How many of the following expressions (quadratics) factor for $n \leq 2018$

Question

I know that $x^2+x-1$ does not factor over the integers but $x^2+x-2$ does (i.e.: (x+2)(x-1)). If I have expressions of the form $x^2+x - n$ $\forall n \leq 2018$, how many of the expressions factor for $n \leq 2018$? I'm trying to notice a pattern because for the first 21 of these expressions I have that:

n = 2, 6, 12, 20 factor but the rest don't. How can I solve this?

Answer 1

The discriminat has to be perfect square, so $$1+4n = a^2$$ for some $a$. Now since $n\leq 2018$ we have $$a^2 \leq 8073\implies a\leq 89$$ and thus since $a$ must be odd it is for $45$ values reducible.

Answer 2

Firstly, you can notice that 2,6,12,20 are all of the form $a(a-1)$ (for $a=2,3,4,5$). Let's prove that this is always the case.

If $x^2+x-n=(x+a)(x+b)$, then $x^2+x-n=x^2+(a+b)x+ab$. Thus,

• $a+b=1 \Rightarrow b=a-1$
• $n=-ab=a\cdot(-b)=a(a-1)$.

So, this quadratic factors over integers iff $n=a(a-1)$ for some integer $a$.

To count for how many $n\leq 2018$ this is the case, observe that

• For $a=45$: $n = 44\cdot 45=1980\leq 2018$
• For $a=46$: $n=45\cdot 46 =2070>2018$

So, for $a=2\dots 45$ the quadric $x^2+x-a(a-1)$ factors in integers, and there are $44$ such quadrics.