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Austria competition problems, 2014

Number theory Difficulty 4.8 AIME Prove it Austria

Determine all solutions of the Diophantine equation
$a^2 = b \cdot (b+7)$
in integers $a \geq 0$ and $b \geq 0$ .

Solution

We have the trivial estimate $a^2 = b \cdot (b+7) \ge b^2$ resulting in $a \ge b$ due to the non-negativity of $a$ and $b$ . On the other hand, the inequality between the arithmetic and the geometric mean implies that
$a = \sqrt{b(b+7)} \le \frac{b+(b+7)}{2} = b + \frac{7}{2}.$
Combining these inequalities shows that $a \in \{b, b+1, b+2, b+3\}$ . Inserting these cases into the original equation only yields solutions for $a = b$ or $a = b+3$ . These are $(a, b) \in \{(0, 0), (12, 9)\}$ .
$\square$

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