# Proving a polynomial has no integer solution

1. Jun 6, 2017

### vishnu 73

1. The problem statement, all variables and given/known data
let p(x) be a polynomial with integer coefficients satisfying p(0) = p(1) = 1999
show that p has no integer zeros

2. Relevant equations

3. The attempt at a solution
$p(x) = \sum_{i= 0}^{n}{a_i x^i}$

using the given information
a0 = 1999( a prime number)
and
$a_n + a_{n-1} ... a_1 = 0$

then rewriting p(x)

$p(x) = a_n(x_n + \frac{a_{n-1}}{a_n} x^{n-1} ... \frac{1999}{a_n})\\ p(x) = a_n(x-r_1)(x - r_2) ....(x - r_n)\\$
i am hoping to do the rest of the proving by contradiction
if i assume the polynomial has integer solution then how can i disprove it

2. Jun 6, 2017

### Buffu

Use Rational roots theorem.

If $p/q$ is a root of the given polynomial then $p$ must be a factor of $a_0$.
Which means $p = 1$ or $p = 1999$.

3. Jun 6, 2017

### vishnu 73

ya i tried doing that what do i get from that

4. Jun 6, 2017

### Buffu

You know that $q$ is a factor of $a_n$. If $p/q$ is a integer then what condition do you get on $q$ for $p = 1$ ?

5. Jun 6, 2017

### vishnu 73

$P(x) = (qx - p)(Q(x))\\ q|a_n \\ p|1999\\ p = \pm 1 or \pm 1999\\$

if p = ±1 then q must also equal ±1
if p = ±1999 then q must equal ±1 or ±1999

6. Jun 6, 2017

### Buffu

So what the possible integer values ?

7. Jun 6, 2017

### vishnu 73

you mean the roots the possible integer roots are ±1 and ±1999

8. Jun 6, 2017

### Buffu

Sorry. Yes I mean roots.

9. Jun 6, 2017

### vishnu 73

you mean the roots the possible integer roots are ±1 and ±1999

10. Jun 6, 2017

### Buffu

$1$ is out because $p(1) = 1999$. Can you eliminate others ?

11. Jun 6, 2017

### vishnu 73

i am sorry i just cant think of ways to eliminate the rest give me another hint
edit:
after much trying managed to eliminate just 1 of the roots
if we a
$a_n + a_{n- 1} ... a_1 = 0\\ a_n + a_{n-1} .... a_2 = -a_1\\ p(-1) = a_n(-1)^n + a_{n-1} (-1)^n-1 .... -a_1 + 1999\\ p(-1) = a_n(-1)^n + a_{n-1} (-1)^n-1 .... a_2 +a_2 +a_3 + .... a_n + 1999\\$
the right hand side will cancel out all the terms with odd subscripts leaving only 2 of each even subscript terms and since they are integers the right hand side is 1 mod 2 thus cannot be zero is this correct?
please give me hints for the next two roots

Last edited: Jun 6, 2017
12. Jun 6, 2017

### Buffu

Yes except I think it should be $p(-1) = a_n(-1)^n + a_{n-1} (-1)^\color{green}{n-1} .... a_2 +a_2 +a_3 + .... a_n + 1999$.

Hint :

Product of roots is $|a_0/a_n|$

Or $\dfrac{|a_0|}{|a_n|} = |r_1r_2 \cdots r_n|$

If one root is $1999$
then,
$|1999| = |a_n||r_2 \cdots r_n| \times 1999 \implies 1 = |a_n||r_2 \cdots r_n|$

13. Jun 6, 2017

### haruspex

If 1999 is a root, how many times might 1999 divide ai1999i? Given that it divides a0 exactly once, what does that tell you about a1? Then, what about the relationship between a2 and a3?

Last edited: Jun 6, 2017
14. Jun 6, 2017

### vishnu 73

@haruspex
i think what @Buffu wrote is correct as he substituted $- a_1 = a_2 + a_3 ... a_n$

@Buffu
following on what you said
$1 = (\mid a_n \mid)(\mid r_2 r_3 ...r_n)$ since $a_n$ is integer$\mid a_n \mid = 1$
then this make $p(x)$ a monic polynomial then its roots can only be integers or irrationals not fractions then from $\mid r_2 r_3 ...r_n\mid = 1$ we get that $r_i$ has to equal 1 but its shown that 1 is not possible hence 1999 is not possible is this valid ?

15. Jun 6, 2017

### haruspex

Ok, I misinterpreted the ......

I'm not following. If $\mid r_2 r_3 ...r_n\mid=\frac 12$ then $(\mid a_n \mid)=2$, etc. How are you ruling that out?

16. Jun 7, 2017

### vishnu 73

oh ya i got over excited and did that and i am stuck again one more hint please is there some kind of restriction on an
should i be writing $\mid a_n \mid$ as $\mid a_1 + a_2 ... a_{n-1} \mid$ does that help

Last edited: Jun 7, 2017
17. Jun 7, 2017

### haruspex

Did you think about my post #13?

18. Jun 7, 2017

### vishnu 73

oh i completely didn't see the post only noticed it after you pointed it out don't know what happened there will take a look now thanks oh wait you edited i forgot to see the edit

19. Jun 7, 2017

### vishnu 73

@ haruspex
are you hinting at geometric series
because i thought of it at first but the fact that a2 + ....an = 0
then i considered all the terms in that to be same but alternating signs but that forces n to be odd this was initial line of thought
i am sorry if am not seeing what you are suggesting

20. Jun 7, 2017

### haruspex

If you plug x=1999 into the polynomial you will get a factor of 1999 in every term of the sum. Some terms will have extra factors of 1999. But in the end, the sum is zero if 1999 is a root. Can there be one term with fewer factors of 1999 than the rest? What does that tell you about a1?