Let $N\in \mathbb{N}$ and $N = pq$ where $p,q\in \mathbb{N}$. I am interested in the functions $y = ax^2 + bx + c$ where $a,b,c$ are chosen so that the parabola passes through the three points $(0,0)$, $(p,N)$, and $(q,N)$. For a fix number $N$ I am interested in the set of parabolas generated by all pairs of factors $N = pq$. Is there any relationship amount the parabolas in this set?

Example for $N = 12$

Given three functions for parabolas:

  • $y = -(x-3.5)^2 + 3.5^2$
  • $y = -(x-4)^2 + 4^2$
  • $y = -(x-6.5)^2 + 6.5^2$

such that they all share a common intersection of points at $y = 12$

is there a way to ascertain a relationship between all three functions?

I have provided a graphic illustrating the intersection along $y=12$ that shows how the functions pass through factors of 12 on a graph.


  • $\begingroup$ I am looking, either through algebra or calculus, a way to ascertain, other than graphically, a common element or functional relationship between all three parabolas on the line where y equals 12. $\endgroup$ – John Edward Law Nov 9 '15 at 23:17
  • $\begingroup$ I am trying to better understand the relationship between these parabolas and how they relate to aligning with the factors of twelve; something related to these three quadratics opening downward and yes -- all passing through (0,0) on their left-side "leg". I find a similar pattern if I look at y = 14 (using functions y = -1(x-4.5)^2 + 4.5^2 and y = -1(x-7.5)^2 + 7.5^2) and y = 16 (using functions y = -1(x-4)^2 + 4^2, y = -1(x-5)^2 + 5^2 and y = -1(x-8.5)^2 + 8.5). Similar parabolas can be formed for odd numbers that have more than just 1 and itself as factors. $\endgroup$ – John Edward Law Nov 9 '15 at 23:40
  • $\begingroup$ Hehe thanks! Yeah, this is what happens when using graphic of parabolas with little to no sleep and starting to see patterns. $\endgroup$ – John Edward Law Nov 10 '15 at 0:30
  • $\begingroup$ As I am not much of a number theorist myself, I am deleting my comments to avoid suggesting that I've provided any other help other than clearing up the question. $\endgroup$ – Paul Sinclair Nov 10 '15 at 0:36

A related set of parabolas would be the ones passing through the points $(0, -N), (p, 0)$ and $(q,0)$. They would have the form $y = -(x-p)(x-q)$. Your parabolas would then be of the form $$y = N - (x-p)(x-q) = (p+q)x - x^2 = -x(x-(p+q))$$

The roots would be $0$ and $p+q$.

The axis of symmetry would be at $x = \frac 12(p + q)$.

The vertices would be at $\left( \dfrac{p+q}{2}, \dfrac{(p+q)^2}{4} \right)$, which means that they would all lie along the line $y = x^2$ in the first quadrant.


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