# Proving a polynomial cannot be factored with integer coefficients

1. Jun 16, 2017

### vishnu 73

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
i tried to do it by writing it as
$a_{1999} x^{1999} + a_{1998} x^{1998} ... a_0 \pm1 = 0$
for 1999 different integer values of x
i am thinking of writing it as
$a_{1999} x^{1999} = -a_{1998} x^{1998} - a_{1997} x ^ {1997} .... a_0 \pm1$
does this lend any information

i also tried to graph it
a 1999 degree polynomial has 1998 stationary points and at most 2000 points at which it is plus minus 1

Last edited: Jun 16, 2017
2. Jun 16, 2017

### Staff: Mentor

I would assume that you can factor it, and then look for a contradiction. What can you say about the value of the factors at these points?

Technical detail: The problem statement should say "non-constant polynomials", otherwise it is wrong: you can write p(x) as 1*p(x).

3. Jun 16, 2017

### Ray Vickson

Use braces: in TeX/LaTeX, if you have a subscript or superscript consisting of more than one character, you need to put it in braces, like this: a_{1999} and x^{1999}. Without braces, a_1999 and x^1999 give you get this: $a_1999, x^1999$; but with braces you get this: $a_{1999}, x^{1999}$.

4. Jun 16, 2017

### vishnu 73

@mfb
sure give me some time i will reply soon
@Ray Vickson
sorry about that still new to letex thanks edited it forgot to check after posting

5. Jun 17, 2017

### vishnu 73

@mfb
this is what i did
assuming it can be factored into a product of two polynomials with integer coeffecients it can be written as
$(a_n x^ n + a_{n-1} x^{n-1} .... a_0)(b_{1999-n}x^{1999-n} + b_{1998-n} x^{1998-n} .... b_0)\\ where \\ a_i , b_i = integers$
hence
$(a_n x^ n + a_{n-1} x^{n-1} .... a_0) = \pm 1\\ (b_{1999-n}x^{1999-n} + b_{1998-n} x^{1998-n} .... b_0) = \pm 1$
for 1999 different values of integer x
if n is greater than or equal to 2
then the degree of the b polynomial is less than 1998 and cannot be plus minus 1 for 1999 distinct values
but if n = 1 the degree of a polynomial is just 1 and certainly can at most be plus minus 1 2 times
hence it cannot be factored like this
is this correct?

6. Jun 17, 2017

### Staff: Mentor

But you have two polynomials: $a(x) + 1$ and $a(x) - 1$. Can't the $1999$ many zeros be spread on these two?

7. Jun 17, 2017

### vishnu 73

@fresh_42
i dont quite get why you are talking about zeros here
the question states the initial polynomial is ±1 for 1999 different integer values of x and since both of the a and b polynomial have only integer terms in them their absolute value must equal 1 also for 1999 values of x

oh wait now that i actually drawn it out i realise i am wrong
because a cubic graph can have 4 points with absolute value equal to 1 and a pentic graph has 8 such points and hence i conjectured for n th degree there will be 2n - 2 such points
so 1998 degree has 3994 such points not 1999 as i initially thought
and now that i think more
1000 degree can have 1998 such points so the b polynomial must be at least of at least 1001 degree and hence a becomes of degree 998 degree which is not possible now correct? i am really hesitant about this method as it seems very vague to me myself but that is the only one i can think of please help

8. Jun 17, 2017

### vishnu 73

no sorry once again after further checking my conjecture was wrong
it turns out that degree 3 graph has 6 such points and degree 5 graph has 10 such points so i now conjecture that number of such points is 2n
so the b polynomial must be atleast 1000 degree now then the a polynomial becomes 999 degree which only has 1998 such points which is not possible

9. Jun 17, 2017

### Staff: Mentor

Correct.

10. Jun 17, 2017

### vishnu 73

really wow thanks

11. Jun 18, 2017

### vishnu 73

12. Jun 18, 2017