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?
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.
To find the values of $a$ that satisfy the given conditions, we need to consider the discriminant of the quadratic equation. The discriminant ($D$) is the expression inside the square root in the quadratic formula and determines the nature of the solutions.
For the quadratic equation $ax^2 + bx + c = 0$, the discriminant is given by $D = b^2 - 4ac$. In our case, $a = 1$, $b = 2a + 1$, and $c = 2a$.
For the quadratic equation to have two integer solutions, the discriminant must be a perfect square ($D = k^2$), where $k$ is an integer.
Substituting the values into the discriminant formula, we get:
$D = (2a+1)^2 - 4(1)(2a)$
$D = 4a^2 + 4a + 1 - 8a$
$D = 4a^2 - 4a + 1$
Now, we need to find the number of values of $a$ such that $D = k^2$ for some integer $k$.
We can simplify the discriminant further:
$D = (2a - 1)^2$
To find the possible values of $a$, we need to find the perfect square values less than or equal to 50.
Checking each square value:
For $k = 1$, $(2a - 1)^2 = 1$ has one solution $a = 1$.
For $k = 2$, $(2a - 1)^2 = 4$ has the solutions $a = 1, 2$.
For $k = 3$, $(2a - 1)^2 = 9$ has the solutions $a = 2, 3$.
For $k = 4$, $(2a - 1)^2 = 16$ has the solutions $a = 3, 4, 5$.
For $k = 5$, $(2a - 1)^2 = 25$ has the solutions $a = 4, 5, 6$.
For $k = 6$, $(2a - 1)^2 = 36$ has the solutions $a = 5, 6, 7$.
For $k = 7$, $(2a - 1)^2 = 49$ has the solutions $a = 6, 7$.
Therefore, the quadratic equation has two integer solutions for $a = 1, 2, 3, 4, 5, 6, 7$. Thus, there are 7 possible values of $a$.