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.
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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