A five-digit square number has all different digits. By swapping its extreme digits, we again get a square number. What are the numbers in question?

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Let $\overline{A B C D E}=x^{2}$, and $\overline{E B C D A}=y^{2}$, where $A0$, so their values are chosen from $1,4,5,6$ and 9, thus $E-A$ can be one of the numbers $1,2,3,4,5$ and 8.

On the other hand, $x$ and $y$ are smaller than the smallest 6-digit number and larger than the square root of the smallest 5-digit number: $100<x< y \leq 316$, so $y+x \leq 631, y-x \leq 215$.

(1) The right side is a multiple of 101, so either $y+x$ is a multiple of it, at most 6 times, or $y-x=101k$, where $k=1$ or 2.

Select the value of $E-A$ among the constraints so far, factorize the left side of (1) into two factors, equate the smaller one to $y-x$ and the larger one to $y+x$, calculate $x$ and $y$, and check in each case if they meet all the requirements. $x$ is half the difference of the two factors, and it is only a 3-digit number if the difference is at least 200.

When $E-A=1$, the possible factorizations of the left side are: $99 \cdot 101$ and $33 \cdot 303$, from which $x=1$, or $x=135$, the latter does not fit because its square does not have all distinct digits.

It cannot be $E-A=2$, because then one of $y-x$ and $y+x$ would be even, the other odd, and thus $x, y$ would not be integers.

When $E-A=3$, the left side of (1) is $297 \cdot 101=99 \cdot 303$. It is enough to examine the latter, but the resulting $x^{2}=102^{2}$ ends in two 0s.

For $E-A=4$, both factors must be even, so we get the double of the results from the $E-A=1$ case, but $x=270$ does not fit because its square ends in two identical digits.

When $E-A=5$, from $495 \cdot 101$ we get $x^{2}=197^{2}=38$ 809, which does not fit; from $505 \cdot 99$ we get $x=203, y=302$, their squares are 41 209 and 91 204, respectively, which fit.

Finally, $E-A=8$ can only result from $E=9, A=1$, so $y \geq 301$ and $x \leq 141$, hence $y-x \geq 160$, and the difference of the factors of the left side of (1) must be at least 320. The only such factorization is $606 \cdot 132$, but from this $y^{2}$ turns out to be a 6-digit number.

There is no other possibility, so only 41209 and 91204 fit.

Tolnay-Knefély Tibor (Budapest, Bláthy O. vill. ip. t. II. o. t.)