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Let $a,b,c,d$ be positive integers such that $a \ge b\ge c\ge d$. Prove That: the equation $x^4- ax^3- bx^2- cx -d=0$ has no integer solution. It is an indian olympiad problem. I am stuck in it from days. Can anyone help me solve this please?

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    $\begingroup$ You must have misstated the problem. If $f(x):=x^4-ax^3-bx^2-cx-d$, $f(0)=-d<0$ while $\lim_{x\to\infty}f(x)=\infty$, so by continuity some $x>0$ satisfies $f(x)=0$. $\endgroup$ – J.G. Sep 14 at 10:58
  • $\begingroup$ No I have written it correctly. You might have seen another problem. $\endgroup$ – Aditya Saran Sep 14 at 11:02
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    $\begingroup$ The challenge cannot possibly be to prove the wrong claim that positive integers $a\ge b\ge c\ge d$ require $x^4-ax^3-bx^2-cx-d=0$ to lack a real root. I suggest you review @FareedAF's edit of your MathJax to ensure the current version of the problem is what you meant. $\endgroup$ – J.G. Sep 14 at 11:19
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    $\begingroup$ Ya it wrote. We have to prove that it does not have integer solution $\endgroup$ – Aditya Saran Sep 14 at 11:26
  • $\begingroup$ About some of your recent questions: please check the question you are asking is correct and is not a duplicate. In addition, please add more context to your questions so that people can answer at the right level for you. $\endgroup$ – Toby Mak Sep 14 at 12:40
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Let's assume we have a positive solution $x_0$. Then $d=x_0d_0$ with $x_0,d_0\geq 1$ and $x_0^4=ax_0^3+bx_0^2+cx_0+d\geq d_0(x_0^4+x_0^3+x_0^2+x_0)$. That's obviously wrong. Maybe for $x_0<0$ you get a similar equation.

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  • $\begingroup$ +1 nice and simple. $\endgroup$ – achille hui Sep 14 at 12:55
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Let $f(x) = x^4 - ax^3 - bx^2 - cx - d$ and $r$ be an integer root for $f(x) = 0$.

  1. $r \ne 0$ because $f(0) = -d \ne 0$.
  2. $r > 0$, otherwise, let $s = -r \in \mathbb{Z}_{+}$, we have

$$f(r) = f(-s) = \underbrace{s^4}_{\ge 1} + \underbrace{(as - b)}_{\ge a - b \ge 0} s^2 + \underbrace{(cs - d)}_{\ge c - d\ge 0} \ge 1$$

  1. $r \le d$, this is because $$f(r) = 0 \implies r(r^3 - ar^2 - br - c) = d \implies r|d$$

  2. Finally for $r \in \mathbb{Z}_{+}$ such that $1 \le r \le d$, we have $$f(r) \le r^4 - d(r^3 + r^2 + r + 1) = \underbrace{(r - 1 - d)}_{\le -1}\underbrace{(r^3 + r^2 + r + 1 )}_{\ge 4} + 1$$ This forces $f(r) \le -4 + 1 = -3 < 0$ and $r$ cannot be an integer root of $f(x)$.

Combine all these, we can conclude $f(x)$ doesn't have any integer as root.

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