If $x;y$ $\in T$($x$ and $y$ can be the same), then $x^2-y \in T $ Prove that : $T = \mathbb Z $

Solution 1:

Set $R(s,t) = s^2-t$

$\textbf{Claim}:$ If $T$ has the numbers $z$, $x_{1},y_{1},...,x_{n},y_n$ then $T$ has the set of numbers $z + \sum_{j=1}^{n}\lambda_{j}(x_{j}^2-y_{j}^2)$ where $\lambda_{j} \in \mathbb{N}$.

$\textbf{Proof}:$ Set $$f_{j}(t)=R(x_{j},R(y_{j},t)) = t+(x_{j}^2-y_{j}^2)$$ and

$$f_{j}^{h+1}(t) = f_{j}(f_{j}^{h}(t))$$

with $$f_{j}^{1}(t) = f_{j}(t).$$

Note

$$f_{j}^{\lambda_{j}}(t) = t+\lambda_{j}(x_{j}^2-y_{j}^2).$$

Thus

$$f_{n}^{\lambda_{n}}(...f_{2}^{\lambda_{2}}(f_{1}^{\lambda_{1}}(z))) = z +\sum_{j=1}^{n}\lambda_{j}(x_{j}^2-y_{j}^2).$$

$\textbf{Corollary}:$ If $T$ has the numbers $z$, $x_{1},y_{1},...,x_{n},y_n$ then $T$ has the set of numbers $z + \sum_{j=1}^{n}\lambda_{j}(x_{j}^2-y_{j}^2)$ where $\lambda_{j} \in \mathbb{Z}.$

$\textbf{Corollary}:$ Set $d = \gcd(\{u^2-v^2:\text{ } u,v \in T \})$. Suppose $z \in T$ then $T$ contains the set of numbers $z + \lambda d$ where $\lambda \in \mathbb{Z}$.

Now we use the numbers $a,b \in T$ which satisfy $\gcd(a,b) = \gcd(a-2,b-2) =1$. Note that $R(b,b) = b^2-b \in T,\text{ }R(a,a) = a^2-a \in T$ and

$$d | \gcd(b^2-a^2, (b^2-b)^2-b^2, (a^2-a)^2-a^2)$$

where $d$ appears in the above corollary.

Simplifying we have

$$d | \gcd(b^2-a^2, b^3(b-2), a^3(a-2))$$

Suppose $p$ is a prime factor that divides $d$ then $p$ does not divide $a$ or $b$ because we have $\gcd (a,b) = 1$ and $p | b^2-a^2$. Thus

$$\gcd(b^2-a^2, b^3(b-2), a^3(a-2))|\gcd(a-2,b-2) = 1.$$

Which means that $d = 1$. By the above corollary we know that $T = \mathbb{Z}$.