=8c=4b=2a$ Hence $d=c$, and $s=lcm(a,b,c,d)=a$. A contradiction
2. $s=7c=4b=2a$ 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

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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.

$$ \begin{array}{cccc|c} a&b&c&d&s\\\hline 4& 3& 3& 2 & 12\\ 4& 4& 3& 1 & 12\\ 5& 2& 2& 1 & 10\\ 6& 4& 1& 1 & 12\\ 9& 6& 2& 1 & 18\\ 10& 5& 4& 1 & 20\\ 12& 8& 3& 1 & 24\\ 15& 10& 3& 2 & 30\\ 21& 14& 6& 1 & 42\\ \end{array} $$