Question

help!

Answer 1

@SithsAndGiggles

Answer 2

Hey!

Answer 3

why to check integer roots of a polynomial modulo is used. I am bit confused with such notion. Anyone please explain the use of modulo precisely in proving that polynomial has no integer roots.

Answer 4

which notation ?

Answer 5

wait! is it about the constant term being divisible by 2. I get this ?!! But why 2

Answer 6

2 has nothing to do with the problem

since 1) the coefficients are integers and 2) you're looking for integer roots,
you are allowed to use modular arithmetic here

Answer 7

consider a quick example :
\[P(x) = x^2-2x +1 = 0 \]

suppose \(n\) is an integer root, then we have :
\[P(n) = n^2-2n+1 = 0\]

we are allowed to take mod both sides because we're dealing with integers (n and all coefficients are integers)

Answer 8

taking mod 2 both sides you get
\[ n^2-2n+1 \equiv 0 \pmod{2}\]

which is same as
\[ n^2+1 \equiv 0 \pmod{2}\]

Answer 9

computations get simpler in modular arithmetic,
observe that middle -2n term has just disappeared in mod2!

Answer 10

they are picking mod2 because it is simpler

Answer 11

It doesn't matter which modulus, \(m\), you pick
it will be true everywhere.

Answer 12

so is below statement clear ?

If \(n\) is an integer root of a polynomial \(P(x)\) with integer coeffcients, then \(P(n) \equiv 0 \pmod{m}\).

Answer 13

yes! now for the question. Can the same technique be used. Like f(x)=(x-a)(x-b)(x-c)(x-d)r(x)+5 and.......?????

Answer 14

what same technique ? do you have a solution ?

Answer 15

The actual problem turns out to be much simpler, we don't need fancy modular arithmetic etc...

Answer 16

okay ! :)

Answer 17

Lets first look at a rough sketch of \(f(x)\)
since \(f(a)=f(b)=f(c)=f(d)=5\), the graph looks something like :