Problem 959

Suppose that $A$ is a perfect square. Then its last digit will be $1, 4, 5, 6$ or 9. But a perfect square cannot end in 05 or 06. Therefore, the number $A$ ends in one of the digits $1, 4, 9$. Let $x$ be the square root of the last digit of the number $A$. Let $k$ be the number of zeros in the number $A - x^2$. (We can assume that $k > 2$.) Since the number $x$ is not divisible by 5, exactly one of the numbers $\sqrt{A} - x, \sqrt{A} + x$ is divisible by 5, and hence by $5^k$. Therefore, one of these numbers is at least $5^k$, and the other is at least $5^k - 6$, so the product of these numbers is at least $5^k (5^k - 6) > 5^k \cdot 9 \cdot 2^k = 9 \cdot 10^k$, which contradicts the $(k+1)$-digit nature of the number $A$. Thus, the number $A$ cannot be a perfect square.