polynomial with relatively prime numbers

[sayantica]

f(x)=(x-n1)(x-n2).....(x-nk)

where n1...nk are relatively prime to each other

find x s.t f(x) is only divisible by either x-n1,x-n2..x-nk

 

[sayantica]

suppose p isn't one of your nks and for contradiction let (x-p) divide f(x)

then (x-p) either divides (x-nk) for some k, or (x-p) is also a factor of f(x).

Show that

a) (x-p) can't divide (x-nk) (this should be trivial); and

b) that (x-p) isn't a factor of f(x) (it's basically stated that it isn't)

and you've shown by contradiction that p must be an nk in order for (x-p) to divide f(x).

We've said nothing at all about x here so this statement holds true for all x

Can u site an example to make it clear.

 

[daon2]

The way the problem is stated makes it false. For example, x fixed, 1 will always be a divisor even if \(\displaystyle 1\neq x-n_i\) for some i.

If you are asking for "nontrivial" divisors, then I still see a problem. Because, for example, if \(\displaystyle 1\neq f(x)=(x-n_1)(x-n_2)\) then \(\displaystyle N=(x-n_1)(x-n_2)\) is a divisor of \(\displaystyle f(x)\) but \(\displaystyle N\neq (x-n_i)\) for any i, as requested by your problem statement.