For $x$ and $y$ are coprime. Prove that $xy$ and $(x+y)$ are coprime. Given That $x$ and $y$ are coprime integers.
Prove that  $xy$ and $(x+y)$ are coprime.
 A: Hint:
If a prime $p$ dvides $xy$, it divides $x$ or $y$ by Euclid's lemma. Can it divide both? Hence, can it  divide $x+y\mkern 2mu$?
A: I found it.
according to Bézout's Lemma (Bézout's Identity), suppose that $x$ and $y$ are coprime integers, there exist $u$ and $v$ relative integers such that $xu + yv = 1$ Therefore:
$$1^2 = (xu + yv)^2 = (xu)^2 + 2xyuv + (yv)^2 = (xu)^2 + 2xyuv + (yv)^2 + xyv^2 + xyu^2 - xyv^2 - xyu^2$$
$$= (xu)^2 + xyv^2 + xyu^2 + (yv)^2 - xyv^2 + 2xyuv - xyu^2 = (x+y)(xu^2 + yv^2) - xy(u-v)^2$$
$$= (x+y)(xu^2 + yv^2) + xy(-(u-v)^2) = 1$$
$(xu^2 + yv^2)$ and $(-(u-v)^2)$ are relative integers, therefore by Bézout's Lemma we conclude that $xy$ and $x + y$ are coprime.  
A: We'll prove the contrapositive.
Suppose $xy$ and $x+y$ are not coprime. Then there is a prime $p$ that divides both $xy$ and $x+y$. We'll show that $x$ and $y$ are not coprime by showing that $p$ divides both $x$ and $y$.
Note that since $p$ divides $xy$, it must divide either $x$ or $y$. Since $p$ divides $x+y$ and one of $x$ and $y$, it must divide both $x$ and $y$. So $x$ and $y$ are not coprime.
