Is this number friends relationship transitive? I am calling two real numbers $a,b$ friends, if there exists a non-constant integer polynomial $p(x)$, such that $p(a)-p(b)=0$. The relation is obviously symmetric and reflexive. Is the relationship also transitive?
( i.e. Can we state for all numbers $a,b,c \in \mathbb R$ that if there is a polynomial $p_1(x) \in \mathbb Z[x]$ and $deg(p_1)>0$ such that $p_1(a)-p_1(b)=0$, and a polynomial $p_2(x) \in \mathbb Z[x]$ and $deg(p_2)>0$ such that $p_2(b)-p_2(c)=0$, does there also exist polynomial $p_3(x) \in \mathbb Z[x]$ and $deg(p_3)>0$ such that $p_3(a)-p_3(c)=0$ )
 A: I'm making another answer since my previous answer is so long ago, it would probably go unnoticed if edited.
The answer is no, the relation is not transitive. 
I'll use your observation that any number $a$ is related to $-a-1$ (take $x(-x-1)$). Through $P(x) = x^2$, any number is related to its negative. I'll just show that not any number $a$ is related to $a+1$.
Suppose it is, through $P(x) = \sum_{k=0}^n c_kx^k$. Then
$$
P(a+1)-P(a)=0= \sum_{k=1}^n c_k \sum_{j=0}^{k-1} {k \choose j} a^j =: Q(a) = c_1 + c_2(1 + 2a) + \ldots + c_n \sum_{j=0}^{n-1} {n \choose j} a^j.
$$
Now $Q$ could be the zero polynomial, but then either $P=0$ (not allowed) or $a$ must take one of finitely many (algebraic) values, e.g. $a = -1/2$ allows $c_2$ to be non-zero. In all other cases, $a$ satisfies a non-zero integer polynomial, so must be algebraic. So whenever $a$ is transcendental, $a \sim -a-1 \sim a+1$, but not $a \sim a+1$. 
A: Here is a partial answer. It involves a little bit of algebraic number theory: mainly, it uses the notions of algebraic and transcendental numbers. Any complex number is either algebraic or transcendental; it is algebraic if it is the root to some polynomial with rational coefficients, and transcendental otherwise.
Suppose $a$ is an algebraic number. If $a \sim b$ for some $b$, then $b$ must also be algebraic:
suppose for a contradiction that $b$ is transcendental. If $P(a) = P(b)$ for some non-constant $P \in \mathbb Z[X]$, then $P(a)$ would be transcendental, since it is a non-constant algebraic function of the transcendental $b$. However, $P(a)$ is an element of the number field $\mathbb Q[a]$, and is hence algebraic; a contradiction. (I am using basic properties here - you can look them up easily if you're unsure about them.)
So let's assume that two numbers $a$ and $b$ are algebraic; say they satisfy non-constant polynomials in $\mathbb Q[X]$. Multiplying out denominators of the coefficients gives us polynomials $P, Q \in \mathbb Z[X]$. Multiply them together: $A := PQ \in \mathbb Z[X]$.
Now $A(a) = A(b)= 0$, so $a \sim b$. So all algebraic numbers are related in this way, and are not related to any transcendental numbers.
The question remains whether transcendental numbers (like $\pi$) are related to anything besides themselves. I couldn't figure this out easily, and after some research, I found out that it appears to be an open problem whether or not (given) transcendental numbers are "algebraically independent", i.e. whether or not you can write one in terms of the other, so to speak. Of course one could still assume that for some distinct transcendental $a$ and $b$, we have $P(a) = P(b)$ and argue from there, but if we don't know whether such $P$ even exists, it may be wasted effort.
A final remark: we end up with only a single equivalence class containing the algebraic numbers, which is a bit boring. The argument I used wouldn't work if you required the polynomials over $\mathbb Z$ to be monic (because we're multiplying out the denominators of the coefficients in $P$ and $Q$). Perhaps this is an interesting thing to look at next.
A: Maybe a wrong answer, dunno yet:
Lets say we have $p(a)-p(b)=0$, then $p(a)=p(b)=u$. Look at the
new polynomial $q(x)$, now thought as from $C[x]$, it must factor:
$$q(x)=p(x)-u=(x-a)(x-b)...$$
Now for two polynomials $p_1,p_2$ and three values $a,b,c,..$ we have:
$$q_1(x)=p_1(x)-u_1=(x-a) (x-b) ...$$
$$q_2(x)=p_2(x)-u_2=(x-b) (x-c) ...$$
We can take the following polynomial:
$$q_3(x)= q_1(x)\,q_2(x) = (x-a) (x-b)^2 (x-c) ...$$
Now for an arbitrary $v$, a polynomial $p_3(x)=q_3(x)+v$ does the job. Problem I have, is showing that we can stay in $p_3(x) \in Z[x]$. Can we?
