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

• October 2nd 2009, 04:58 PM
Harry1W
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:

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

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.
• October 3rd 2009, 06:14 AM
awkward
Quote:

Originally Posted by Harry1W
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:

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

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.
• October 3rd 2009, 12:34 PM
Hello Harry1W
Quote:

and by comparing coefficients we see that:

$\begin{array}{rcl}
A D & = & 4 \\
A E + B D & = & 2 \\
A F + B E + C D & = & 2 \\
C F & = & 1998
\end{array}$
.
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?