Let's determine those two-digit numbers, for which, by swapping the tens and units in their square, we arrive at the square of a number that is 1 greater.
I. Solution: Let the sought number be $10 x+y$, then $(10 x+y)^{2}=100 x^{2}+10(2 x y)+y^{2}$.

According to the problem, assuming that $2 x y$ and $y^{2}$ are single-digit numbers,

$$
100 x^{2}+10 y^{2}+2 x y=[10 x+(y+1)]^{2}
$$

After rearrangement, we obtain the following quadratic equation for $y$

$$
9 y^{2}-(18 x+2) y-(20 x+1)=0
$$

from which

$$
\begin{gathered}
y=\frac{2(9 x+1) \pm \sqrt{4(9 x+1)^{2}+36(20 x+1)}}{18}= \\
=\frac{9 x+1 \pm \sqrt{81 x^{2}+18 x+1+180 x+9}}{9}=\frac{9 x+1 \pm \sqrt{81 x^{2}+198 x+10}}{9}= \\
=\frac{9 x+1 \pm \sqrt{(9 x+11)^{2}-111}}{9}
\end{gathered}
$$

For $y$ to be an integer, a necessary (though not necessarily sufficient) condition is that the quantity under the square root must be a perfect square, that is,

$$
(9 x+11)^{2}-111=z^{2}
$$

or

$$
(9 x+11)^{2}-z^{2}=(9 x+11+z)(9 x+11-z)=111=3 \cdot 37
$$

Since the assumption $9 x+11+z=111$ and $9 x+11-z=1$ leads to $z=5, z=55$ and thus $y=1$, we have

$$
9 x+11+z=37, \quad 9 x+11-z=3
$$

From the sum of these two equations,

$$
18 x+22=40, \quad \text { from which } \quad x=1
$$

and thus

$$
y=\frac{9+1 \pm \sqrt{20^{2}-111}}{9}=\frac{10 \pm 17}{9}
$$

but only the value $y=\frac{10+17}{9}=3$ is valid.

Therefore, the sought number is 13. Indeed, $13^{2}=169,196=14^{2}$.

Considering negative numbers as well, -14 also fits.

Lajos Szász (Balassagyarmat, Balassa g. IV. o. t.)

Remark: This solution only shows that there is no other solution where $2 x y$ and $y^{2}$ are single-digit, apart from the ones found. In fact, more is true. If $y^{2}$ contains $v$ tens, and $2 x y+v$ contains $u$ tens, then the last two digits of $(10 x+y)^{2}$ are $y^{2}-10 v$ and $2 x y+v-10 u$, so

$$
(10 x+y)^{2}=100\left(x^{2}+u\right)+10\{2 x y+v-10 u\}+\left\{y^{2}-10 v\right\}
$$

and the expressions in the \{\} brackets give the last two digits. According to the conditions of the problem,

$$
(10 x+y+1)^{2}=100\left(x^{2}+u\right)+10\left(y^{2}-10 v\right)+2 x y+v-10 u
$$

From this, by rearrangement,

$$
\begin{aligned}
& 9\left(y^{2}-10 v\right)=9(2 x y+v-10 u)+20 x+2 y+1= \\
& \quad=9(\{2 x y+v-10 u\}+2 x)+2(x+y)+1
\end{aligned}
$$

follows, so $2(x+y)+1$ is divisible by 9. Assuming positive single-digit numbers, this is only true for $x+y=4$ and $x+y=13$. If $x+y=13$, then $x$ is at least 4, and thus,

$$
9 \cdot(\{2 x y+v-10 u\}+2 x)+2(x+y)+1 \geqq 9 \cdot 8+27=9 \cdot 11
$$

so $y^{2}-10 v$ could not be a single-digit number, contradicting the determination of $v$.

If $x+y=4$, then $2 x y$ and $y^{2}$ are single-digit, and thus the above solution gives the sought numbers.

If negative numbers are also allowed, and these are considered such that the digits are negative, then the values $x+y=-5$, $x+y=-14$ are also possible. The latter can be excluded, while the former leads to the pair of numbers $-14,-13$.

II. Solution: Let the sought number be $x$.

The last digit of any square number, in order of the last digit of the base, is

$$
0,1,4,9,6,5,6,9,4,1
$$

According to the problem, the last digit of $(x+1)^{2}$ is equal to the tens digit of $x^{2}$ and vice versa, so - using (1) - the last two digits of $-x^{2}$ can only be:

$$
10,41,94,69,56,65,96,49,14,01
$$

Numbers ending in 10, 94, and 14 are even but not divisible by 4, and the number ending in 65 is divisible by 5 but not by 25, so such numbers cannot be square numbers. Therefore, the last two digits of the squares of the sought two consecutive numbers can only be 69, 96. Since the tens digits of the square numbers are the same, the difference $(x+1)^{2}-x^{2}=2 x+1$ gives 27 and -27, from which the possible solutions $x=13$, and $x=-14$ are obtained, which also fit.

Remark: In this solution, we did not use that $x$ is two-digit. Therefore, this condition can be omitted from the problem statement.

Imre Csiszár (Bp. I., Petófi g. II. о. t.)