Factoring $\frac{n(n+1)}2x^2-x-2$ for $n\in\mathbb Z$

Question

I was factoring quadratic polynomials for high-school practice and I noticed a pattern: $$\begin{align} x^2-x-2 &=(x+1)(x-2) \\ 3x^2-x-2 &=(x-1)(3x-2) \\ 6x^2-x-2 &= (2x+1)(3x-2) \\ 10x^2-x-2 &= (2x-1)(5x+2) \\ 15x^2-x-2 &= (3x+1)(5x-2) \\ 21x^2-x-2 &= (3x-1)(7x+2) \\ &\vdots\end{align}$$ So it seemed that, for any integer $n$, we have:

$$\frac{n(n+1)}2x^2-x-2=\Big(\left\lceil{\frac n2}\right\rceil x+(-1)^{n+1}\Big)\Big(\left\lceil{n+\frac{(-1)^n}2}\right\rceil x + 2(-1)^n\Big).$$

where the pattern I showed above begins with $n=1$ and ends with $n=6$.

I am not sure how to prove this (assuming it is true). I know that $-2=(-1)^{n+1}\cdot 2(-1)^n$ but I don't know how to prove that: $$\left\lceil{\frac n2}\right\rceil\cdot \left\lceil{n+\frac{(-1)^n}2}\right\rceil=\frac{n(n+1)}2\tag{$\star$}$$ and $$2(-1)^n\left\lceil{\frac n2}\right\rceil+(-1)^{n+1}\left\lceil{n+\frac{(-1)^n}2}\right\rceil=-1$$ which I believe is necessary to prove this conjecture. Since $n$ is an integer, I was thinking of letting $n=\left\lceil\frac k2\right\rceil$ for any real $k$ or something, and I'd assume that $$\left\lceil{\frac n2}\right\rceil=\left\lceil{\frac{\left\lceil{\frac k2}\right\rceil}2}\right\rceil=\left\lceil{\frac k4}\right\rceil$$ but I'm not sure how "round-off arithmetic" works (informally speaking). Any help is appreciated.

---

Edit:

Thanks to @lone_student's comment, I have shown $(\star)$ to be true for all $n\in\mathbb Z$ by considering $n$ even and odd.

Lemma:$$\left\lceil {\frac nm}\right\rceil=\left\lfloor{\frac{n-1}m+1}\right\rfloor\tag1$$

Here, $m=2$.

Also: $$n-\left\lfloor\frac n2\right\rfloor=\left\lceil\frac n2\right\rceil\tag2$$ Using these, we can show that: $$\frac{n(n+1)}2=\left\lceil\frac n2\right\rceil\cdot\left\lceil{n+\frac{(-1)^n}2}\right\rceil=\underbrace{\Big(n-\left\lfloor\frac n2\right\rfloor\Big)}_{\text{By } (2)}\cdot\underbrace{\left\lfloor\frac{2n+(-1)^n-1}2+1\right\rfloor}_{\text{By (1)}}$$

When $n=2k\in\mathbb Z$ we have $$k(2k+1)=\big(2k-\left\lfloor k\right\rfloor\big)\lfloor2k+1\rfloor=k(2k+1)$$ since $k$ and $2k+1$ are integers, and, by definition, $\lceil \alpha\rceil =\lfloor \alpha\rfloor = \alpha$ iff $\alpha\in\mathbb Z$.

Similarly, when $n=2k-1$, we have: $$k(2k-1)=\Big(2k-1-\left\lfloor k-\frac 12\right\rfloor\Big)\lfloor 2k-1\rfloor = k(2k-1)$$ since obviously $\left\lfloor k-\frac 12\right\rfloor = k-1$ for $k\in\mathbb Z$.

---

I believe there was some confusion towards my question: did I mean to factorise the quadratic in terms of $n$ or did I mean to prove specifically the ceiling-function product identity? I did intend to ask a question on the latter subject, but I had falsely assumed that the case-by-case polynomial pattern I showed above could only be represented through the ceiling functions. This was wrong.

Answer 1

Hint

To get factors, write $$\frac{n (n+1)}{2}x^2 - x- 2 = \frac{n (n+1)}{2} \left(x^2 -\frac{2}{n(n+1)}x - \frac{2}{n} \cdot \frac{2}{n+1} \right) = \frac{n (n+1)}{2} \left(x^2 -(\frac{2}{n} -\frac{2}{n+1})x - \frac{2}{n} \cdot \frac{2}{n+1} \right) = \frac{n (n+1)}{2} \left( (x- \frac2n)(x+ \frac{2}{n+1}) \right)$$

Can you proceed?

Answer 2

Alternative factoring method:

When $B^2 - 4AC \geq 0,$ with $A > 0$, then $Ax^2 + Bx + C$ factors into

$A\left(x^2 + \frac{B}{A}x + \frac{C}{A}\right)$

$= A \left[\left(x + \frac{B}{2A}\right)^2 - \frac{B^2}{4A^2} + \frac{C}{A}\right]$.

$= A \left[\left(x + \frac{B}{2A}\right)^2 - \frac{B^2 - 4AC}{4A^2}\right]$

$= A \left[\left(x + \frac{B}{2A}\right)^2 - \left(\frac{\sqrt{B^2 - 4AC}}{2A}\right)^2\right]$

Under the assumptions, this factors into the difference of two squares as

$$= A\left[\left(x + \frac{B}{2A} + \frac{\sqrt{B^2 - 4AC}}{2A}\right) \times \left(x + \frac{B}{2A} - \frac{\sqrt{B^2 - 4AC}}{2A}\right)\right].\tag1$$

With the posted problem, you have that
$A = \frac{n(n+1)}{2}$
$B = -1$
$C = -2$.

From this you can immediately conclude (since $A>0, C<0$) that $B^2 - 4AC > 0$. Therefore, the formula in (1) above applies.

$\frac{B}{2A} = \frac{-1}{n(n+1)}.$
$\frac{\sqrt{B^2 - 4AC}}{2A} = \frac{\sqrt{1 + 4n(n+1)}}{n(n+1)} = \frac{(2n+1)}{n(n+1)}.$

Therefore, $Ax^2 + Bx + C$ factors into

$\frac{n(n+1)}{2}\left[\left(x + \frac{-1}{n(n+1)} + \frac{(2n+1)}{n(n+1)}\right) \times \left(x + \frac{-1}{n(n+1)} - \frac{(2n+1)}{n(n+1)}\right)\right].$

This simplifies into
$$\frac{1}{2n(n+1)} ~\left\{~ \left[~n(n+1)x + 2n\right] ~\times ~\left[~n(n+1)x - 2(n+1)~\right]~ \right\}.$$

This further simplifies to $$\frac{1}{2} \times \left[(n+1)x + 2\right] \times \left[nx - 2\right].\tag2$$

In (2) above, either $(n+1)$ or $n$ will be even, thus allowing the factor of $\frac{1}{2}$ to be cleared.

Therefore, you can forgo any consideration of the floor or ceiling functions, and simply divide the formula mentioned in (2) above into two cases: either $n$ is odd or $n$ is even.

Finally, while I showed the derivation of the formula in (2) above, my answer could have been significantly shorter, if I had instead provided a (sanity-checking) verification of the formula in (2) above.

That is, if you manually multiply the factors in (2) above, the product will be $Ax^2 + Bx + C$, where $A,B,C$ are as specified in your original question.