1
$\begingroup$

Written in a more formal way, proof that there are at the most $2$ numbers $n$ of six digits, that

$$n^2 \equiv n \mod 10^6$$

Research effort:

if $n^2 \equiv n \mod 10^6$ this means $10^6\mid n^2-n \rightarrow10^6\mid n(n-1)$

Given that $5^6\mid 10^6$ then $5^6\mid n(n-1)$ whitch means $5^6\mid n$ or $5^6\mid n-1$

Same for $2^6$, Given that $2^6\mid 10^6$ then $2^6\mid n(n-1)$ whitch means $2^6\mid n$ or $2^6\mid n-1$

By the chinese remainder theorem I got assured $4$ solutions, but I tried to solve the systems and it's not a very happy thing to do, so I think I did something wrong somewhere.

What do you think?

$\endgroup$
2
  • 4
    $\begingroup$ Two of the solutions are $0$ and $1$. You are looking for the other two. $\endgroup$ Commented Mar 6, 2014 at 17:12
  • $\begingroup$ Note that if you have $2^6|n$ and $5^6|n+1$ then you have also $2^6|10^6-n$ and $5^6|10^6-n-1$ giving you a second pair of consecutive numbers which will solve the problem. $\endgroup$ Commented Mar 6, 2014 at 17:28

1 Answer 1

Reset to default
0
$\begingroup$

If $\displaystyle2^6|(n-a)$ and $5^6|(n-a),$

$\displaystyle$ lcm$(2^6,5^6)|(n-a)$

As $(2,5)=1,(2^6,5^6)=(2,5)^6=1$

$\implies$ lcm$(2^6,5^6)=(2^6\cdot5^6)=10^6\iff n\equiv a\pmod{10^6}$

Here $a=0,1$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.