a 2 distance set has an upper bound for number of its elements

Can anyone help me with this problem?

We call a set $A\subseteq \mathbb R^n$ a $2$-distance set if for each $v_i,v_j$ in $A$, $i\neq j$, $|v_i-v_j|=r$ or $s$. Find an upper bound for the number of the elements of $A$.

Our Teacher proved that $|A|\le \frac{n^2+5n+4}{2}$ using linear algebra, but I read somewhere that $|A|\le \frac{n^2+3n+4}{2}$. Any good upper bound is welcomed!

The best known general upper bound for the size of a 2-distance set in $\mathbf{R}^n$ is $(n+1)(n+2)/2$. However, it can be improved for small $n$. For $n=1,2,3,4,5,6,7,8$ the exact upper bound is $3,5,6,10,16, 27,29,45$, respectively. (See P. Lisonek, New maximal two-distance sets, Journal of Combinatorial Theory, series A, 77 (1997), pp.318-338). A recent paper on the subject is A. Barg and O. Musin, Bounds on sets with few distances. Journal of Combinatorial Theory, series A, 118 (2011), pp. 1465–1474.