Disproving a Polynomial with Integer Coefficients: Elementary Math Proof

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A polynomial with integer coefficients evaluated at odd integers 1 and 3 must yield results of the same parity. Since both inputs are odd, if one output is even, the other must also be even, contradicting the premise that f(1) is even and f(3) is odd. The difference f(3) - f(1) must also be even, reinforcing that both outputs cannot differ in parity. Therefore, it is impossible to have a polynomial that satisfies the given conditions. The conclusion is that no such polynomial exists.
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Homework Statement


Disprove the following: There exists a polynomial f(x) with integer coefficients such that f(1) is even and f(3) is odd.

Homework Equations

The Attempt at a Solution


It's a little bit intuitive.
Proof
1 and 3 have the same parity. They are both odd
so if(odd)=odd then f(1)=odd and f(3)=odd
or if(odd)=even then f(1)=even and f(3)=even

is that right?
 
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lolo94 said:

Homework Statement


Disprove the following: There exists a polynomial f(x) with integer coefficients such that f(1) is even and f(3) is odd.

Homework Equations

The Attempt at a Solution


It's a little bit intuitive.
Proof
1 and 3 have the same parity. They are both odd
so if(odd)=odd then f(1)=odd and f(3)=odd
or if(odd)=even then f(1)=even and f(3)=even

is that right?

No, it is not correct, because you are essentially assuming what you want to prove. In order to be able to assert that f(odd) = odd, for example, you need to show that it is not possible to have f(odd1) = odd but f(odd2) = even, etc.
 
Hint: if there was such a polynomial ##f(3)-f(1)## would be both even and odd.
 
geoffrey159 said:
Hint: if there was such a polynomial ##f(3)-f(1)## would be both even and odd.
why would they both be even and odd?
 
lolo94 said:
why would they both be even and odd?
What do you know about ##f(3) - f(1)##?
 
lolo94 said:
why would they both be even and odd?
Let the polynomial be Σpnxn. What does f(3)-f(1) look like?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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