Math 115. A Rephrased Theorem 3.10

This handout gives a more detailed variant of Theorem 3.10.

Theorem. Let

f(x, y) = ax^{2} + bxy + cy^{2}

be a nonzero binary quadratic form with integer coefficients, and let d = b^{2} − 4ac be its discriminant. (a). If d is a perfect square (including 0 ), then f can be factored into two linear

factors with integer coefficients (note Exercise 9). (b). If d is not a perfect square, then f cannot be factored into linear factors

with rational coefficients.

Proof. (a). Assume that d is a perfect square.

Case I: a = 0 . Then f(x, y) = bxy +cy^{2} = (bx+cy)y can be factored, and d = b^{2} is a perfect square.

Case II: a ≠ 0 . By algebra we know that az^{2} +bz +c has rational roots r_{1} and r_{2} (possibly equal), so f(x, y) = a(x − r_{1}y)(x − r_{2}y) . Write r_{1} = h_{1}/k_{1} and r_{2} = h_{2}/k_{2} as fractions in lowest terms. Then

f(x, y) = \frac{a}{k_{1}k_{2}} (k_{1}x − h_{1}y)(k_{2}x − h_{2}y) ,

where

\frac{a}{k_{1}k_{2}} (−k_{1}h_{2} − h_{1}k_{2}) = b ∈ ℤ  and  \frac{a}{k_{1}k_{2}} h_{1}h_{2} = c ∈ ℤ .

It will suffice to show that k_{1}k_{2} | a, since then \frac{a}{k_{1}k_{2}} ∈ ℤ, and the factor \frac{a}{k_{1}k_{2}} (k_{1}x−h_{1}y) has integer coefficients.

Let p be a prime and let α, β be such that p^{α} ∥ k_{1} and p^{β} ∥ k_{2} . We want to show that p^{α+β} | a.

Case IIa: p | h_{1} and p | h_{2} . Then α = β = 0 , so there is nothing to prove.

Case IIb: p ∤ h_{1} and p ∤ h_{2} . Since c ∈ ℤ, we have k_{1}k_{2} | ah_{1}h_{2} , hence p^{α+β} | ah_{1}h_{2} . This implies p^{α+β} | a (since p ∤ h_{1}h_{2} ), so we are done.

Case IIc: p | h_{1} but p ∤ h_{2} . Then p ∤ k_{1} , so p ∤ k_{1}h_{2} . But p | h_{1}k_{2} , so p ∤ (−k_{1}h_{2} − h_{1}k_{2}) , therefore (since b ∈ ℤ) k_{1}k_{2} | a(−k_{1}h_{2} − h_{1}k_{2}) , and therefore p^{α+β} | a(−k_{1}h_{2} − h_{1}k_{2}) , so p^{α+β} | a.

Case IId: p ∤ h_{1} but p | h_{2} . This is similar to Case IIc (interchange the indices 1 and 2 ).

Therefore f factors into linear factors with integer coefficients.

(b). We will prove the contrapositive: If f can be factored into linear factors with rational coefficients, then d is a perfect square.

First, if a = 0 then d = b^{2} is a perfect square.

Otherwise, by assumption f(x, y) = (k_{1}x+h_{1}y)(k_{2}x+h_{2}y) with k_{1}, h_{1}, k_{2}, h_{2} ∈ ℚ, and k_{1}k_{2} ≠ 0 since a ≠ 0 . Therefore the quadratic polynomial az^{2}+bz+c has rational roots −h_{1}/k_{1} and −h_{2}/k_{2} , so \sqrt{d} ∈ ℚ. As was noted earlier (in class on September 3), this implies that the integer d = b^{2} − 4ac is the square of an integer. □