Determine all solutions of the equation

$$
a^{2}=b \cdot(b+7)
$$

with integers $a \geq 0$ and $b \geq 0$.

W. Janous, WRG Ursulinen, Innsbruck
Due to $a^{2}=b^{2}+7 b$ and $b \geq 0$, it follows that $a^{2} \geq b^{2}$ and since $a \geq 0$, it follows that $a \geq b$.

Thus, there exists a natural number $k$ with $a=b+k$. Substituting this into the given equation yields

$$
(b+k)^{2}-b(b+7)=0
$$

and thus

$$
b=\frac{k^{2}}{7-2 k}
$$

Since $b$ must not be negative, $k$ must be 0, 1, 2, or 3. Only for $k=0$ or 3 is $b$ an integer, namely 0 or 9. For $b=0$, we get $a=0$ and for $b=9$, we get $a=12$. This gives the solution pairs $(a, b)=(0,0)$ and $(12,9)$.

(R. Henner)