MHB Prove an equation has no integer solution

[anemone]

Let $p,\,q,\,r,\,s$ be positive integers such that $p\ge q\ge r \ge s$.

Prove that the equation $x^4-px^3-qx^2-rx-s=0$ has no integer solution.

 

[Albert1]

Let $p,\,q,\,r,\,s$ be positive integers such that $p\ge q\ge r \ge s$

Prove that the equation $x^4-px^3-qx^2-rx-s=0----(1)$ has no integer solution.

Spoiler

by" Rational zero theorem"
if m is the integer solution of (1)
then : s is a multiple of $\mid m\mid $

but :$m^4-pm^3-qm^2-rm-s\neq 0$
for $p\ge q\ge r\ge s \ge\mid m \mid>0$
and $p,q,r,s\in N$

 

[anemone]

Spoiler

by" Rational zero theorem"
if m is the integer solution of (1)
then : s is a multiple of $\mid m\mid $

but :$m^4-pm^3-qm^2-rm-s\neq 0$
for $p\ge q\ge r\ge s \ge\mid m \mid>0$
and $p,q,r,s\in N$

Hi Albert,

Your concept is correct, thanks for participating. :)

Solution of other that based on the same principle:

Spoiler

Suppose that $m$ is an integer root of $x^4-px^3-qx^2-rx-s=0$ . As $s\ne 0$, we have $m\ne 0$. Suppose now that $m>0$, then $m^4-pm^3=qm^2+rm+s>0$ and hence $m>p\ge s$. On the other hand, $s=m(m^3-pm^2-qm-r)$ and hence $m$ divides $s$, a contradiction.

If $m<0$, then writing $n=-m>0$, we have $n^4+pn^3-qn^2+rn-s=n^4+n^2(pn-q)+(rn-s)>0$, a contradiction. This proves that the given polynomial has no integer roots.