jmc

If the quadratic equation $x^2+(2a+1)x+a^2=0$ has two integer solutions, then $$x=\frac{-2a-1\pm \sqrt{(2a+1{)}^2-4a^2}}{2}$$is an integer, so it follows that the discriminant $(2a+1{)}^2-4a^2=4a+1$ must be a perfect square. Also, $1\le a\le 50$, so it follows that $5\le 4a+1\le 201$. Clearly $4a+1$ can only be the square of an odd integer; conversely, the square of any odd integer $(2n+1{)}^2$ is of the form $4n^2+4n+1=4(n^2+n)+1$ and so can be written as $4a+1$. The odd perfect squares from $5$ to $201$ are given by $9=3^2,5^2,7^2,9^2,11^2,169=13^2$, so it follows that there are $\boxed{6}$ such values of $a$.