How to do this kind of question?

Suppose p(x) is a polynomial with integer coefficients. Show that if p(a) = 1 for some integer a then p(x) has at most two integer roots.

Question

How to do this kind of question?

Suppose p(x) is a polynomial with integer coefficients. Show that if p(a) = 1 for some integer a then p(x) has at most two integer roots.

ganeshie8 · Answer

in other words :

$\large  p(x) =   1 \mod  a$

then show that $p(x)$ has atmost 2 roots in mod a

nincompoop · Answer

this is going to be a long proof

anonymous · Answer

What is mod? How to do that?

ganeshie8 · Answer

yes nin... 
oh, this question is not from number theory ?

anonymous · Answer

I do not know.
I have hw as this question.

Can you teach please?
I don't know any mod.

ganeshie8 · Answer

looks @nincompoop  has something :)
im still thinking....

anonymous · Answer

ok

nincompoop · Answer

can we use intermediate value theorem for this one?

anonymous · Answer

What is that!?

nincompoop · Answer

probably not, because IVT only guarantees the existence of one value

nincompoop · Answer

I am sorry, dude. I was just learning last night how to do this via perturbation theorem so I won't be able to help you

anonymous · Answer

i have an idea of an approach but i did not finish it

anonymous · Answer

if $p(x)=a_nx^n++a_0$ 
and $p(r)=1$ then $q(x)=a_nx^n+...+a_0-1$ has root $r$ 
since $r$ is an integer, it divided the constant $a_0-1$

anonymous · Answer

@nincompoop no probs.

@satellite73 Can you explain a little?

anonymous · Answer

actually now i see i was wrong because it does not follow that $r$ divides $a_0-1$
gotta think some more
what class is this for?

anonymous · Answer

i think the approach is to look at the zeros of $p(x)-1$ since you know $p(a)=1$ but frankly i am not sure where to go from there

nincompoop · Answer

I was just reading this but it is beyond me http://web.mit.edu/yisun/www/notes/polynomials.pdf

ganeshie8 · Answer

First observation :

since we're looking for integer roots, then by rational root theorem,
if $n$ is a root, then   $n$ must divide the constant term $a_0$

ganeshie8 · Answer

Second observation :

from the hypothesis, $p(a) = 1$

$\implies  a | (a_0-1)  $

nincompoop · Answer

does rational root theorem necessarily have to be f(x) = 0
I was thinking of this approach earlier but declined to proceed since I thought the root pertained to zeroes.

nincompoop · Answer

is it going to be f(x) - 1 = 0 then?

ganeshie8 · Answer

yeah and next step is the conclusion

ganeshie8 · Answer

you have a point... rational root theorem gives possible roots,
not the acctual roots :|

ganeshie8 · Answer

however out first two observations are still valid
$a | a_0-1$ is still a necessary condition for $P(x)$ to leave  to leave a remainder $1$, when divided by $a$

akashdeepdeb · Answer

Hey! We have the same surname! If that is your surname. :P

I have SOME KIND OF solution to it. i'll post it in a minute.

ganeshie8 · Answer

yes :)

akashdeepdeb · Answer

I am not sure if it is correct though. :/

ganeshie8 · Answer

looks good to me xD
but il give a  good thought tomorrow :))

akashdeepdeb · Answer

Okay. Thanks!

ganeshie8 · Answer

@nincompoop  plz see .. 
il have a good look again in the morning :)

ikram002p · Answer

:o