Lowest form of rational number Suppose $\frac pq$ is a positive rational in its lowest form, prove that ${\frac1q}+{\frac{1}{p+q}}$ is also in the lowest form  I tried with the Least common multiple of the denominators and it was reduced to $\frac{p+2q}{p(p+q)}$  Further I do not know how prove that their GCD is 1
 A: Note that the fraction reduces to $$\frac {p+2q}{q(p+q)}$$
Now note that a common factor of $p+2q$ and $q$ is also a factor of $(p+2q)-2\cdot q =p$, therefore must be $1$.
Also a common factor of $p+2q$ and $p+q$ is a factor of both $(p+2q)-(p+q)=q$ and $2\cdot(p+q)-(p+2q)=p$ hence must be $1$.
A: Hint If $\frac{p+2q}{q(p+q)}$ is not in the lowest form, the denominator and numerator have a common prime divisor. 
Now if $d$ is prime and divides $q(p+q)$ then $d$ divides either $q$ or $p+q$. 
You also know that $d$ also divides $p+2q$. Prove that this implies that $d$ divides $p$ and $q$.
A: Hint $\ (a,b)=1\,\Rightarrow\,\dfrac{1}a+\dfrac{1}b = \dfrac{a\!+\!b}{ab}\,$ reduced, by $\,(a,a\!+\!b)= \overbrace{(a,b)}^{\large =\,1}=(b,a\!+\!b)\overset{\rm Euclid}\Rightarrow\!\!\,(ab,a\!+\!b)=1$
A: Let $d\in\mathbb{Z}$ be maximal such that $d\mid p+2q$ and $d\mid q^2+pq$ (that is, $d=\gcd(p+2q,q^2+pq)$). 
Now consider the equalities:
$$(p+2q)^2 - 4(q^2+pq) = p^2$$
$$q(p+2q) - (q^2+pq) = q^2$$
So $d\mid p^2$ and $d\mid q^2$. By Bézout's Lemma, $\gcd(p,q)=1\Rightarrow \exists x,y \in \mathbb{Z}$ such that $px+qy=1$. We now show that $\exists x',y' \in \mathbb{Z}$ such that $p^2x' + q^2y' = 1$.
For this, first notice that $p(x+kq)+q(y-kp) = px + kpq + qy - kpq = 1, \forall k \in \mathbb{Z}$. Now, we affirm that $\exists k \in \mathbb{Z}$ such that $p\mid x+kq$ and $q\mid y-kp$. Indeed, let $\tilde p, \tilde q$ be the inverses of $p,q$ modulo $q,p$, respectively (they certainly exist since $p,q$ are relatively prime).  
We set $k=p\tilde p\cdot y\tilde p-q\tilde q\cdot x\tilde q$, which is an integer, and we get $$x+kq\equiv x -qx\tilde q\equiv 0 \pmod p$$
$$y-kq\equiv y - py\tilde p \equiv 0 \pmod q$$
So just as we wanted, $x+kq=px'$ for some $x'\in \mathbb{Z}$ and $y-kp=qy'$ for some $y'\in \mathbb{Z}$. Then:
$$p(x+kq)+q(y-kp) = p^2x' + q^2y' = 1$$
And therefore, by Bézout's Lemma again, $d$ must be equal to $1$. 
