Vtyjkyuk

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?

Everything is fine upto

(m+1)(m-1)=1001x

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

Try looking at it this way

$m^2-1 = 1001x$ if and only if
$m equiv pm 1 pmod7$ ,
$m equiv pm 1 pmod11$, and
$m equiv pm 1 pmod13$

beginarrayr
n & mod 7 & mod11 & mod13 \
hline
11cdot 13 = 143 & 3 & 0 & 0 \
7 cdot 13 =91 & 0 & 3 & 0 \
7 cdot 11=77 & 0 & 0 & 12 \
hline
-2 cdot 143 = -286 & 1 & 0 & 0 \
4 cdot 91 = 364 & 0 & 1 & 0 \
-1 cdot 77 = -77 & 0 & 0 & 1 \
hline
endarray

So $m equiv pm 286 pm 364 pm 77 pmod1001$

beginarrayr
m & m pmod1001 & m^2 \
hline
286 + 364 + 77 & 727 & 528529 \
286 + 364 - 77 & 573 & 328329 \
286 - 364 + 77 & 0 & 0 \
286 - 364 - 77 & 846 & 715716 \
-286 + 364 + 77 & 155 & 24025 \
-286 + 364 - 77 & 1 & 1 \
-286 - 364 + 77 & 428 & 183184 \
-286 - 364 - 77 & 274 & 75076 \
hline
endarray

Solve using Chinese remainder theorem

as $1001=13cdot7cdot11, $

$$mequivpm1pmod13$$

$mequivpm1pmod7$

$mequivpm1pmod11$

But $mnotequivpm1pmod1001$