Prove that $\gcd(abc + abd + acd + bcd, abcd) = 1$

Question

Let $a, b, c, d \in \mathbb Z$. Prove that $\gcd(abc + abd + acd + bcd, abcd) = 1$ if and only if $a, b, c, d$ are pairwise relatively prime.

I am very confused as to how I should even start this problem. How would I show that $a$, $b$, $c$, $d$ are relatively prime such that this could be solved?

Answer 1

$(\Rightarrow)$

$\gcd(abc+abd+acd+bcd, abcd)=1\Rightarrow \exists{m,n}\in Z$ such that:

$m(abc+abd+acd+bcd)+n(abcd)=1\Rightarrow a(mbc+mbd+mcd+nbcd)+m(bcd)=1$ hence

$\gcd(a,bcd)=1\Rightarrow \gcd(a,b)=\gcd(a,c)=\gcd(a,d)=1$

You can prove it completely in the similar manner for other pairs.

$(\Leftarrow)$

Let $a,b,c$ and $d$ be pairwise prime integers. Then we have:

$\gcd(abc,d)=1$

$\gcd(abd,c)=1$

$\gcd(acd,b)=1$

$\gcd(bcd,a)=1$

So we can claim that:

$\gcd(abc+(ab+ac+bc)d,d)=1$

$\gcd(abd+(ab+ad+bd)c,c)=1$

$\gcd(acd+(ac+ad+cd)b,b)=1$

$\gcd(bcd+(bc+bd+cd)a,a)=1$

hence we have:

$\gcd(abc+abd+acd+bcd,d)=1$

$\gcd(abc+abd+acd+bcd,c)=1$

$\gcd(abc+abd+acd+bcd,b)=1$

$\gcd(abc+abd+acd+bcd,a)=1$

hence

$\gcd(abc+abd+acd+bcd,abcd)=1$

Answer 2

Let $S = ab(c{+}d) + cd(a{+}b)$, you want to check when $$\gcd(S,abcd)=1$$ If a prime $p$ divides one of $a,b$ and one of $c,d$, then it clearly divides both summands of $S$ and the $\gcd$ can't be $1$. If it divides both $a$ and $b$, then it also divides $a{+}b$ and hence both summands of $S$. Similarly for $p$ dividing $c$ and $d$.

For the other case suppose every prime $p$ divides at most one of $a,b,c,d$. If it divides $a$ or $b$, then it can't divide $c, d,$ and $a+b$, so it can't divide $S$. Similarly for $p$ dividing $c$ or $d$. Therefore the $\gcd$ is $1$.

Answer 3

$\Rightarrow$:

$gcd(abc+abd+acd+bcd,abcd)=1 \Rightarrow a,b,c,d$ relatively prime.

Add $a,b,c,d$ not relatively prime as additional premise.

So $\exists i\in Z$ such that $i|a$ and $i|b$. So $\exists k,j\in Z$ such that $a=ik,b=ij$.

So $gcd(abc+abd+acd+bcd,abcd)=\\gcd((ik)(ij)c+(ik)(ij)d+(ik)cd+(ij)cd,(ik)(ij)cd)=i$

$\therefore F_0$ (contradiction)

$\Leftarrow:$

$gcd(abc+abd+acd+bcd,abcd)\neq 1 \Rightarrow a,b,c,d$ not relatively prime. (proof by contraposition)

$\exists i,k,j \in Z$ such that $ab(c+d)+cd(a+b) = ik, abcd=ij$. So because $ab(c+d)+cd(a+b)= ik$, $i$ must divide $a,b,c$ and $d$ and hence they are not relatively prime.