Sketch of the Solution:
Claim 1: There cannot be a negative integer solution. Suppose other wise. If possible $x= -k$ (k positive) be a solution.
Then we have $ (k^4 + ak^3 +ck = bk^2 +d)$. Clearly this is impossible as $ (a\ge b , k^3 \ge k^2 )$ and $ (c \ge d )$.
Claim 2: 0 is not a solution (why?)
Claim 3: There cannot be a positive integer solution. Suppose other wise. If possible x=k (k positive) be a solution.
Then we have $ (k^4 = a k^3 + b k^2 + c k + d)$
This implies that the right hand side is divisible by k which again implies that d is divisible by k (why?).
Let $d=d'k$
Now $(c\ge d) \implies (c \ge d'k) \implies (c \ge k)$.
Thus $(a \ge c \ge k ) \implies (a \cdot k^3 \ge k \cdot k^3 )$.
Hence the equality $ (k^4 = a k^3 + b k^2 + c k + d)$ is impossible.
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