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Consider a cubic equation with integer coefficients (coefficient of $x^3$ being $1$). In hit and trial method, we assume each divisors of the last term as the root and check if it satisfies the equation. If none of the divisors of the last term satisfy the equation, we are sure that the equation has no integer roots.

Euler also concluded that, no fractional roots can exist in the above case. Can someone please explain me how?

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  • $\begingroup$ Consider $(2x-3)^3$ as a counterexample. It's of degree $3$, it has no integer roots, yet it has fractional roots. $\endgroup$ – Jakobian Aug 17 '18 at 12:19
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I am a bit confused by your phrasing (no offense meant), but I think you are referring to the rational root test. Hopefully, this link is helpful:

https://en.wikipedia.org/wiki/Rational_root_theorem

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It follows immediately from the rational root theorem that if $$ax^3+bx+c=0,$$ with $a,b,c\in\mathrm Z$ has a rational root $p/q$ with $p$ and $q$ coprime, then $p$ divides $c$ and $q$ divides $a.$

Therefore, if $a=1,$ it is clear that if no root of the corresponding cubic divides $c,$ then there can be no rational roots, since all the rational roots in this case (if they exist) must be integers.

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Let $P(x)$ be a polynomial of degree $d$ with integer coefficients and let $x$ be an integer root. Then

$$P(x)=0=p_dx^d+p_{d-1}x^{d-1}+\cdots p_0=p_0\mod x$$

and $x$ must divide $p_0$.

The theorem holds for rational $x=\dfrac ab$ where the fraction is irreducible. Considering

$$b^dP\left(\frac ab\right)=0$$ you will conclude that $a$ divides $b^dp_0$, hence $p_0$.

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