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.

*Related*