1. ## Polynomial problem

Stuck on a problem...
Can anyone help?

I am after any polynomial with integer coeff that passes through the points ( 1,3) and (3,2).
I started with quadratics and i think its not possible but i can't prove that either?

Hint/help would be much appreciated!

2. ## Re: Polynomial problem

Originally Posted by rodders
Stuck on a problem...
Can anyone help?

I am after any polynomial with integer coeff that passes through the points ( 1,3) and (3,2).
I started with quadratics and i think its not possible but i can't prove that either?

Hint/help would be much appreciated!
Two points determine a line (surely you learned that in geometry?)
The graph of a linear equation is of the form y= ax+ b. Since it passes through the point (1, 3) we must have 3= a(1)+ b. Since it passes through (3, 2) we must have 2= a(3)+ b. Solve those two equation for a and b.

You say "I started with quadratics and i think its not possible". It certainly is possible to find a parabola that passes through any two points- the problem is that there are an infinite number! We can write a quadratic as y= ax^2+ bx+ c. Since it is to pass through (1, 3) we must have 3= a(1)+b(1)+ c. Since it is to pass through (3, 2) we must have 2= 9a+ 3b+ c. That is only two equations with three unknowns. If we subtract a+ b+ c= 3 from 9a+ 3b+ c= 2 we get 8a+ 2b= 1. We can write that as b= (1/2)- 4a. Then a+ (1/2)- 4a+ c= -3a+ (1/2)+ c= 3. From that c= 3+ 3a- 1/2= 3a+ 5/2. Take a to be any number and those b and c will give a parabola passing through those two points. (If you take a= 0, you get the previous answer.)

3. ## Re: Polynomial problem

My problem is not trying to find any quadratic/polynomial but one with integer coefficients?
Thats where i am stuck!

4. ## Re: Polynomial problem

Hey rodders.

Can you try starting by setting up a couple of congruence relationships for your polynomial?

5. ## Re: Polynomial problem

Not sure what you mean? Like Diophantine equations?

6. ## Re: Polynomial problem

Look at this pattern:

(1, 3) & (3, 2)

ax + b = f(x)

3a + b = 2
a + b = 3
------------
2a = -1

An even left-hand side equal to an odd integer means the variable a cannot be equal to an integer.

$ax^2 + bx + c = f(x)$

9a + 3b + c = 2
a + b + c = 3
------------------
8a + 2b = -1

An even left-hand side equal to an odd integer means at least one of the variables cannot be an integer. *

Cubic:

$ax^3 + bx^2 + cx + d = f(x)$

27a + 9b + 3c + d = 2
a + b + c + d = 3
--------------------------
26a + 8b + 2c = -1

See * above.

And so on.

After subtracting the two equations for any increasing degree polynomial, the left-hand side will always be an
even-valued expression, while the right side equals 1.

So, there is no such polynomial function with integer coefficients that passes through those two points.

7. ## Re: Polynomial problem

Originally Posted by rodders
My problem is not trying to find any quadratic/polynomial but one with integer coefficients?
Thats where i am stuck!
Do you understand that if you get an equation with fraction coefficients,, you can immediately get one with integer coefficients by multiplying the entire equation by the least common denominator of the fractions?

8. ## Re: Polynomial problem

To make Greg's argument a bit clearer we can translate our curve so that the first point is at (0,0).

This is a translation of (-1,-3) and applied to the second point we get (2,-1).

Any polynomial without a constant term will pass through (0,0) so we are after

$\displaystyle{\sum_{k=1}^n}~c_k 2^k = -1$

$2\displaystyle{\sum_{k=1}^n}~c_k 2^{k-1} = -1$

$\displaystyle{\sum_{k=1}^n}~c_k 2^{k-1} = -\dfrac 1 2$

$c_k \in \mathbb{Z}$

$2^{k-1} \geq 1 ,~k=1,\dots n \Rightarrow 2^{k-1} \in \mathbb{Z}$

there is no way that the sum on the left ends up equaling a rational number as the integers are closed under multiplication and addition.

thus there is no polynomial with integer coefficients that passes through these two points and thus no polynomial with integer coefficients that passes through the original two points.

9. ## Re: Polynomial problem

Many thanks All! I sort of suspected this but i like the conclusive proof!