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### Re: Problem 027

Posted: Thu Dec 19, 2019 10:31 am
All the solutions on a parabola are related to each other in a very simple way. Given that the solution to the problem lies on such a parabola, it would be a spoiler to discuss it too much.

### Re: Problem 027

Posted: Thu Dec 19, 2019 5:54 pm
Thanks jaap...

If you'd care to elaborate on such a simple relationship @ PE.net,
I would sure appreciate it, as I imagine
Lucas-C would too...

Happy Holidays
"Irregardless"

### Re: Problem 027

Posted: Fri Dec 20, 2019 5:23 am
kenbrooker wrote: Thu Dec 19, 2019 5:54 pmIf you'd care to elaborate on such a simple relationship @ PE.net,
I would sure appreciate it, as I imagine
Lucas-C would too...
Take a look at hk's post on the first page of the problem's discussion forum. It shows the relationship between two solutions. If you plot all the related solutions (i.e. for all p as used in his post) you get the parabola.

### Re: Problem 027

Posted: Fri Dec 20, 2019 9:15 am
Thanks Much jaap...

I will do that next, even at 1am here on the USA's West Coast; and,
my compliments to you for finding that and to

Happy Polynomial Holidays,
glasshopper

### Re: Problem 027

Posted: Sun Dec 22, 2019 2:45 am
Speaking of PE27 and parabolas, came across this definition of a parabola's equation --

The standard form is (x - h)^2 = 4p (y - k), where the focus is (h, k + p) and the directrix is y = k - p.
If the parabola is rotated so that its vertex is (h,k) and its axis of symmetry is parallel to the x-axis,
it has an equation of (y - k)^2 = 4p (x - h), where the focus is (h + p, k) and the directrix is x = h - p.

Sure enough, there's hk right in the thick of it!
Maybe "Royalties" are in order?

Happy Parabolic Holidays to
hk and All other vertices by
first and last initials...

### Re: Problem 027

Posted: Tue Dec 24, 2019 5:46 am
In summary of/to Lucas-C's challenge --

In a graph of a,b pairs such that prime P = N^2 + aN + b for -100 <= a <= 100 and 0 <= b <= 2000 and
such that the number of Ps for consecutive values of N is greater than 15, for example...
Why is an apparent parabola displayed?

I used the same graphical approach and found b = f(a), indeed a parabola...

Thanks to jaap and hk, I derived the same b = f(a) algebraically...

However, I don't see that finding b = f(a) -- a parabola -- explains
Why is b = f(a) a parabola?
Is that not at all...
Surprising???

Or, maybe I get it -- WHO CARES?! Maybe as moot as --
If you want a headlight with
parallel rays, use a
reflector that's...
Parabolic!!

Welcome 2020 --
THE Year of...
VISION!!

Lucas-C I sent you a PM...