# An Olympiad problem in algebra

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?

• 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 $f(x)=0$. – J.G. Sep 14 at 10:58
• No I have written it correctly. You might have seen another problem. – Aditya Saran Sep 14 at 11:02
• 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. – J.G. Sep 14 at 11:19
• Ya it wrote. We have to prove that it does not have integer solution – Aditya Saran Sep 14 at 11:26

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.

• +1 nice and simple. – achille hui Sep 14 at 12:55

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.