Prove that a function has no integer roots

[anemone]

Let $p, q, r, s \in \mathbb{N}$ such that $p \ge q \ge r \ge s$. Show that the function $f(x)=x^4-px^3-qx^2-rx-s$ has no integer root.

 

[Opalg]

Let $p, q, r, s \in \mathbb{N}$ such that $p \ge q \ge r \ge s$. Show that the function $f(x)=x^4-px^3-qx^2-rx-s$ has no integer root.

[sp]Suppose that $n$ is an integer such that $f(n) = 0$. Then $n\ne0$, because $s\ne0$.

Suppose that $n>0$. Then the equation can be written $n^4 = pn^3+qn^2+rn+s$. We must have $p<n$, otherwise the right side would be greater than the left. Hence $p,q,r,s$ are all $\leqslant n-1$. Therefore $n^4 \leqslant (n-1)n^3 + (n-1)n^2 + (n-1)n + (n-1) = n^4-1.$ That is a contradiction, so there cannot be any positive solutions.

Suppose that $n<0$. Then the equation can be written $n^4 + p|n|^3 +r|n| = qn^2 + s$. But $n^4>0$, $p|n|\geqslant q$ and $r|n|\geqslant s$. So the left side of the equation is greater than the right, and again we have a contradiction.

Therefore there are no integer solutions.[/sp]