## Task 19/86

Find, without using any aids, all (in the decimal system truly) four-digit natural numbers that are squares of a natural number and for which both the first two digits and the last two digits are equal.
The sought natural numbers $n$ are of the form

$$
n=1100 a+11 b=11(100 a+b)=k^{2}
$$

with $a ; b ; k \in N, a ; b \leq 9 ; a ; k>0$.

Obviously, $n$ is divisible by 11. Then $k$ is also divisible by 11, and thus $n=k^{2}$ is even divisible by $11^{2}$. This further implies the divisibility of $100 a+b$ by 11. According to the divisibility rule for the number 11, $a+b$ is also divisible by 11, and since $a ; b \leq 9 ; a>0$, it follows that $a+b=11, b=11-a$. Therefore,

$$
n=11(100 a+b)=11(100 a+11-a)=11(99 a+11)=11^{2}(9 a+1)=k^{2}
$$

Thus, $9 a+1=m^{2}$ with $m \in N, m>0$. Trying the possible values of $a$ $(a=1 ; 2 ; \ldots ; 9)$, one finds that only $a=7$ yields a perfect square: $m^{2}=9 \cdot 7+1=64=8^{2}$. Therefore, there exists exactly one number $n$ with the required property: $n=7744=88^{2}$.