Suppose $a+b+c+d=0$. If one of $a,b,c,d$ is zero then others are found easily and it can be check that $a+b+c+d\not =0$. So, we can suppose that all are nonzero. Add each two equalities to get 6 other $(abc-d)+(bcd-a)=3 \Longleftrightarrow (bc-1)(a+d)=3$ $(ad-1)(b+c)=-3$ .... But $(bc-1)(a+d)=3=- (-3)=-(ad-1)(b+c)=(ad-1)(-b-c)=(ad-1)(a+d)$. Thus $bc=ad$ and similiarly $bd=ac$, $cd=ab$. Then $a^2=b^2=c^2=d^2$ (*). Multiply initial equalities by $d,a,b,c$ respectively to get \[abcd-d^2=d, \ \ abcd-a^2=2a, \ \ abcd-b^2=3b , \ \ abcd-c^2=-6c\]So, $2a=3b=-6c=a$, which contradicts to (*).

We can easily eliminate the cases where either of the numbers is zero (actually, the only possibility is $(a,b,c,d)=(-2,-3,0,-1)$) and see that $a+b+c+d\neq 0$ is satisfied in those cases. So assume $abcd\neq 0$. The equations yield $abc+bcd+cda+dab=a+b+c+d$, hence $a,b,c,d$ are the roots of the polynomial $P(x)=x^4-Mx^3+Nx^2-Mx+K$, where $M,N,K$ are some constants. Assume $M=0$. Then $P(x)=x^4+Nx^2+K$, hence $\{a,b,c,d\}=\{p,-p,q,-q\}$ for some real numbers $p,q$. Therefore ${p^2q^2\over p}-p=\alpha\land{p^2q^2\over -p}+p=\beta$, where $\alpha, \beta$ are some distinct elements of $\{1,2,3,-6\}$. However, the last condition implies $\alpha=-\beta$, which can't be satisfied. Therefore $M\neq 0\iff a+b+c+d\neq 0$

Suppose $a+b+c+d=0$. Adding the original equations, $abc+bcd+cda+dab=0$. Substituting $d=-a-b-c$, we get $-b^2c-bc^2-a^2c-ac^2-abc-a^2b-ab^2-abc=0$ $-(a+b)(b+c)(c+a)=0$ So we must have either $a=-b$, $a=-c$, or $a=-d$ (since $b+c=0 \implies a+d=0$) $a=-b \implies bcd+b=2$ and $-bcd-b=3$, contradiction $a=-c \implies bcd+c=2$ and $-bcd-c=-6$, contradiction $a=-d \implies bcd+d=2$ and $-bcd-d=1$, contradiction So $a+b+c+d \neq 0$.

Suppose, for the sake of contradiction, that $a+b+c+d=0$. Adding up the 4 equations gives that $abc+abd+acd+bcd=0$ as well, so $(x-a)(x-b)(x-c)(x-d)=x^4+\alpha x^2+\beta$, for some real $\alpha$ and $\beta$. Therefore one of $b$, $c$, or $d$ is the additive inverse of $a$, and the numbers other than $a$ and its additive inverse are the additive inverses of each other. Let $a$, $b$, $c$, and $d$ be $p$, $q$, $-p$, and $-q$, in some order, for some reals $p$ and $q$. Note that if one of our variables is 0, then $x^2|(x-a)(x-b)(x-c)(x-d)$, which invalidates one of our original equations, so $p$ and $q$ are nonzero. However, this implies that one of $abc-d$, $bcd-a$, or $abd-c$ is the negatie of $acd-b$, which is a contradiction, so $a+b+c+d\neq 0$. I feel like this solution isn't rigorous enough...

1=2 wrote: $(x-a)(x-b)(x-c)(x-d)=x^4+\alpha x^2+\beta$, for some real $\alpha$ and $\beta$. Therefore one of $b$, $c$, or $d$ is the additive inverse of $a$. How do you know this? I know $(x-p)(x+p)(x-q)(x+q)$ is of the form $x^4+ \alpha x^2 + \beta$ but why is it the only possibility?

functions f(x)=(x-a)(x-b)(x-c)(x-d)

My solution abc-d=1 (1) bcd -a=2 (2) cda - b=3 (3) dab-c=-6 (4) Suppose a+b+c+d=0 We have equality abc+bcd+cda+dab-(a+b+c+d)=1+2+3-6 abc+bcd+cda+dab=0 bc(a+d)+ad(c+b)=0 bc(a+b)+ad(c+b-a-b-c-d)=0 bc(a+d)-ad(a+d)=0 (a+d)(bc-ad)=0 Must be a+b=0 or bc=ad (*) Prove a+b><0 Suppose a+b=0 From (1)+(2) we get abc+bcd-(a+d)=1+2 abc+bcd=3 bc(a+d)=3 0=3 contradiction Prove bc><ad If bc=ad From (a+b+c+d=0)*a a^2+ab+ac+ad=0 a^2+ab+ac+bc=0 a(a+b)+c(a+b)=0 (a+b)(a+c)=0 Must be a+b=0 or a+c=0 It prove same (*) and we prove $[b]a+b+c+d\not=0[/b]$

if $a+b+c+d=0$,adding four equations yields$\sum abc=0$hence$\sum\frac{1}{a}=0$ if there are 3 (or similarly,one) positives in a,b,c,d,then $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1$(where d is the only negative),which violates Carlson's ineq. so,two positives. then let $s=-c,t=-d$,then $a+b=s+t,\frac{1}{a}+\frac{1}{b}=\frac{1}{s}+\frac{1}{t}$ hence$(a,b,c,d)=(a,b,-b,-a) or (a,b,-a,-b)$ solving it reaches a contradiction.

We suppose $a+b+c+d=0$. By summing, we have : $abc+bcd+cda+dab=0\Rightarrow bc(a+d)+ad(b+c)=0\Rightarrow (a+d)(bc-ad)=0$. It's simple to prove $a+d\neq 0$ (if $a+d=0$, we have : $(abc-d)+(bcd-a)=3\Rightarrow (a+d)(bc-1)=3$, false), so $ad=bc$ and analogues. But $ab=cd\Rightarrow (ad)*(ab)=(bc)*(cd)\Rightarrow a^2=c^2$. We have $|a|=|b|=|c|=|d|\Rightarrow a=\pm x$, where $x=|a|$, so $abc=\pm x^3$. Among by the numbers $abc-d;bcd-a;cda-b;dab-c$, we have two equal numbers or two opposite numbers, false!

Suppose that $a+b+c+d=0$, then we have $abc+bcd+cda+dab=0$, notice that $a,b,c,d$ are roots of the equation \[x^4+sx^2+p=0\]where $s=ab+bc+ca+ad+bd+cd$ and $p=abcd$, $s,p\in \mathbb{R}$. Since if $\alpha $ is a solution to the equation, then $-\alpha $ is also a solution to the equation, so WLOG $a=-b, c=-d$, then from original equation, we have \[-a^2c+c=1\]\[ac^2-a=2\]\[-ac^2+a=3\]\[a^2c-c=-6\]which clearly contains no solutions. $\square$

Thanks Farenhajt for the finish as I was getting a little stuck.

FTSOC assume $a+b+c+d=0$ we get $abc+bcd+cda+dab=0\implies ab(c+d)-cd(c+d)=0\implies (ab-cd)(c+d)=0\implies ab=cd$ or $c=-d$ if $c+d=0\implies a+b=0 \qquad \qquad (\spadesuit)$ then we have: $bcd+cda=5\implies cd(a+b)=5$ which is a contradiction if $ab=cd$ then: we clearly have $a \neq 0\implies (a+b+c+d)a=0\implies a^2+cd+ac+ad=0\implies (a+c)(b+d)=0$ then we proceed similarly as in $(\spadesuit)$ to get contradiction hence we get that our assumption was wrong and hence we must have $a+b+c+d \neq 0$ $\blacksquare$

Suppose FTSOC $a+b+c+d=0$. If WLOG d=0 then $abc=1, -a=2, -b=3, -c=-6$, which means $a=-2, b=-3, c=6$, absurd that a+b+c+d=0, and also abc is not 1. Now if none of them are 0, add each permutation of one equation with another (i.e. 1st with 2nd and 3rd with 4th, 1st with 3rd and 2nd with 4th, and 1st with 4th and 2nd with 3rd) to get $(abc-d)+(bcd-a)=3 \Longleftrightarrow (bc-1)(a+d)=3$ $(ad-1)(b+c)=-3 ...$ But $(bc-1)(a+d)=3=- (-3)=-(ad-1)(b+c)=(ad-1)(-b-c)=(ad-1)(a+d)$. Thus $bc=ad$ and similiarly $bd=ac$, $cd=ab$. Then multiplying any two equations together out of the three gives us three equalities $a^2=b^2=a^2=c^2=a^2=d^2$ (*). Multiply initial equalities by $d,a,b,c$ respectively to get \[abcd-d^2=d, \ \ abcd-a^2=2a, \ \ abcd-b^2=3b , \ \ abcd-c^2=-6c\]So, $2a=3b=-6c=a$, which contradicts to $a^2=b^2=c^2=d^2$.