Determine all four-digit numbers $\overline{abcd}$ which are perfect squares and for which the equality holds: $\overline{ab}=3 \cdot \overline{cd} + 1$.
Problem
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Tags: number theory
SCLT
10.08.2020 18:54
Since all perfect squares are congruent to $0$ or $1$ $\pmod{4}$ the last two digits must must also satisfy this congruence. This leaves us with the following list of 15 candidates to check:
$1304, 1605, 2508, 2809, 3712, 4013, 4916, 5217, 6120, 6421, 7324, 7625, 8528, 8829, 9732$
By manual inspection, only $2809 = 53^2$ is a perfect square.
Solar Plexsus
10.08.2020 19:18
According to the given information there is an integer $x \in [32,99]$ s.t. $(1) \;\; \overline{abcd} = x^2$, $(2) \;\; \overline{ab} = 3 \cdot \overline{cd} + 1$. According to (1) we have $100 \cdot \overline{ab} + \overline{cd} = x^2$, which combined with (2) give us $(3) \;\; 7 \cdot 43 \cdot \overline{cd} = (x - 10)(x + 10)$. Since the two smallest positive integers satisfying $301 | x \pm 10$ are $x=10$ and $x=291$ (contradicting $32 \leq x \leq 99)$, we must have $7 \mid x \pm 10$ and $43 \mid x \mp 10$, yielding $301 \mid x \mp 53$. Hence $x=53$ is the only possibility, which give us $\overline{abcd} = 53^2 = 2809$ and $3 \cdot \overline{cd} + 1 = 3 \cdot 9 + 1 = 27 + 1 = 28 = \overline{ab}$. Conclusion: The only four digit number satisfying conditions (1)-(2) is $\overline{abcd} = 2809$.