Show the indivisibility $\,xy\nmid x^2+y^2\,$ if $\, \gcd(x,y)= 1$ There are an integer $x,y$ where $\gcd(x,y)=1$ show the indivisibility of $\frac{x^2+y^2}{xy}$ where $x \neq y \neq 1$
I tried to transform $\frac{x^2+y^2}{xy}=\frac{(x+y)^2-2xy}{xy}=\frac{(x+y)^2}{xy}-2$
but I don't know if it is a good direction.
 A: By Euclid's Lemma (http://en.wikipedia.org/wiki/Euclid%27s_lemma), since $x\nmid x^2 + y^2$ since $x\nmid y$, and similarly $y\nmid x^2 + y^2$, then $xy\nmid x^2 + y^2$
A: A few sketched proofs, by $(1)$ Unique Factorization, $(2)$ Euclid's Lemma, $(3)$ Bezout's Lemma.
$(1)\,\ \ \ \rm \color{#0A0}{prime}\ p\mid\color{#C00}{x\mid xy\mid x^2\!+y^2}\Rightarrow\, p\mid y^2 \color{#0A0}\Rightarrow\: p\mid y\,\Rightarrow\: p\mid(x,y)=1\,\Rightarrow\Leftarrow$
$(2)\,\ \ \ \rm\color{#C00}{x\mid xy\mid  x^2\!+y^2} \,\Rightarrow\, x\mid y^2\:$ contra $\,\ \rm (x,y)=1 \overset{Euclid}\Rightarrow (x,y^2) = 1$
$(3)\,\ \ \rm\ 1\! \overset{Bezout}=\! (ax\!+\!by)^2 = (a^2\!-b^2)\color{#C00}x^2\!+b^2\color{#C00}{(x^2\!+y^2)}+2ab\color{#C00}xy,\,$  so $\rm\ \color{#C00}{x\mid rhs}\, \Rightarrow\ x\mid lhs\!=\!1\, \Rightarrow\Leftarrow$
Remark $\, (2)$ and $(3)$ are in fact equivalent to the uniqueness of prime factorizations of integers. These and many other such equivalent properties are listed in this post.
