Question: If $p\_i$ is prime then $p\_i \Bbb Z \cap p\_j \Bbb Z= \emptyset \ \forall i,j \in \Bbb N, i\neq j$. Where $a \Bbb Z=\{x\in \Bbb Z: x=0 \mod a\}$.

I ran into some problem where having this lemma proved would really simplify it, is there any really really simple (to understand) proof of this?

Answer: 
If $x+y\lt 1$ then $(x+y)^2\lt 1$, and therefore $x^2+y^2\lt 1$.

But from $0\lt x^2+y^2\lt 1$, we conclude that $(x^2+y^2)^2 \lt x^2+y^2$, and therefore $x^4+y^4\lt x^2+y^2$.
let $x+y=u,xy=v^2 \implies u^2 \ge 4v^2 \\ x^2+y^2=u^2-2v^2,x^4+y^4=(x^2+y^2)^2-2(xy)^2 =(u^2-2v^2)^2-2v^4=u^4-4u^2v^2+2v^4 \implies \\
u^4-(4v^2+1)u^2+2v^4+2v^2=0 \\ u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2}$ 

$u^2=\dfrac{4v^2+1-\sqrt{8v^4+1}}{2} \le 2v^2$ 

$v^2\ge 0 \implies 
u^2=\dfrac{4v^2+1+ \sqrt{8v^4+1}}{2} \ge \dfrac{1+1}{2}=1$ when $v^2=0$