Select Page
3     Let $(a,b,c,d \in \mathbb{N})$ such that $(a \ge b \ge c \ge d)$. Show that the equation $(x^4 - ax^3 - bx^2 - cx -d = 0)$ has no integer solution.

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.