# Polynomial with integer coefficients of degree d

Let $P(X)$ be a polynomial with integer coefficients of degree $d>0$.

$(a)$ If $\alpha$ and $\beta$ are two integers such that $P(\alpha)=1$ and $P(\beta)=-1$ , then prove that $|\beta - \alpha|$ divides $2$.

$(b)$ Prove that the number of distinct integer roots of $P^2(x)-1$ is atmost $d+2$.

Let $P(X)=a_0+a_1x+a_2x^2+... +a_dx^d$where $a_0, a_1,..., a_d$ are inegers . Then $P(\alpha)-P(\beta)=a_1(\alpha-\beta)+a_2(\alpha^2-\beta^2)+... a_d(\alpha^d-\beta^d)$. It is clear that $\alpha-\beta$ divides $P(\alpha)-P(\beta)$. I am not able to solve second part. Any ideas? Thanks.

As commented, we consider the roots of $P(x)=1$ and $P(x)=-1$. If the roots are from solely one of the equations, then we are done. Now suppose that the roots are from both equations. We can imagine that one solution of the first equation as a white(W) ball, one solution of the other equation as a black(B) ball. Notice $|\alpha-\beta|$ divides 2, if a solution B exsits, there is at most 4 solutions W exsit, namely B-2,B-1,B+1,B+2, otherwise the distance is lager than 2. We have the following cases:

1. All 4 solutions W exist. In this case, we can not find another solution B, since the distance of some W and the new B will always be larger than 2. Therefore we have at most $d+1$ solutions, namely $d$ from the first and 1 from the second.

2. 3 solutions W exist. There are two case: WWBW* or WWB*W, where * stands for that this number is no solution of the both equaitons. In the first case, it is impossible to give a new B, and we have at most $d+1$ solutions; in the second case, we have at most one extra solution $B-1$, and we can not find anymore solutions B. Therefore we have at most $d+2$ solutions.

3. 2 solutions W exist. There are 3 possibilities: $*WBW*, W*BW*,W*B*W,WWB**$. In the first and third case, we can not add one more B solution, therefore we have at most d+1 solutions. In the second case, we can add at most one more B solution, and we have at most d+2 solutions. In the fourth case, we have at most one more B solution, namely BWWB, and we can not add anymore B solution. We have then at most d+2 solutions.

4. 1 solution W exist. It is then WB. We now have 3 possibilities to add new solution B: $B*WB*,*BWB*,**WBB$, but this go back to previous, and we can add at most 2 more W solutions, and therefore we have at most d+2 solutions. This completes the proof.

• Thanks for your answer.but I couldn't get the cases.I mean how does numbers of solutions of first equation affect the total number of solutions.Can you please explain a bit more – Navin May 12 '17 at 14:44
• Lets say, you have a solution $x$ for the equation $P(x)=1$, then the solution for the equation $P(x)=-1$ can only be one element of the set $\{x-2,x-1,x+1,x+2\}$, since their difference is a divider of $2$. The same holds vice versa if $x$ is a solution of $P(x)=-1$. Therefore we obtain a restriction on the numbers of the solutions. – ehochix May 12 '17 at 15:07
• And for instance, for the first case, we have $x-2,x-1,x,x+1,x+2$, where $x$ is a solution of the second equation $P(x)=-1$, and the others are solutions of the first equation $P(x)=1$. Now if we have one more solution $y$ for the second equation $P(x)=-1$, then it must be less or greater than $x$. For the first case, its distance to $x+2$ is larger than $2$, and for the second case, its distance to $x-2$ is larger than $2$, so both situations are not allowed due to your result in the first problem (a). – ehochix May 12 '17 at 15:12
• Yes in your first case if for a particular solution for the second equation only four solutions of first equation are possible.Therefore there should be 5 solutions in total.I couldnot get when you mention that there are atmost $d+1$ solutions namely d from the first and 1 from the second. – Navin May 12 '17 at 15:49
• Notice $|\alpha-\beta|$ divides $2$ is valid for all solutions $\alpha$ and $\beta$. If we have at least two solutions of the second case, then these two solutions should all satisfy this restriction. But from our construction we can see this is impossible. Therefore we can have at most one solution for the second equation. – ehochix May 12 '17 at 15:54

Hint: $$P^2(x)-1=(P(x)+1)(P(x)-1)$$ $$(P(x)+1)(P(x)-1)=0$$

$$a_0-1+a_1x+a_2x^2+....+a_dx^d=0\tag{d roots possible}$$ $$a_0+1+a_1x+a_2x^2+....+a_dx^d=0\tag{d roots possible}$$

• This gives at most $2d$ roots, not $d+2$. – lhf May 11 '17 at 10:56