PRMO 2018 question 9

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9. Suppose $a, b$ are integers and $a + b$ is a root of
$x^2 + ax + b = 0$. What is the maximum possible value of $b^2$?
[PRE-RMO – 2018]

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We are given that $a + b$ is a root of the quadratic equation:

$$x^2 + ax + b = 0$$

Substituting $x = a + b$ into the equation:

$$(a + b)^2 + a(a + b) + b = 0$$

Expanding the terms:

$$a^2 + 2ab + b^2 + a^2 + ab + b = 0$$ $$2a^2 + 3ab + b^2 + b = 0$$

This is a quadratic equation in $a$:

$$2a^2 + 3ab + (b^2 + b) = 0$$

To ensure that real values of $a$ exist, the discriminant $D$ of this quadratic must be a perfect square. The discriminant is:

$$D = (3b)^2 - 4 \cdot 2 \cdot (b^2 + b) = 9b^2 - 8(b^2 + b)$$ $$D = 9b^2 - 8b^2 - 8b = b^2 - 8b$$

Let $D = p^2$, where $p \in \mathbb{Z}$. Then:

$$b^2 - 8b = p^2 \Rightarrow (b - 4)^2 - p^2 = 16$$

This is a difference of squares:

$$(b - 4 + p)(b - 4 - p) = 16$$

Since $b, p \in \mathbb{Z}$, both factors are integers. The possible factorizations of 16 are:

Case 1:

$$b - 4 + p = 16,\quad b - 4 - p = 1$$

Adding the equations:

$$2(b - 4) = 17 \Rightarrow b = \frac{25}{2}$$

Not an integer ⇒ Discard.

Case 2:

$$b - 4 + p = 8,\quad b - 4 - p = 2$$

Adding:

$$2(b - 4) = 10 \Rightarrow b = 9$$

Case 3:

$$b - 4 + p = 4,\quad b - 4 - p = 4$$

Adding:

$$2(b - 4) = 8 \Rightarrow b = 8$$

Thus, the integer values of $b$ satisfying the condition are $b = 8$ and $b = 9$.
The maximum value of $b$ is:

$$\boxed{9}$$

Hence, the maximum value of $b^2$ is:

$$\boxed{81}$$

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Note: Before finalizing the answer, put the value of b in the discriminant to check whether 'a' comes as an integer.