Show relations are Diophantine Show that the following relations are diophantine.
(a) $x_3$ is the remainder when $x_1$ is divided by $x_{2}+1$. 
(b) $x_3$ is the integer part when $x_1$ is divided by $x_{2}+1$
I'm not sure how to approach this question. The book I'm going through isn't about diophantine equations or sets but is using it as an illustrative example, so I'm unfamiliar with this topic.  Is $x_3$ supposed to be the parameter of the equation or also just one of the unknowns? 
For the first question I just assumed that we'd need an equation in the form $a=qb+r$ thus $x_1=y(x_2+1)+x_3$ where $y$ is the parameter. Is this wrong?
 A: Ordinarily in this game the variables range over the non-negative integers.   There is a trick (the Lagrange four-square theorem) that allows us to let variables range over the integers.
A relation $R(x_1,\dots, x_k)$ is Diophantine if there exists a polynomial $P(x_1,\dots,x_k,y_1,\dots,y_n)$ with integer coefficients such that
for all $x_1,\dots,x_k$ we have
$$R(x_1,\dots,x_k)\quad\text{if and only if}\quad \exists y_1\cdots\exists y_n(P(x_1,\dots,x_k,y_1,\dots,y_n)=0).$$
Here, as mentioned above, the $y_i$ range over the non-negative integers. Alternately, we use positive integer coefficients only, and replace $P=0$ by $P_1=P_2$.
(a) We want to say that $x_3$ is less than $x_2+1$ and there is a $q$ such that $x_1=q(x_2+1)+x_3$. We can say that $x_3$ is less than $x_2+1$ by saying there is a $y$ such that $x_3+y=x_2$.  
So we want to say that there is a $q$ and a $y$ such that
$$(x_1=q(x_2+1)+x_3)\land (x_3+y=x_2).$$
We can say this "Diophantinely" by writing
$$\exists q\exists y((x_1-q(x_2+1)-x_3)^2+(x_3+y-x_2)^2=0).$$
For a sum of squares is $0$ precisely if each term is $0$.
If the rules say we cannot use minus, we can expand the above polynomial, and bring all negative terms to one side.
(b) Now you have some tools. Please try to solve the second problem. I will help in comments or by adding to the answer if help is needed.
