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 $x$
Now, $n=1000x+(x+1)$
Because $n$ is a perfect square,
$$m^2=1001x+1$$
Using the difference of two squares identity:
$$(m+1)(m-1)=1001x$$
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.
Hence, $x=999$
Now, here the problem arises.
If $x=999$,
then the last $3$ digits should be $999+1=1000$ making the number:
$9991000$
However, this is a $7$ digit number, making the initial condition false.
Where did I go wrong?
Best Answer
Everything is fine upto
After that, you asume that it implies $$(m+1=1001 \land m-1 = x) \lor (m+1=x \land m-1=1001)$$
This is not true, for instance:
11*9=99=33*3, yet, we don't have 33=11 nor 9=3