Prove that polynomial $x^3+ax^2+17x+3b$ doesn't have same roots.

Question

All roots of polynomial $x^3+ax^2+17x+3b$, $a,b\in \Bbb Z$ are integers. Prove that this polynomial doesn't have same roots.

My plan was to find all three solutions and then compare them.

I don't know how to find the first root - tried to use cubic formula to find it but I got a large expression, just a few numbers canceled. Any advices? Thanks.

Answer 1

Say it has, so $x_1=x_2 = m$ and $x_3=k$, then we have by (second) Vieta formula $$m^2+2mk =17$$ and (third) $$m^2k =-3b$$

So $m$ or $k$ is divisible by $3$. Clearly $m$ is not (since else $3\mid 17$), so $3\mid k$. But then we have $m^2\equiv 2\pmod 3$, a contradiction.

Answer 2

Hint: There is a theorem that $f$ and $f'$ are relatively prime iff the polynomial $f$ has no repeated roots in a splitting field. .

Answer 3

Let $f(x)=x^3+ax^2+17x+3b$

$f'(x)=3x^2+2ax+17$

Putting $f'(x)=0$ we have $$x=\frac{-a\pm\sqrt{a^2-51}}{3}$$

If $f(x)=0$ has integer roots and if it has repeated roots then roots of $f'(x)=0$ will

also have integer roots. So $a^2-51$ must be perfect square of an integer which is

possible only for $a=\pm 10,a=\pm26$. Correspondingly the integer roots obtained are $\pm1,\pm17$(neglecting the non-integer roots)

The third root can be obtained using Vieta's theorem

As $\alpha\beta+\beta\gamma+\gamma\alpha=17$

Setting $\alpha=\beta=1$ we obtain $\gamma=8$

Similarly,setting $\alpha=\beta=-1$ we obtain $\gamma=-8$

Similarly,setting $\alpha=\beta=17$ we obtain $\gamma=-8$

Similarly,setting $\alpha=\beta=-17$ we obtain $\gamma=8$

So the product of roots turns out to be $3b=-8,8,-2312,2312$

which is not possible as the value of $b$ obtained is not an integer

which is obviously a CONTRADICTION

Answer 4

A polynomial has a multiple root iff its discriminant is zero.

The discriminant of $x^3+ax^2+17x+3b$ is $-12 a^3 b + 289 a^2 + 918 a b - 243 b^2 - 19652$.

Mod $3$, this discriminant reduces to $a^2 + 1$, which is never zero.

Answer 5

Hint $ $ the roots are distinct $\bmod 3\!:\ f \equiv x(x^2\!+\!ax\!-1)\not \equiv x(x\!-\!r)^2\,$ by $\,r^2\not\equiv 1$