Let's find those square numbers which, when divided by 11 with a remainder, the quotient is a prime number and the remainder is 4.
Let this square number be $n^{2}$, where $n>0$ (it cannot be $n=0$). According to the condition, $n^{2}=11 \cdot p+4$, for some prime number $p$. From this, we get that $11 \cdot p=n^{2}-4=(n-2)(n+2)$. Since the prime factorization is unique in the domain of positive integers, there are four possibilities for the product on the right side: either one of the factors is 1, or one of the factors is 11 and the other factor is the prime number $p$.

Since $n>0$, the case $n+2=1$ is impossible from the start. In the case $n-2=1$, $n+2=5$, which cannot match $11 \cdot p$. In the case $n-2=11$ and $n+2=p$, $p=11+4=15$, which is not a prime number. The only possibility left is $n+2=11$ and $n-2=p$, from which $p=7$ follows. Indeed, we have obtained a prime number. (Indeed: $81=11 \cdot 7+4$).

Note. In the solution, we considered only natural numbers as prime numbers. In the domain of integers, the negative of a prime number must also be considered a prime number. Regardless of this, $n$ can still be taken as positive. If $p<0$ based on the condition, only $n=1$ and $n=2$ could come into question. In the first case, $1^{2}=1$, and dividing this by 11 does not give a remainder of 4. In the second case, $2^{2}=4$. Dividing this by 11 does give a remainder of 4, but the quotient is 0, which is not a prime number.