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Math Help - Proof by Contradiction: Polynomial cannot be written as the product of two quadratics

  1. #1
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    Proof by Contradiction: Polynomial cannot be written as the product of two quadratics

    I've been working through Chapter 1 of D. L. Johnson's "Elements of Logic via Numbers and Sets" and have become stumped on the following problem, amongst others.

    "Prove that the polynomial f(x) = x^4 + 2x^2 + 2x + 1998 cannot be written as a product of two quadratic polynomials with integer coefficients."

    The preceding section (and indeed the other questions in this set) was concerned with 'proof by contradiction', so I assume that I am intended to solve this via this method.

    Therefore, for contradiction, assume that f(x) can be written as previously described, i.e.

     f(x) = x^4 + 2x^2 + 2x + 1998 = (Ax^2 + Bx +C)(Dx^2 + Ex + F)

    where  A, B, C, E and F are integers. Thus,

     x^4 + 2x^2 + 2x+1998 = AD x^4 + (AE + BD) x^3 + (AF + BE + CD) x^2 + (BF + CE) x + CF ,

    and by comparing coefficients we see that:

     \begin{array}{rcl}<br />
A D & = & 4 \\<br />
A E + B D & = & 2 \\<br />
A F + B E + C D & = & 2 \\<br />
C F & = & 1998<br />
\end{array} .

    I suppose for the purposes of the proof then, I should aim to show that there are no integer solutions to these set of simultaneous equations - but I have no idea about how to approach this. Any suggestions would be much appreciated.
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  2. #2
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    Quote Originally Posted by Harry1W View Post
    I've been working through Chapter 1 of D. L. Johnson's "Elements of Logic via Numbers and Sets" and have become stumped on the following problem, amongst others.

    "Prove that the polynomial f(x) = x^4 + 2x^2 + 2x + 1998 cannot be written as a product of two quadratic polynomials with integer coefficients."

    The preceding section (and indeed the other questions in this set) was concerned with 'proof by contradiction', so I assume that I am intended to solve this via this method.

    Therefore, for contradiction, assume that f(x) can be written as previously described, i.e.

     f(x) = x^4 + 2x^2 + 2x + 1998 = (Ax^2 + Bx +C)(Dx^2 + Ex + F)

    where  A, B, C, E and F are integers. Thus,

     x^4 + 2x^2 + 2x+1998 = AD x^4 + (AE + BD) x^3 + (AF + BE + CD) x^2 + (BF + CE) x + CF ,

    and by comparing coefficients we see that:

     \begin{array}{rcl}<br />
A D & = & 4 \\<br />
A E + B D & = & 2 \\<br />
A F + B E + C D & = & 2 \\<br />
C F & = & 1998<br />
\end{array} .

    I suppose for the purposes of the proof then, I should aim to show that there are no integer solutions to these set of simultaneous equations - but I have no idea about how to approach this. Any suggestions would be much appreciated.
    Hi Harry1w,

    I haven't worked this out, but notice that since AD = 4 = 2^2, there are only a few possibilities for A. (We can assume A and D are positive-- otherwise just multiply both factors by -1.) Either A = 1, A= 2, or A=4, with corresponding values for D. There are also only a few possibilities for C and F, since 1998 = 2 \times 3^2 \times 37. It shouldn't be hard to work through all the possibilities.
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  3. #3
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    Hello Harry1W
    and by comparing coefficients we see that:

     \begin{array}{rcl}<br />
A D & = & 4 \\<br />
A E + B D & = & 2 \\<br />
A F + B E + C D & = & 2 \\<br />
C F & = & 1998<br />
\end{array} .
    I think you mean:

    Coefficient of x^4:AD = 1.

    Coefficient of x^3:AE+BD=0.

    Without loss of generality, we may assume A=D=1, and so

     E+B=0

    Do you want to take another look at it now?

    Grandad
    Last edited by Grandad; October 4th 2009 at 12:29 AM. Reason: Correction
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