Two relations involving the gcd This is part of a bigger problem I am solving. Let $k\ge 2$ be a fixed positive integer. Is it possible to find an integer $v, v>k$ such that $$(k-1,v-1)=\left(\frac{k(k-1)}{2},\frac{v(v-1)}{2}\right)=1\;?$$ If so, how do I choose such a $v$?
 A: If you just make $v$ large in the obvious way, this is always possible (in fact there are infinitely many such $v$).  Here is my solution.
For syntactical neatness, we will instead solve the equivalent problem where $k \ge 1$ and we need to find $v > k$ such that
$$
(k, v) = \left( \frac{k(k+1)}{2} , \frac{v(v+1)}{2} \right) = 1
$$
For any positive integer $a$, just let $v = ak(k+1) + 1$.
Note that by Euclidean algorithm,
\begin{align*}
(k,v) &= (k, ak(k+1) + 1) = 1 \\
(k+1,v) &= (k+1, ak(k+1) + 1) = 1 \\
(k,v+1) &= (k, ak(k+1) + 2) = (k, 2) \\
(k+1,v+1) &= (k+1, ak(k+1) + 2) = (k+1,2)
\end{align*}
Now, a nice property of the GCD (which is fairly easy to verify) is that for any $a,b,c$ integers, $(ab,c) \bigg| (a,c)(b,c)$.
Using this, for $a,b,c,d$ integers, we have
$$
(ab,cd) \bigg| (a,cd)(b,cd) \bigg| (a,c)(a,d)(b,c)(b,d)
$$
Therefore, 
\begin{align*}
\left( k(k+1) , v(v+1) \right) &\bigg| (k, v)(k+1,v)(k,v+1)(k+1,v+1) \\
&= 1 \cdot 1 \cdot (k,2) \cdot (k+1, 2) = 2
\end{align*}
Therefore,
$$
\left( \frac{k(k+1)}{2} , \frac{v(v+1)}{2} \right) = \frac{\left( k(k+1) , v(v+1) \right)}{2} = 1
$$
We have shown both the desired GCDs equal 1, so $v = ak(k+1) + 1$ works for any $a$.
There are other ways to approach this problem, but one of my favorite ways to deal with GCDs (and in many ways one of the nicest ways) is to use their algebraic properties such as those employed above.
