Proving a polynomial has no integer solution

[Dick]

wow that was easy thanks a lot
is this is a standard trick in solving such problems
but can you help me with proving 1999 is not a root of the polynomial

I don't know that it's a standard anything. It's just something I thought of thinking about the problem. Showing there are no integer roots pretty much rules out 1999 being a root. If you mean an alternative way like haruspex and Buffu were suggesting, I'm not sure. Maybe they would like to elaborate...

 

[Buffu]

I don't know that it's a standard anything. It's just something I thought of thinking about the problem. Showing there are no integer roots pretty much rules out 1999 being a root. If you mean an alternative way like haruspex and Buffu were suggesting, I'm not sure. Maybe they would like to elaborate...

I think I have to say sorry because now I don't know how to prove $1999$ is not a root. When I saw this question I scribbled something on paper which looked convincing to me at that point as a proof but now I look at it again I think I am missing one key element. I am really sorry.

Here is my approach, I think one key information is missing,

Using vieta's formula,

$\dfrac{|a_1|}{|a_n|} = |r_2 ... r_n + r_1 r_3 ... r_n + ... + r_1 r_2... r_{n-1}|$

$|a_1| =r_1 \left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$

I thought it was easy enough to see that $a_1$ is not a integer if $r_1 = 1999$ but now that I look at the proof again I see that I don't have any convincing argument to prove that $\left| \dfrac{1}{r_1} + \dfrac{1}{r_2} + \cdots + \dfrac1{r_n}\right|$ is not a integer.

I think @haruspex have a proof that 1999 is not root.

 

[haruspex]

I think @haruspex have a proof that 1999 is not root

Like you, I thought I did, but it didn't pan out. I could show (if 1999 is a root) that a₁=-1 and that 1999 must divide a₂ exactly once more than it divides a₃. After that it petered out.
An interesting corollary of Dick's observation is that if a polynomial has integer coefficients then the gradient between two integer values of x is also an integer.

can you help me with proving 1999 is not a root of the polynomial

Haven't you proved that in post #28?

 

[timetraveller123]

yup that happened to me too at first before i posted here happens to every one but thanks a lot
@haruspex
how did you show if r₁ is 1999 then a₁ = -1 please tell me
@Dick
what i want to know how did you just pull out such a thing like how did you notice about the formula you said that's what i want to know that's why i asked if it is a standard thing and it is astonishing to me that you randomly thought of that please tell me thanks every one

 

[Dick]

yup that happened to me too at first before i posted here happens to every one but thanks a lot
@haruspex
how did you show if r₁ is 1999 then a₁ = -1 please tell me
@Dick
what i want to know how did you just pull out such a thing like how did you notice about the formula you said that's what i want to know that's why i asked if it is a standard thing and it is astonishing to me that you randomly thought of that please tell me thanks every one

Well, it wasn't exactly 'random'. I thought if there is a root then $f(n)=0$ for some $n$. Ok, so $f(n)-f(0)=f(n)-f(1)=-1999$ and 1999 is prime. Must be something wrong with that. Must be a factor. Then look at the differences and notice things like $n^i-1^i$ have $n-1$ as a factor. That's all. It's not that big a leap if you think along the right lines. Getting stuck trying to get the result using the rational roots theorem appears to be a bit of a trap.

 

[timetraveller123]

oh now i see i am too used to seeing polynomials in the standard form thanks a lot to everyone
@haruspex @Buffu @Dick