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Re: Problem 027
Posted: Thu Dec 19, 2019 10:31 am
by jaap
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
by kenbrooker
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
by jaap
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
by kenbrooker
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
hk for pre-addressing Lucas-C's question!!
Happy Polynomial Holidays,
glasshopper
Re: Problem 027
Posted: Sun Dec 22, 2019 2:45 am
by kenbrooker
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
by kenbrooker
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...
Re: Problem 027
Posted: Wed Jun 23, 2021 2:30 am
by benbradley
Part of this problem states "The product of the coefficients, −79 and 1601, is −126479." Does this have any relevance whatsoever to the problem or the solution? It sure seems not.
ETA: Oh, never mind, I see at the end where one is to find the product of the coefficents of the appropriate polynomial to enter this product as the numerical answer. The statement above didn't make sense by itself.