มีรากตรรกยะอย่างมาก1ตัว

[polsk133]

c เป็นจำนวนตรรกยะ จงพิสูจน์ว่า

$x^3-3cx^2-3xc+c=0$ มีรากตรรกยะได้อย่างมาก1ค่า

ช่วยทีครับ

[Anton]

Problem. Let $c$ be a rational number. Prove that the cubic equation $$x^3-3cx^2-3cx+c=0$$ has at most one rational solution.

The case $c=0$ is a counterexample. All three roots of this cubic polynomial are rational numbers (i.e., they are all zero).
But let's ignore that, since the asker may not count multiplicities of roots. So, we assume $c\neq 0$.

The discriminant (with respect to $x$) of the cubic polynomial $x^3-3cx^2-3cx+c$ is $27c^2(7c^2+10c-1)$. If this polynomial has at least two rational roots, then all roots are ratioals. Thus, $27c^2(7c^2+10c-1)$ must be a perfect square of a rational numbers. Hence,
$$t:=\frac{27c^2(7c^2+10c-1)}{(3c)^2}=3(7c^2+10c-1)$$
is a perfect square of a rational number.

Let $v_p$ denote the $p$-adic valuation for each prime natural number $p$. Now, if $v_3(c)>0$, then $v_3(t)=1$ is odd. If $v_3(c)=0$, then $v_3(t)=1$ as well. On the other hand, if $v_3(c)<0$, then $v_3(t)=1+2v_3(c)$ is also an odd integer. Thus, $t$ cannot be a perfect square.