The quartic polynomial $f(x) = x^4 + a x^3 + b x^2 + c x + d$ is such that $ad$ is odd and $bc$ is even. Prove that $f(x)$ does not has all rational roots.

My attempt:

Clearly, $f(x)$ will have either 0 or 2 or 4 rational roots. Let us assume all the four roots of the equation are rational and they be $p_1/q_1, p_2/q_2, p_3/q_3$ and $p_4/q_4$. As the question suggests: a is odd, d is odd, b is even/odd, c is even/odd, but, atleast one of b or c must be even. $(p_1/q_1)*(p_2/q_2)*(p_3/q_3)*(p_4/q_4)=d$ (odd integer) --(1) $(p_1/q_1)+(p_2/q_2)+(p_3/q_3)+(p_4/q_4)=-a$ (odd integer) --(2) if --(1) is true, the denominator of a root must be contained in the numerator of the remaining roots. This means, these numerators cannot have denominators containing a factor of their numerator. So, there will be atleast two denominators whose gcd will be 1. Thus, we cannot find any four rational numbers which add to give an integer because rational numbers only give an integer if their denominators have atleast one common factor. So, the equation cannot have 4 rational roots. Same can be explained for 2 rational roots.

So, I just wanted to ask am I correct in my approach or am I making any mistake. Further, I am not able to understand why there is a distinction of ad and bc as odd and even?? Please help me out..

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    $\begingroup$ Why can't $f$ have exactly one rational root? $\endgroup$ – Gerry Myerson Jun 3 '16 at 13:35
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    $\begingroup$ Also, it may help to notice that the rational roots of this polynomial must be integers (I am assuming $a,b,c,d$ are meant to be integers). $\endgroup$ – Gerry Myerson Jun 3 '16 at 13:37
  • $\begingroup$ If f has a rational root, then the other roots will be irrational or complex which can only exist in pairs.. $\endgroup$ – Utkarsh Jun 3 '16 at 13:43
  • $\begingroup$ Even if the roots are integers, q1, q2, q3 and q4 will be 1. That won't make a difference to my solution. integers are also rational numbers. $\endgroup$ – Utkarsh Jun 3 '16 at 13:45
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    $\begingroup$ @GerryMyerson yep, you are correct. It's an equation that comes under Casus irreducibilis. $\endgroup$ – Utkarsh Jun 30 '16 at 7:54

For $(a,b,c,d)=(1,2,3,1)$ we have $ad=1$ and $bc=6$, but nevertheless the polynomial $$ x^4+ax^3+bx^2+cx+d=x^4 + x^3 + 2x^2 + 3x + 1=(x+1)(x^3+2x+1) $$ has a rational root. Did I misunderstand something ?

Edit: Suppose that $$ x^4+ax^3+bx^2+cx+d=(x-a_1)(x-a_2)(x-a_3)(x-a_4) $$ where we may assume that $a,b,c,d$ and all $a_i$ are integers. Then we obtain, by mulriplying out, $$ ad:=-(a_1 + a_2 + a_3 + a_4)a_1a_2a_3a_4. $$ However, this shows that $ad$ is always even, because every $a_i$ in the product must be odd, in order to have $ad\equiv 1\bmod 2$, but then $a_1+\ldots + a_4$ is even. This is a contradiction, so that not all four roots can be rational.

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  • $\begingroup$ Sorry, I made a mistake in the question. the question says all four roots cannot be rational. $\endgroup$ – Utkarsh Jun 3 '16 at 13:55
  • $\begingroup$ @Utkarsh I edited my answer. Now I do not need that $bc$ must be even. Is there a mistake ? $\endgroup$ – Dietrich Burde Jun 3 '16 at 14:56
  • $\begingroup$ you have proved it for integers, but how to prove it by taking any rational roots(excluding integers)? $\endgroup$ – Utkarsh Jun 3 '16 at 15:23
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    $\begingroup$ If you have a rational root $x=u/v$ with $gcd(u,v)=1$ you obtain $u^4+au^3v+bu^2v^2+cuv^3+dv^4=0$, hence $v\mid u$. But then, since $u$ and $v$ are comprime, $v=\pm 1$. Hence we can assume that $x$ is integral. $\endgroup$ – Dietrich Burde Jun 3 '16 at 15:27
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    $\begingroup$ @Utkarsh Look again at the equation $u^4+au^3v+bu^2v^2+cuv^3+dv^4=0$. Which terms are divisible by $v$ ? You should try to find the answer now yourself ! $\endgroup$ – Dietrich Burde Jun 3 '16 at 18:24

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