For all integers b, c, and d, if x is rational such that x^2+bx+c=d, then x is an integer Prove or disprove the following statment: For all integers b, c,and d, if x is a rational number such that $x^2+bx+c=d$, then x is an integer.
This is a homework question from the book Discrete Mathematics for Computer Scientists by Stein, Drysdale and Bogart.
I since x is rational I thought I could start off with:
${(\frac{m}{n})}^2+b\frac{m}{n}=d-c$
But I don't know where to go from here.
Or I could try using the quadratic formula
$x=\frac{1}{2}\left(\pm\sqrt{b^{2}-4c+4d}-b\right)$
but I am very weak with elementary number theory that I don't know where to go.  I am thinking that regardless of if $\sqrt{b^{2}-4c+4d}$ is an integer or not, the fact that I have 
$x=\frac{1}{2}*\pm$ SomeNumber
means that x is not an integer.
I am new to writing proofs, and unfortunately, I don't really know how to prove this.  Any hints would be appreciated.
Thank you.
Edit:  By plugging in simple numbers, for example x=1, b=1, c=1 and d=3 I can see that x is probably an integer, for all integers b,c,and d - so that means my thinking about the quadratic formula is not correct.  I will still work on this.
2nd Edit:  Now I plug in more numbers and don't get integers. For example $x^2+2x+3=4$.  I am also new to this site, so I am not sure if I should continue to edit the post or write in the comments sections anytime I think of something new.  Please advise.
3rd Edit: I think I know what to do.  The last section of the book covered universal quantifiers.  I believe the authors are are trying to get me to realize that they are saying $\forall b, c, d \in Z$ and I only need to give one one example for which the assertion is untrue.  And in my previous edit, b=2, c=3, and d=4 did not result in x being an integer.
 A: This is simply the monic quadratic case of the Rational Root Test. You could specialize that proof, or else proceed similarly to various irrationality proofs for square-roots. $\: $ E.g. $\:$  below is a proof that I discovered in high-school. First I present the proof for square-roots - where the idea is clearer.
Theorem $\ $ For $\rm\: c\in \mathbb Z,\:$ any rational root $\rm\:r\:$ of $\rm\ x^2 = c\ $ is am integer.
Proof $\  $ Put $\rm\ \color{#0a0}{r = m/n}\ $ with $\rm\:(m,n) = 1.\:$ Then $\rm\ \color{#c00}{jm-kn =1}\;$  for some $\:\rm j,k \in \mathbb{Z}\,$ by Bezout.
Hence $\,\rm \color{#0a0}{0 = (m-nr)}\:(k+jr) = mk-njc + (\color{#c00}{jm-kn}) r  \ \Rightarrow\ r = -mk+njc \ \in\ \mathbb{Z}\ \ \ $ QED
This proof easily extends to the root of a general monic quadratic as follows.
Theorem $\ $ For $\rm\:b,c\in\mathbb Z,\,$ any rational root $\rm\:r\:$ of $\rm\ x^2 = \color{#90f}{b\ x + c}\ $ is an integer.
Proof $\ $ Put $\rm\ \color{#0a0}{r = m/n},\ (m,n)=1,\,$ so $\rm\,(m\!-\!nb,n)=1\ $ so $\rm\, \exists\ j,k\in \mathbb Z\!:\ \color{#c00}{1 = j(m\!-\!nb)\!-\!kn} $
Hence $\rm\, \color{#0a0}{0 = (m\!-\!nr)}(k\!+\!jr)\ =mk\! +\! (jm\!-\!kn)r\!-nj(\color{#90f}{br\!+\!c}) = mk\!-\!njc + (\color{#c00}{j(m\!-\!nb)\!-\!kn})r$

The same proof easily extends to higher degree polynomials that are monic (lead coef $=1).$
If you learn about denominator ideals then you'll see that the above proof simply says that the denominator ideal of $\rm\:r\:$ contains $\rm\:n\:$ and $\rm\:nr = m,\:$ so it contains their gcd $\rm\:(n,m) = 1,\,$ so $\rm\ r\in \mathbb Z.$ Using Dedekind's notion of conductor ideal, the proof easily generalizes to higher degree monic polynomials, yielding that PIDs are integrally closed.
A: Hint:  First, you can combine $c$ and $d$ as you only care about their difference.  There is the theorem that the square root of a positive integer is either integer or irrational.  You are right that one example disproves the assertion "for all b,c, and d". 
Added:  If you look at the quadratic formula, the only threat to x being integral (if it is rational) is the division by 2.  But if the square root is integral, it must have the same parity as b.
