# Proof that a square quadratic discriminant implies that integers suffice for factorisation

I want to prove the following statement.

Suppose $$\phi(x) = ax^2 + bx + c$$ where $$a,b,c\in\mathbb Z$$, and the quadratic discriminant $$\Delta = b^2 - 4ac$$ satisfies $$\Delta \in \{n^2:n\in\mathbb Z\}$$. Then $$\phi(x)$$ can be written as $$K(mx-s)(nx-t)$$ where $$K,m,s,n,t\in\mathbb Z$$.

Here is my attempted proof.

If $$\Delta \in \{n^2 : n\in\mathbb Z\}$$, then $$\Delta = k^2$$ for some $$k \in \mathbb Z$$, $$k\geq 0$$, so that $$k = \sqrt \Delta$$. The roots $$\alpha, \beta$$ of $$\phi$$ are then given by $$x=\frac{-b\pm k}{2a}$$.

Now we claim that $$\phi(x)=\gcd\{a,b,c\}\left(\tfrac{2a}{\gcd\{-b-k,2a\}} x - \tfrac{-b-k}{\gcd\{-b-k,2a\}} \right)\left(\tfrac{2a}{\gcd\{-b+k,2a\}} x - \tfrac{-b+k}{\gcd\{-b+k,2a\}} \right).$$ Notice that the fact that the coefficients are all integers is obvious, because each denominator is a divisor of the numerator. Thus we simply need to show that this is in fact equal to $$\phi(x)$$, and the proof will be done. Indeed, \begin{align*} &\gcd\{a,b,c\}\left(\tfrac{2a}{\gcd\{-b-k,2a\}} x - \tfrac{-b-k}{\gcd\{-b-k,2a\}} \right)\left(\tfrac{2a}{\gcd\{-b+k,2a\}} x - \tfrac{-b+k}{\gcd\{-b+k,2a\}} \right)\\[10pt] =&\frac{\gcd\{a,b,c\}}{\gcd\{-b-k,2a\}\gcd\{-b+k,2a\}}(2ax-(-b-k))(2ax-(-b+k))\\[10pt] =&\frac{\gcd\{a,b,c\}}{\gcd\{-b-k,2a\}\gcd\{-b+k,2a\}}(4a^2x^2+4abx+(-b-k)(-b+k))\\[10pt] =&\frac{\gcd\{a,b,c\}}{\gcd\{-b-k,2a\}\gcd\{-b+k,2a\}}(4a^2x^2+4abx+4a^2(\tfrac{-b-k}{2a})(\tfrac{-b+k}{2a}))\\[10pt] =&\frac{4a\gcd\{a,b,c\}}{\gcd\{-b-k,2a\}\gcd\{-b+k,2a\}}(ax^2+bx+c), \end{align*} since $$\alpha\beta = (\tfrac{-b-k}{2a})(\tfrac{-b+k}{2a}) = c/a$$. Thus all we need to show is that $$\frac{4a \gcd\{a,b,c\}}{\gcd\{-b-k,2a\}\gcd\{-b+k,2a\}} = 1$$. Indeed, \begin{align*} \frac{4a\gcd\{a,b,c\}}{\gcd\{-b-k,2a\}\gcd\{-b+k,2a\}} &= \frac{4a\gcd\{a, -a(\alpha+\beta),a\alpha\beta\}}{\gcd\{2a\alpha,2a\}\gcd\{2a\beta,2a\} }\\ &= \frac{4a^2\gcd\{1, -(\alpha+\beta),\alpha\beta\}}{2a\gcd\{\alpha,1\}2a\gcd\{\beta,1\} }\\ &= \frac{4a^2}{4a^2} = 1, \end{align*} as required. $$\square$$

I think my proof is correct. Is there a simpler way to go about this? I appreciate any feedback.

Your way looks very nice.

Here is another way.

There exists a non-negative integer $$k$$ such that $$k^2=b^2-4ac$$.

Since $$b$$ and $$k$$ have the same parity, we see that both $$b-k$$ and $$b+k$$ are even.

Since $$\frac{(b-k)(b+k)}{4a}\in\mathbb Z$$, there exist even numbers $$s,t$$ such that $$\frac{b-k}{s}\in\mathbb Z,\qquad\frac{b+k}{t}\in\mathbb Z\qquad\text{and}\qquad st=4a$$ Therefore, we can write $$\phi(x)=a\left(x-\frac{-b-k}{2a}\right)\left(x-\frac{-b+k}{2a}\right)=\left(\frac s2 x-\frac{-b-k}{t}\right)\left(\frac t2x-\frac{-b+k}{s}\right)$$

• This is a nice way! Mine is longer but explicitly determines $s$ and $t$. – Luke Collins Nov 11 '18 at 14:21