Roots and stationary points of cubics and integer values

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apple2357

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This cubic has a nice property because it has three distinct roots, such that roots and stationary points all lie on points with integer coordinates..
y= x(x+9(x-15).

1608202652516.png


I want to generate other cubics with this property. Is it obvious that translating this curve horizontally will retain this property? it seems to work but whilst it is obvious it will keep integer values for the intercepts i can't quite explain why the stationary points will remain as integers. Is it obvious?
 
If you translate the graph horizontally, all points will move the same distance. If the coords of the stationary points are both integers to start with, they will remain so if you translate the graph horizontally by an integer value. Or vertically for that matter.
 
But vertically will not keep the intercepts at integer values though. Neither stretching seems to do that either.
 
Yes, it is obvious that translating horizontally an integer distance will preserve the fact that turning points have integer coordinates. Translating horizontally an integer distance guarantees that the x coordinates remain integer and the y coordinates don't change at all.

(There is an old story of a math professor saying to the class "Now, it is obvious that...", stopping and saying "Why is that obvious?", sitting down at his desk, scribbling furiously for ten minutes, then standing up and announcing "Yes, it is obvious!")
 
So i tried to find other cubics that have this property and seeing if i could generalise the problem. I know the above generates infinite number but only through translating and thinking about it graphically. I went back to the algebra:

I started with f(x) = x(x+a)(x-b)
found f'(x) and set it equal to zero. Solving this equation using the quadratic formula i am left with trying to make the following an integer:

{(b-a) +- sqrt ( a^2+b^2+ab)} /3

so b-a will need to be multiple of 3 which means b and a will need to have a gap of 3.
And the second bit a^2+b^2+ab will need to be a multiple of 9

By trial and error a=9 and b =15 satisfies the above ( b-a =6 and a^2+b^2+ab = 441= 9*49)

I cant see a way to generate other solutions, any thoughts or suggestions i could play with are most welcome!
 
y=ax^3 + bx^2 + cx + d
y' = 3ax^2 + 2bx + c
y'=0 when x= {-b +-sqrt(b^2-3ac)}/(3a)
y"= 6ax+2b
y"= 0 when x = -b/3a

So you want {-b +-sqrt(b^2-3ac)}/(3a) and -b/3a to be integers.
 
Jomo said:
y=ax^3 + bx^2 + cx + d
y' = 3ax^2 + 2bx + c
y'=0 when x= {-b +-sqrt(b^2-3ac)}/(3a)
y"= 6ax+2b
y"= 0 when x = -b/3a

So you want {-b +-sqrt(b^2-3ac)}/(3a) and -b/3a to be integers.
Click to expand...

so shoudl i just pick some values a=1, b=3 say and see what c needs to be generate an integer ? What about d?
 
apple2357 said:
But vertically will not keep the intercepts at integer values though. Neither stretching seems to do that either.
Click to expand...
I wonder about that. If you make the graph twice (or triple or quadruple or ....) was wide why would the x-values not double?
 
Jomo said:
I wonder about that. If you make the graph twice (or triple or quadruple or ....) was wide why would the x-values not double?
Click to expand...
Stretching SF 2 will work for example:

1608824652110.png
But stretch SF 1/2 will not as the roots/ turning point coords may stop being integers..? For example:

1608824761347.png
 
All these are based on the original y= x(x+9)(x-15)... I am still searching for other non related functions
 
apple2357 said:
y= x(x+9(x-15).
Click to expand...

Don't forget reflection y= x(x-9)(x+15). But similar to integer magnification, and translation, it doesn't give a very "interesting" set of additional solutions.

I wrote a computer program to perform a search. There are many more interesting solutions. There's THREE distinct solutions of the form:-

y= x(x-24)(x-b) where abs(b)<24
15 -9

24 -21
24 9
24 15

45 21
45 24

48 -15

63 15
63 48

72 -33

99 -21

105 -39
105 33
105 72
...

However it would be nice if there was a better way to find these more mathematically rather than using brute force.
 
Cubist said:
Don't forget reflection y= x(x-9)(x+15)
Click to expand...

Note the above is a double reflection about both X and Y axes.

For a reflection about Y axis only (which is what I actually intended!) then y= -x(x-9)(x+15)
 
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