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?

lisyarus · Answer 1 · 2018-10-29T20:43:27.667

1

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.

edited Oct 29 '18 at 20:43

answered Oct 29 '18 at 19:22

lisyarus

15,517

where are you getting the 44 from? – user8358234 Oct 29 '18 at 19:39
shouldn't it be for $a = 44, n = 44*43$? – user8358234 Oct 29 '18 at 19:42
@user8358234 Indeed, thank you. I've fixed the answer. – lisyarus Oct 29 '18 at 20:44

score 1 · Accepted Answer · answered Oct 29 '18 at 19:26

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.

answered Oct 29 '18 at 19:26

nonuser

90,026

how did you get 8073? – user8358234 Oct 29 '18 at 19:37
and hw are you getting 45? shouldn't a have to be even? – user8358234 Oct 29 '18 at 19:38
$4\cdot 2018+1 = 8073$ and $n+1$ is odd so $a$ is – nonuser Oct 29 '18 at 19:39
oh i see, so then you get $a\leq 89$ but then how do you arrive at $45$? I know that half the numbers must be even, and the other half odd but I don't see how it's relevent to the question – user8358234 Oct 29 '18 at 19:40
i guess my question is what is the significance of 45? – user8358234 Oct 29 '18 at 19:40
The question was, for how many values $n$ the expresion is reducilbile. The answer is it is reducibile iff it has integer root and that is when the discriminat is perfect square and so for 45 values. – nonuser Oct 29 '18 at 19:43
but I thought it was a perfect square for 89 values? – user8358234 Oct 29 '18 at 19:44
because you have plus or minus the number – user8358234 Oct 29 '18 at 19:44
Yes but, $a$ must be odd since 4n+1 is odd – nonuser Oct 29 '18 at 19:44
oh i see, but how do you know that it's not 43? because half of 90 is 45 so half of 89 is less than 45? – user8358234 Oct 29 '18 at 19:45
1

odd numbers are of the form 2k-1, if $k=1$ we get $2k-1=1$ and if $k=45$ we get $2k-1 =89$ – nonuser Oct 29 '18 at 19:48
Do you understand? – nonuser Oct 29 '18 at 20:25
Yes thank you very much for your help – user8358234 Oct 29 '18 at 21:45

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

2 Answers2