## Problem 027

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Comments, questions and clarifications about PE problems.

### Re: Problem 027

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.

_{Jaap's Puzzle Page}

- kenbrooker
**Posts:**137**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

### Re: Problem 027

*Thanks*

If you'd care to elaborate on such a simple relationship @ PE.net,

I would sure appreciate it, as I imagine

Happy Holidays

"Irregardless"

**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"

"

*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."Experience comes from Bad Judgment

### Re: Problem 027

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.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-Cwould too...

_{Jaap's Puzzle Page}

- kenbrooker
**Posts:**137**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

### Re: Problem 027

*Thanks Much*

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

**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

"

*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."Experience comes from Bad Judgment

- kenbrooker
**Posts:**137**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

### Re: Problem 027

*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 (

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

Maybe "Royalties" are in order?

Happy Parabolic Holidays to

first and last initials...

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 byfirst and last initials...

"

*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."Experience comes from Bad Judgment

- kenbrooker
**Posts:**137**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

### Re: Problem 027

In summary of/to

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...

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

Thanks to

However, I don't see that finding b = f(a) -- a parabola -- explains

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**'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..."

*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."Experience comes from Bad Judgment