Squares, a riddle

In fact, at least one of the last two digits of a square number greater than $9$ is always even.

First proof: list all possible pairs of final digits of odd squares (there are only $11$ of them) and check: $01,09,25,49,81,21,69,89,61,41,29$

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Second proof: all even squares end in an even digit; we don't need to worry about those. So consider an odd square, $s=(2n+1)^2$ .

We know that $s-1=4n^2+4n$ is a multiple of $4$ . Also, the last digit of $s$ is one of $1,5,9$ . So the last digit of $s-1$ is one of $0,4,8$ . Hence (by divisibility rules for $4$ ) the tens digit must be even.

Since the number's square whose all digits are odd, the number is a odd number.

Let the number be $10A + u$ where $A$ is any positive integer, $u$ is odd number $u \in \{1, 3, 5, 7, 9\}$

• $1^2 = 1$
• $3^2 = 9$
• $5^2 = 25$
• $7^2 = 49$
• $9^2 = 81$

All carry forward digits $c_{u^2}$ are even.

$(10A+u)^2 = 100A^2 + 20Au + u^2$

The tens digit $= 20\cdot A \cdot u + 10 \cdot c_{u^2} \pmod{10} = 2 \cdot A \cdot u + c_{u^2} \pmod{10} =$ even

$\therefore$ no such perfect numbers whose digits are odd.