Here's the problem I am working on:-
Find,showing your method, a six-digit integer n with the following properties:
$(1)$ $n$ is a perfect square
$(2)$ the number formed by the last three digits of $n$ is exactly one greater than the number formed by the ﬁrst three digits of $n$. (Thus $n$ might look like $123124$; although, this is not a square.)"
Here's my approach:
Consider the first $3$ digits of $n$ as $x$
Because $n$ is a perfect square,
Using the difference of two squares identity:
Now $1001=m+1$ as if $1001$ is $m-1$, then $1003=m+1=x$, and this is not possible since $x$ is a $3$-digit number.
Now, here the problem arises.
If $x=999$, then the last $3$ digits should be $999+1=1000$ making the number:
However, this is a $7$ digit number, making the initial condition false.
Where did I go wrong?