There exists an integer $n$ such that $\theta(x-n)=0$; indeed, just take $n=\theta(x)$. We claim that $\theta(p(x))=p(n)$ for all polynomials $p(x)$. Indeed, note that for $m\neq n$ we have $n-m=\theta(x-m)\mid \theta(p(x)-p(m))=\theta(p(x))-p(m)$. Note that $n-m\mid p(n)-p(m)$ so $n-m\mid \theta (p(x))-p(n)$. Note that the right hand side is fixed for all $m$, while the absolute value of the left hand side can be arbitrarily large, so we are forced to have $\theta(p(x))-p(n)=0$ as desired. $\square$

Lemma: For constant polynomials $p\equiv c,$ $\theta(c)=c.$ Proof: We show that $\theta(0)=0.$ Indeed, $c=\theta(c) \mid \theta(0)$ for all but at one $c$ (that is equal to $0$). This shows that $\theta(0)=0,$ and the lemma follows. Now the key idea is that for any polynomial $p,$ \[x-c \mid P(x)-P(c)\implies \theta(x)-c=\theta(x-c) \mid \theta(P(x)-P(c))=\theta(P(x))-P(c)\]for all but at most $1$ value of $c$ (namely, $c=\theta(x)$). For these values of $c,$ we see that \[\theta(P(x))-P(c)\equiv \theta(P(x))-P(\theta(x)) \mod \theta(x)-c,\]which implies that $\theta(P(x)) = P(\theta(x))$ since we can vary $c.$ Now if $\theta(x)=a,$ we see that $\boxed{\theta(p)\equiv p(a)}$ is our solution set (this is easily checked to work).

I like this problem a lot more than I normally would Posting my solution that's not at all new v_Enhance wrote: As usual, let ${\mathbb Z}[x]$ denote the set of single-variable polynomials in $x$ with integer coefficients. Find all functions $\theta : {\mathbb Z}[x] \to {\mathbb Z}$ such that for any polynomials $p,q \in {\mathbb Z}[x]$, $\theta(p+1) = \theta(p)+1$, and if $\theta(p) \neq 0$ then $\theta(p)$ divides $\theta(p \cdot q)$. Evan Chen and Yang Liu Observe that $x-N \mid p(x)-p(N)$ and $\theta(x)-N=\theta(x-N) \mid \theta(p(x))-\theta(p(N))$ for all $p \in \mathbb{Z}[x]$. Now $\theta(x)-N \mid p(\theta(x))-p(N)$. Now observe that for some integer $a$ we have $\theta(a)=0$ so $\theta(n)=n-a$ for all integers $n$. Let $\theta(x)=c$ then $\theta(x)-N \mid p(\theta(x))-\theta(p(x))-a$ so $c-N \mid p(c)-a-\theta(p)$; but for $N \rightarrow \infty$ this yields $p(c)-a=\theta(p)$. Now $\theta(p+N) \mid \theta(0)$ so $\theta(0)=0$ and $a=0$ (easy to prove). Thus, we conclude $\theta(p)=p(c)$ for some fixed $c \in \mathbb{Z}$. All of these are solutions.

We will show that $P(\theta (x))= \theta (P(x))$ for all $x\in \mathbb{Z}$. Let $\theta (x)=a$ and $\theta (P(x))=b$. For all $n\in \mathbb{Z}$ that $n<a$, we get $\theta (x-n)=\theta (x)-n=a-n>0$. And also $\theta (P(x)-P(n)) =\theta (P(x))-P(n)=b-P(n)$. Since $P(x)-P(n)=(x-n)\cdot q$ for some $q\in \mathbb{Z}[x]$, we get $a-n\mid b-P(n)$ for all $n<a$. It's clear that $P(n)=(n-a) r +P(a)$ for some $r\in \mathbb{Z}$. Hence, $a-n\mid b-P(a)$ for all $n<a$. This clearly gives $b-P(a)=0\implies P(\theta (x)) =\theta (P(x))$. This means $\theta (P)=P(c)$ for some integer constant $c$, which is a solution for all $c\in \mathbb{Z}$.

Question:Can we put constant things to $p $?

tenplusten wrote: Question:Can we put constant things to $p $? Yes, but if $p$ is a constant polynomial, then it must be an integer.

I don't understand the question can someone explain

Plops wrote: I don't understand the question can someone explain For each polynomial $p$ with integer coefficients, we assign an integer, call it $\theta(p)$. This assignment satisfies the properties given in the problem. Find all possible assignments.

Here is a much more streamlined solution than the one I submitted on the contest (which was really messy). We claim that $\theta(p)=p(y)$ for some fixed $y$. It is easy to see that this works. First, we show that $\theta(n)=n$ for all constant polynomials $n$. It is obvious that $\theta(n)=n+\theta(0)\equiv n+a$ by induction. Note that $\theta(1)\mid\theta(2)$, so $a+1\mid 2a+1$ if $a\not=-1$, so $a+1\mid 1$, so $a+1=\pm 1$, so $a=-2$ or $a=0$. We also have $\theta(-1)\mid\theta(-2)$, so $a-1\mid 2a-1$, or $a-1\mid 1$, or $a=0$ or $a=2$. Thus, $a\in\{-1,-2,0\}\cap\{1,0,2\}=\{0\}$, so $\theta(0)=a=0$. Let $y=\theta(x)$. We will now show that $\theta(f)=f(y)$ for all $y\in\mathbb{Z}[x]$. We see that $x-b\mid f(x)-f(b)$ for any integer $b$, so $y-b\mid \theta(f)-f(b)$. But $y-b\mid f(y)-f(b)$, so we must have $y-b\mid \theta(f)-f(y)$ for all integers $y$. Thus, we can make $b$ sufficiently large to derive a contradiction, unless $\theta(f)=f(y)$, so we must have $\theta(f)=f(y)$, as desired.

Beautiful problem!

Tintarn stated: Quote: We claim that all the solutions are given by $\theta(p)=p(n)$ for some fixed integer $n$. Clearly, by shifting the argument of the polynomials we get another solution so we may w.l.o.g. assume $\theta(x)=0$ and we shall prove that $\theta(p)=p(0)$ for all $p$. By the first condition, this is equivalent to proving $\theta(p)=0$ for all $p$ with $p(0)=0$. So fix some $p$ with $p(0)=0$. For any natural number $a$ the polynomial $p-p(a)$ is divisible by $x-a$ and hence $\theta(p-p(a))=\theta(p)-p(a)$ is divisible by $a$. But $p(a)=p(a)-p(0)$ is also divisible by $a$ and hence $\theta(p)$ is divisible by $a$. But $a$ was chosen arbitrarily and we conclude that $\theta(p)=0$. Done. Can someone explain the last two sentences of the proof

Doctor Math said: Quote: There exists an integer $n$ such that $\theta(x-n)=0$; indeed, just take $n=\theta(x)$ Can someone please explain this as well.

@above Its not hard to see that for integer $n$, $\theta(f+n)=\theta(f)+n$. Indeed its just rule 1 applied $n$ times (if $n<0$ apply it backwards). Thus, $\theta(x-n)=\theta(x)-n$, so $n=\theta(x)$ works.

I see. Thanks

Since $\theta(p+1)=\theta(p)+1$, the range of $\theta$ has infinitely many distinct elements. Plugging in $q=0$ into the second condition gives $\theta(p)$ divides $\theta(0)$ for $\theta(p)\not =0$. Since there are infinitely many $p$ such that $\theta(p)\not =0$, we must have $\theta(0)=0$. Then, by the first condition, $\theta (n)=n+\theta(0)=n$ for all $n \in \mathbb{Z}$. Note that by the first condition, we pull out any constants outside $\theta$ (for example, $\theta(2x+3)=\theta(2x)+3$). Let $b\in \mathbb{Z}$. Since $x-b\mid p(x)-p(b)$, then $\theta(x-b) \mid \theta(p(x)-p(b)) = \theta(p(x))-p(b)$. Let $\theta(x)=c$. So $c-b\mid \theta(p(x))-p(b)$. But $c-b\mid p(c)-p(b)$, so $c-b\mid \theta(p(x)) - p(c)$. But $\theta(p(x))-p(c)$ is a constant, and infinitely many integers divide it. Hence, it is 0, so $\theta(p(x))=p(c)$. Therefore, the functions $\theta$ are $\theta(p)=p(c)$ for any $c\in \mathbb{Z}$.

We claim that any function $\theta$ must take polynomials $P$ to $P(c)$, where $c$ is some integer. Clearly, any solution of this form will work. Claim: For any polynomial $p$ and integer $a$, we have that $\theta(p) + a = \theta(p + a)$. Proof: We apply induction to show this for all positive $a$. Base Case: We are given that $\theta(p + 1) = \theta(p) + 1$ for all polynomials $p$. Induction Step: Given that the claim holds for $a = k$, we will show that it holds for $a = k+1$. We note that \[\theta(p+k+1) = \theta(p+1+k) = \theta(p+1)+k = \theta(p)+1+k = \theta(p) + k+1,\]so we are done. Now, for negative $a$, we note that $\theta(p+a)-a = \theta(p+a-a) = \theta(p)$, so $\theta(p) + a = \theta(p+a)$. $\blacksquare$ Now, we set $\theta(x) = c$. We take any prime $p$ and any polynomial $P$ with $\deg P \ge 1$ and divide $x-c-p$ into $P$ to get \[P = Q \cdot (x-c-p) + R\]for polynomials $Q$ and $R$. Since $x-c-p$ is a linear function, $R$ must just be an integer; in particular, it is $P(p+c)$. Now, we observe that \[\theta(x-c-p) = c-c-p = -p\]divides $\theta(Q \cdot (x-c-p)$, so \[\theta(P) \equiv \theta(Q \cdot (x-c-p) + P(p+c)) \equiv \theta(Q \cdot (x-c-p)) + P(p+c) \equiv \]\[P(p+c) \equiv P(c) \qquad (\mbox{mod } p).\]This applies for every prime $p$, so $\theta(P) = P(c) = P(\theta(x))$ for all polynomials $P$ with $\deg P \ge 1$. Now, we will deal with constant polynomials $K$. Let $a = \theta(0)$. We claim that $a = 0$. We note that any integer $k$ can be achieved by taking $\theta(-a+k)$. Then, if $a+1$ is nonzero, we have that $a+1 \mid k$ because $1 \mid -a + k$. This can only work for an integer $a$ if $a \in \{0,-1,-2\}$. But we can similarly take $a-1$ to see that $a \in \{2,1,0\}$, forcing $a = 0$, as desired. $\blacksquare$

for reference

Let Me know if there is any mistakes in my solution. I loved this problem . At first I $\textcolor{red}{\text{claim}}$ that $\theta (P+k)=\theta(P)+k$ for all $k\in \mathbb{Z}$. Suppose $k$ is positive integer. Then, \[\theta (P+k) =\theta (P+k-1)+1=\theta (P+k-2)+2\]. Inductively we are done.If $k$ is negetive then assume $k =-k_0$ and it's easy to show $\theta (P-k_0)+k_0=\theta(P)$. Now I should find the value of $\textcolor{red}{\theta(0)}$. Since for any $q\in {\mathbb Z}[x]$ , $\theta (P) \mid \theta(P.q)$ Now,put $q=0$. ,to get $\theta (p) \mid \theta (0)$ now we can choose $\theta(p)$ very large . So only possibility is $\boxed{\theta(0)=0}$. $\textcolor{red}{\text{claim(2)}}$ For a prime $p$ ,$\theta(p)=p$. Now let's put $P$( constant polynomial) $=p$ (a prime) and $q(x)=x$ then we get , $\theta (p) \mid \theta(px) =\theta(p+p(x-1)) =\theta(p) +p(x-1)$. $\theta (p)\mid p(x-1)$ . Now we can vary $x$ so for instance put $x=2$ to get , $\theta(p) \mid p$ this should give either$\theta(p)= p$ or $\theta (p)=1$. If $\theta(p)=1$ for all prime $p$ then we must get $\theta(3)=1$ it's not possible because we previously get , \[\theta(3)=\theta(0)+3=3\]. So ,$\theta(p)=p$ for all prime $p$. Now for any constant $c$ ,$N =P(c)$ can be represented as $ap+r$ . \[\theta (P(c))=\theta(ap+r)=\theta(ap)+r\]. \[\theta (P(c))=\theta(p+(a-1)p)+r= \theta (p)+(a-1)p+r =P(c) \]. In this way we can cover all polynomials.

For convenience of typesetting, I'll just write f instead of \theta. For a constant y, we have that $$x-y\mid p(x)-p(y)\implies f(x)-y=f(x-y)\mid f(p(x)-p(y))=f(p(x))-p(y)\stackrel{f(x)-y\mid p(f(x))-p(y)}{\implies}f(x)-y\mid f(p(x))-p(f(x)),$$which, for arbitrarily large or small y LHS>RHS, whence RHS=0 so $f(P)=f(p(x))=p(f(x)):=p(c)$ for some constant c, and this is the answer.

We claim that $\theta$ is the valuation at $n$ for some $n$, for which the result holds. It follows from the first condition that $\theta$ is linear on the integers, and it is follows that $\theta(z) = z$ holds for $z \in {\mathbb Z}$. Then, let $\theta(x) = n$. It immediately follows that $\theta(x + z) = n + z$ for every $z \in {\mathbb Z}$. From here, we claim that for every $p \in {\mathbb Z}[x]$, that $\theta(p) = p(n)$. Note that for every integer $z \in {\mathbb Z}$, that \[ \theta(x + z) \mid \theta(p) - \theta({p(-z)}) \]Note that this holds for $p(n)$ substituted for $\theta(p)$ as well. Then, set $z$ such $\theta(x + z) > \theta(p) - p(n)$. Then we have that \[ \theta(x + z) \mid (\theta(p) - \theta(p(-z))) - (p(n) - \theta(p(-z))) = \theta(p) - p(n) \]which forces $\theta(p) = p(n)$

We will show that all functions $\theta$ are of the form $\theta(p)=p(k)$ for some integer $k.$ These clearly work. First, consider $\theta(1).$ If it is nonzero, it must divide all integers, since clearly the range of $\theta$ is all integers. Thus $\theta(1)=1,0,-1.$ Similarly, $\theta(-1)=1,0,-1.$ Since $\theta(1)=2+\theta(-1)$ we find $\theta(1)=1,$ and so $\theta(n)=n$ for any integer $n.$ Next, take $\theta(x)=k.$ Then we get $\theta(x-c)=k-c$ for any integer $c.$ Now set $\theta(p(x))=h.$ Since $p(c)-p(c)=0$ we get $p(x)-p(c)$ is divisible by $x-c,$ so $\theta(p(x)-p(c))=h-p(c)$ is divisible by $k-c$ for all $c \ne k.$ Now, notice that $p(c)-p(k)$ is divisible by $c-k$ by a well-known property of integer-coefficient polynomials. Thus, $h-p(c)+p(c)-p(k)=h-p(k)$ is divisible by $k-c$ for all $c \ne k.$ However, $h-p(k)$ is constant and must be divisible by infinitely many integers, so we get $h-p(k)=0,$ so $h=p(k)$ and we are done.

Why is $\theta(p(x))$ not a constant that's just equal to x? isn't $p(x)$ a constant

We want to show that $\theta(P(x))=P(\theta (x))=P(c)$ for some $c \in \mathbb{Z}$ which obviously work. We know that $x-n \mid P(x)-P(n) \forall n \in \mathbb{Z}$ and since both sides are in $\mathbb{Z}[X]$ we have: $$c-n= \theta(x) - n =\theta(x-n) \mid \theta (P(x)-P(n)) = \theta(P(x))-P(n)$$ So $c-n \mid \theta(P(x))-P(n)$, combining this with $c-n \mid P(c)-P(n)$ we conclude that $c-n \ \theta(P(x)) - P(c)$. Finally, taking $N$ large enough proves the desired claim. $$\mathbb{Q.E.D.}$$

qwerty123456asdfgzxcvb wrote: Why is $\theta(p(x))$ not a constant that's just equal to x? isn't $p(x)$ a constant No, $P(x) \in \mathbb{Z}[X]$ i.e. $P(x)$ is a polynomial with integer coefficients. So the value of $\theta(P(x))$ doesn't depend on the "constant" you plug in $P(x)$. It only depends on the coefficients.

Our answer is $\boxed{\theta(p(x)) = p(c), c \in \mathbb{Z}}$. In particular, $c=\theta(x)$. We proceed with induction on the degree of $p$, starting with our base cases: Degree 0: The first condition tells us $\theta(p(x)) = x+\theta(0)$. The second condition tells us, for all $x$, we have \[\theta(x) = x+\theta(0) \mid 2x+\theta(0) = \theta(2x) \implies x+\theta(0) \mid \theta(0) \implies \theta(0) = 0.\] Degree 1: We have $\theta(ax+b) = \theta(ax) + \theta(b) + \theta(0) = ac + b$. Otherwise, if $\operatorname{deg} p \ge 1$, we can assume there exists infinitely many integers $k$ such that $p(c) \neq p(k)$. Our inductive hypothesis then tells us \[\frac{p(c)-p(k)}{c-k} = \theta \left(\frac{p(x)-p(k)}{x-k}\right) \mid \theta(p(x)) - p(k) \implies \frac{p(c)-p(k)}{c-k} \mid \theta(p(x)) - p(c).\] Since the LHS can achieve infinitely many values by varying $k$, the RHS must be 0, as desired. $\blacksquare$

Solved with Orthogonal.. We claim the only functions are $\theta(p)\equiv p(c)$ for some $c\in\mathbb{Z}$. It is easy to check that these work. Note that $\theta(p+n)=\theta(p)+n$ for all $p\in\mathbb{Z}[x]$ and $n\in\mathbb{Z}$. Thus $\theta(n)=n+\theta(0)$. By the second condition so have $\theta(n)\mid\theta(2n)$ so $n+\theta(0)\mid 2n+\theta(0)-2(n+\theta(0))=-\theta(0)$. Since all integers divide $\theta(0)$, it follows that $\theta(0)=0$ so $\theta(n)=n$. Fix $p\in\mathbb{Z}[x]$. Since $x-y\mid x^\alpha-y^\alpha$ for all $\alpha\in\mathbb{N}$, it follows that $x-y\mid p(x)-p(y)$. Thus $\theta(x-n)\mid\theta(p(x)-p(n))$ so \[ \theta(x)-n\mid\theta(p(x))-p(n)-(p(\theta(x))-p(n))=\theta(p(x))-p(\theta(x)) \]for all $n\in\mathbb{Z}$. Since all integers divide $\theta(p(x))-p(\theta(x))$, we must have $\theta(p(x))=p(\theta(x))$, as desired. $\square$

We clearly get that for any constant polynomial $c$, $\theta(c)=c$. Now let $P(x)=x$, thus suppose $\theta(P)=k$, thus we get that for any linear polynomial $p$, $\theta(p)=p(a)$ for some fixed $a$. Now suppose all polynomials $P$ of degree $\leq n-1$ have the property $f(P)=P(a)$ for a fixed $a$, now let $Q$ be a polynomial of degree $n$, clearly we have that due to remainder factor theorem whatever its called we can add some values $k$ to $Q$ so that $Q+k=PR$ where $R$ is linear, thus we get that $\theta(Q)+k=\theta(Q+k)=\theta(P)\theta(R)=P(A)R(A)=Q(A)+k$, thus we get by induction that $\theta(P)=P(a)$ for some fixed $a$ for all polynomials $P\in {\mathbb Z}[x]$.

The solutions are of the form $\theta (P) = P(c)$, where $c$ is some integer constant. We prove this in stages: Constant $P$: The first condition implies that $\theta(P)$ is surjective over constants $P$, so $\theta(-1)$ and $\theta(1)$ must be in the set $\{ -1, 0, 1 \}$. It follows that $\theta(-1) = -1$ and $\theta(1) = 1$, so $\theta (P) = P$ for constant polynomials $P$. Linear, monic $P$: Let $K := \theta (x)$. Clearly, if $\theta$ is a solution, for any horizontal shift $\sigma$, we have that $\theta \circ \sigma$ is also a solution. So, from hereon, WLOG, assume that $\theta(x) = 0$. Then, $\theta (x+b) = b$. All other $P$: it suffices to show that $\theta (x Q(x)) = 0$ for all polynomials $Q$. Let $p$ be a prime and let $P(x) = xQ(x) - p Q(p)$, so that $\theta (x-p) \mid \theta(P(x))$. But we have \[\theta(P(x)) = \theta(xQ(x)) - pQ(p),\]so $p \mid \theta(xQ(x))$ for all primes $p$. This implies $\theta(xQ(x)) = 0$, as desired.

Notice that $\theta$ is surjective over constant polynomials. We also have that \[\theta(c)\mid \theta(0), \ \forall c\in \mathbb Z\]Hence $\theta(0)=0$ so $\theta(k)=k$ for any integer $k$. Now the main idea is to use $a-b\mid P(a)-P(b)$. It's also not hard to see that the quotient of these is an integer polynomial. Fix a real number $x$ and pick $c$ at random. We have \[\theta(x-c)\mid \theta(P(x)-P(c))\implies \theta(x)-c\mid \theta(P(x))-P(c)\]Now notice that $\theta(x)-c\mid P(\theta(x))-P(c)$ so \[\theta(x)-c\mid\theta(P(x))-P(\theta(x))\]Therefore, by varying $c$ we get that $\theta(P(x))=P(\theta(x))$, which is indeed a solution. $\blacksquare$

Heavily hinted , , but really nice FE . Observe $\theta(a) \mid \theta (0)$ for infinitely integers $a$ this combined with condition one implies $\theta(0)=0$ and $\theta(k)=k$ for all integers $k$. Now take $a \neq \theta (x)$, and some polynomial $P(x)$ then since $x-a \mid P(x) - P(a)$ (an idea i've been failing to come up with by myself recently oops ) then we have $$\theta(x)- a = \theta(x-a) \mid \theta(P(x)-P(a))= \theta(P(x))-P(a)$$$$\implies \theta(P(x)) \equiv P(a) \pmod{\theta(x)-a}$$Now by infinite CRT accross all values of $a$ we get $P(x) = P(\theta(x))$. Our solution set is $\theta(P(x))= P(\theta(x))$ for any integral $\theta(x)$. Which works. Remark: Yeah I better see the x-a | P(x)-P(a) when it shows up again bc of failing usemo p4 and this.

The only solutions are of the form $\theta(P(x)) = P(c)$ for fixed $c \in \mathbb{Z}.$ These solutions can easily be confirmed to work. First, we will show that $\theta(a) = a$ for all $a \in \mathbb{Z} \subset \mathbb{Z}[x].$ Let $b = \theta(0).$ By the given assertion, we have that if $\theta(1) = b + 1 \ne 0,$ then $\theta(1) \mid \theta(2),$ or $b+1 \mid (b+1) + 1 = b+2.$ Thus $b + 1 \mid 1,$ so either $b = 0, b = -2,$ or $b = -1.$ The second case fails since $\theta(5) = 3$ does not divide $\theta(10) = 8,$ and similarly the third case fails since $\theta(4) = 3$ does not divide $\theta(8) = 7.$ Therefore, we have $b = \theta(0) = 0,$ implying $\theta(a) = a$ for all $a \in \mathbb{Z}.$ Now we will show that $\theta(P(x)) = P(\theta(x))$ for all polynomials $P(x).$ Let $d$ be an arbitrary integer. Then $\theta(P(x)) - P(d) = \theta(P(x) - P(d)).$ We know that $P(x) - P(d)$ is divisible by $x - d,$ so $\theta(x-d) = \theta(x) - d$ divides $\theta(P(x)) - P(d).$ Moreover, we know that $\theta(x) - d$ also divides $P(\theta(x)) - P(d),$ so $\theta(x) - d$ divides $$(\theta(P(x)) - P(d)) - (P(\theta(x)) - P(d)) = \theta(P(x)) - P(\theta(x)).$$Varying $d,$ we see that $\theta(P(x)) - P(\theta(x))$ is divisible by every positive integer, implying that $\theta(P(x)) = P(\theta(x))$ for all polynomials $P(x).$ Letting $\theta(x) = c,$ we discover that $\theta(P(x)) = P(c)$, which is the form of the solution described at the beginning, so we are done. lolzers i saw $x - a \mid P(x) - P(a)$ today but not in usemo skul

Here we go: We claim that the solutions are $\theta(P(x))=P(c)$ where $c \in \mathbb Z$ is some constant. Note that: they indeed work. Now, we will prove that, these are the only solutions. Shifting the numbers by sufficient constants (which is possible due to the condition (i)), WLOG assume that $\theta(x) = 0$. We will prove that: $\theta(p(x)) = p(0)$ for all $p \in \mathbb Z[x]$. It suffice to show that: $\theta(p(x)) = 0$ whenever $p(0)=0$ (by shifting the constant term of the polynomials due to condition (i)). Consider an integer $a \in \mathbb Z$: Notice that: $x-a|p(x)-p(a)$ which implies $\theta(x-a)|\theta(p(x)-p(a)) = \theta(p(x)) - p(a)$. Thus: $a|\theta(p(x)-p(a)) = \theta(p(x)) - p(a)$. Note that: $p(0)=0$ which implies $a|p(a)$. Thus: $a|\theta(p(x))$ for every integer $a \in \mathbb Z$. Therefore: $ \theta(p(x)) = 0$ and we are done.