Problem 2 · The perfect-square triple
$n + 2n = 3n$ being a square forces $n = 3k^{2}$.
Integer answer, at most 4 digitsHow many two-digit integers are such that the sum of the number and its double is a perfect square?
Copa Cangur · SCM
Easy
Closed answer
Reasoned solution
Key idea: the sum is $n + 2n = 3n$. If $3n$ is a square, the factor $3$ must appear twice: $n$ must be $3$ times a square.
Write $n = 3k^{2}$ (then $3n = 9k^{2} = (3k)^{2}$ ✓). With $10 \le n \le 99$:
$$k=2: n=12, \quad k=3: n=27, \quad k=4: n=48, \quad k=5: n=75.$$
Check: $36 = 6^{2}$, $81 = 9^{2}$, $144 = 12^{2}$, $225 = 15^{2}$ ✓. Four numbers.
Answer: 4