MHB Prove an equation has no integer solution

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The discussion focuses on proving that the polynomial equation \(x^4 - px^3 - qx^2 - rx - s = 0\) has no integer solutions for positive integers \(p, q, r, s\) where \(p \geq q \geq r \geq s\). Participants confirm the validity of the approach and share insights on similar proofs that utilize the same principles. The conversation emphasizes the importance of understanding polynomial behavior and integer properties in the context of this equation. Ultimately, the consensus is that the equation cannot yield integer solutions under the specified conditions. The proof relies on analyzing the structure of the polynomial and the constraints imposed by the integer values of \(p, q, r, s\).
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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.
 
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anemone said:
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.
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 $p,q,r,s\in N$
 
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Albert said:
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:

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.
 
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