Diminishing Coefficients

$\large \sum_{cyclic}^{} p = a+b+c+d , \sum_{cyclic}^{}pq = -(a+b+c) , \sum_{cyclic}^{}pqr = (a+b) , pqrs = -a$

Where $p,q,r,s$ denote the roots of the given expression.

We have $pqrs=-a$ , So we may conclude that either 1 or 3 of the roots are negetive out of p,q,r,s $\ne0$ .

Now ,

$\sum_{cyclic}^{}p + \sum_{cyclic}^{}pq + \sum_{cyclic}^{}pqr + pqrs + 1 = (a+b+c+d) - (a+b+c) + (a+b) - a +1$

$(p+1)(q+1)(r+1)(s+1) = b+d+1$ .................................... (1)

We have RHS which is a positive integer, so LHS must be. Also RHS > 0 , So $LHS\ne0$ & > 0 . $p,q,r,s\ne-1$

We must have any one or three of p,q,r,s <-1.

In either case any one or three of (p+1),(q+1),(r+1),(s+1) is negetive as any one or three of p,q,r,s< -1 .

But in that case we can't have the RHS positive as in eqn (1) .

Contradiction, Thus it must not have integral roots.

We see a, b ,c and d as are roots (by vieta's) . Now we see that bcd=-1.Then we have $(b,c,d)$ as $(1,1,-1),(-1,1,1),(1,-1,1),(-1,-1,-1)$ . Thus we find a . But none satisfies.