We can factor the quadratic as \[x(x + 2a) + (2a + 1)x + 2a = (x + 2a)(x + 1) + (x + 2a).\]Therefore, the quadratic has two integer solutions if and only if $(x + 2a)(x + 1)$ divides $x + 2a$. This means \[x + 2a = -1, -2a, 0.\]We can rule out $x + 2a = 0,$ or $x = -2a,$ because $a$ is positive. Then the original equation becomes \[(-2a - 1)(-2a + 1) + 2a = 0.\]Simplifying gives \[4a^2 - 1 - 2a + 2a = 0,\]so $4a^2 = 1,$ or $a^2 = \frac14.$ Since $a$ is an integer, the only such value of $a$ is $a = \frac12,$ which is not a positive integer.
Public Service Announcement: Be extremely cautious when dividing both sides of an equation by a sum or difference that could be $0.$ We have divided both sides by $x + 2a,$ but the case $x = -2a$ must be discarded.
For how many values of $a$ is it true that:
(1) $a$ is a positive integer such that $a \le 50$.
(2) the quadratic equation $x^2 + (2a+1)x + 2a = 0$ has two integer solutions?
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1 answer