## Task 5/65

Which three-digit natural numbers have the following property:

If their square, represented in the decimal system, is split into two three-digit segments, the right segment is 1 larger than the left.
If we denote the number we are looking for with $n$ and the digits of $n^{2}$ with $a ; b ; c ; a ; b ;(c+1)$, then

$$
n^{2}=10^{5} a+10^{4} b+10^{3} c+10^{2} a+10 b+c+1=\left(10^{3}+1\right)\left(10^{2} a+10 b+c\right)+1
$$

or $n^{2}-1=1001\left(10^{2} a+10 b+c\right)$. This means that $n^{2}-1$ is divisible by 1001.

Since $n^{2}-1=(n+1)(n-1)$ and $1001=7 \cdot 11 \cdot 13$, we need to determine $n$ such that $(n+1)(n-1)$ is divisible by $7 ; 11 ; 13$. There are 8 possibilities:

1. $n+1$ is divisible by $7 ; 11 ; 13$
2. $n+1$ is divisible by $7 ; 11 ; n-1$ is divisible by 13
3. $n+1$ is divisible by $7 ; 13 ; n-1$ is divisible by 11
4. $n+1$ is divisible by $11 ; 13 ; n-1$ is divisible by 7
5. $n-1$ is divisible by $7 ; 11 ; 13$
6. $n-1$ is divisible by $7 ; 11 ; n+1$ is divisible by 13
7. $n-1$ is divisible by $7 ; 13 ; n+1$ is divisible by 11
8. $n-1$ is divisible by $11 ; 13 ; n+1$ is divisible by 7

For the calculation of each case:

1. $n+1=7 \cdot 11 \cdot 13 \cdot k=1001 k$; for $k=0$ we get $n=-1$, for $k \geq 1$ we get $n>1000$, these numbers are excluded as they do not meet the conditions of the problem.
2. $n+1=7 \cdot 11 \cdot k=77 k$; $77 k-2$ must be divisible by 13. This is the case for $k=11$. Therefore, $n_{1}=846$.
3. $n+1=7 \cdot 13 \cdot k=91 k$; $91 k-2$ must be divisible by 11. This is true for $k=8$. Thus, $n_{2}=727$.
4. $n+1=11 \cdot 13 \cdot k=143 k$; $143 k-2$ must be divisible by 7. This is the case for $k=3$. Hence, $n_{3}=428$.
5. $n-1=7 \cdot 11 \cdot 13 \cdot k=1001 k$; the remark for 1. applies.
6. $n-1=7 \cdot 11 \cdot k=77 k$ yields analogously $n_{4}=155$.
7. $n-1=7 \cdot 13 \cdot k=91 k$; yields $n_{5}=274$.
8. $n-1=11 \cdot 13 \cdot k=143 k$; yields $n_{6}=573$.

Thus, the following results $\left(n ; n^{2}\right)$ have been found: $(846 ; 715716) ;(727 ; 528529) ;(428 ; 183184)$; $(155 ; 024025) ;(274 ; 075076) ;(573 ; 328329)$.