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 $$\phi(x)=a\left(x-\frac{-b-k}{2a}\right)\left(x-\frac{-b+k}{2a}\right)=\left(\frac s2 x-\frac{-b-k}{t}\right)\left(\frac t2x-\frac{-b+k}{s}\right)$$

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

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