If $(1+3+5+\dots +p)+(1+3+5+\dots +q)=(1+3+5+\dots +r)$

where each set of parentheses contains the sum of 

consecutive odd integers as shown, the smallest possible value of $p+q+r,$ (where $p>6$)

is

We know that $1+3+5+\dots +(2k-1)=k^2$

Thus, the given equation can be written as

$\begin{array}{l} {\left(\frac{p+1}{2}\right)}^2+{\left(\frac{q+1}{2}\right)}^2={\left(\frac{r+1}{2}\right)}^2\\ \Rightarrow (p+1{)}^2+(q+1{)}^2=(r+1{)}^2\end{array}$

Therefore, $(p+1, q+1, r+1)$ forms a Pythagorean triplet. As $p>6,p+1>7.$

The first Pythagorean triplet containing a number > 7 is (6, 8, 10).

$\therefore$ We may take

$\Rightarrow p+q+r=21.$