The number $\overline{x y z t}$ is a perfect square, and so is the number $\overline{t z y x}$, and the quotient of the numbers $\overline{x y z t}$ and $\overline{t z y x}$ is also a perfect square. Determine the number $\overline{x y z t}$. (The overline indicates that the number is written in the decimal system.)





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I solution: The quotient of two 4-digit numbers can only be a 1-digit number, so the question's quotient can only be 1, 4, or 9.

If the quotient were 1, then both square numbers would be in the form $\overline{x y y x}$. However, this is divisible by 11 according to the divisibility rule for 11. Therefore, only $33^{2}=1089, 44^{2}=1936, 55^{2}=3025, 66^{2}=4356, 77^{2}=5929$, $88^{2}=7744, 99^{2}=9801$ could meet our conditions, but none of these are in the form $\overline{x y y x}$.

$\overline{x y z t}=4 \cdot \overline{t z y x}$ is also impossible, because the left side, being an even square number, can only end in 0, 4, or 6. However, for $t \geq 3$, the right side would be a 5-digit number; and for $t=0$, $z=0$ and thus $100 \overline{x y}=4 \overline{y x}$, which would require $996 x=-60 y$ to hold.

Therefore, only

$$
\overline{x y z t}=9 \cdot \overline{t z y x}
$$

can be considered.

In this case, for $t \geq 2$, the right side would no longer be a 4-digit number, so it must be

$$
t=1, \quad \text { and thus } \quad x=9,
$$

which means

$$
9000+100 y+10 z+1=9(1000+100 z+10 y+9)
$$

from which

$$
y-89 z=8
$$

The only solution to this equation in non-negative single-digit integers is

$$
z=0, \quad y=8
$$

Thus, the only pair of numbers that can satisfy our problem is 9801 and 1089; and indeed, this pair does satisfy the condition, because

$$
9801=99^{2}=9 \cdot 1089=3^{2} \cdot 33^{2}
$$

Zoltán Szatmáry (Bp. VIII., Piarist g. II. o. t.)

II. solution: After determining that the quotient can only be 9, it follows that the upper limit for $\overline{t z y x}$ is $\frac{10000}{9}$, or

$$
1000<\overline{t z y x} \leq 1111
$$

But only the squares of 32 and 33 fall between 1000 and 1111. $32^{2}=1024$ does not fit, because 4201 is not a square number, but $33^{2}=1089$ does fit, because $9801=99^{2}=9 \cdot 1089$.

Gyula Surányi (Bp. I., Toldy Ferenc g. II. o. t.)