Automorphic numbers 
Problem. We say that the $n$-digit number $x$ is automorphic iff $x^2\equiv x \mod(10^n)$. Prove that if $x$ is $n$-digit automorphic number then $(3x^2-2x^3)\mod(10^{2n})$ is $2n$-digit automorphic number. Hint: use Chinese reminder theorem to find the necessary and sufficient condition for number to be automorphic.

So from Chinese reminder theorem we have that $x$ is automorphic iff: $$ \begin{cases} x(x-1)\equiv 0 \mod(2^n)\\ x(x-1)\equiv 0 \mod(5^n)\end{cases} $$ which gives us four systems of equations:
$$ \begin{cases} x\equiv 0 \mod(2^n)\\ x\equiv 0 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 1 \mod(2^n)\\ x\equiv 1 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 0 \mod(2^n)\\ x\equiv 1 \mod(5^n)\end{cases} $$
$$ \begin{cases} x\equiv 1 \mod(2^n)\\ x\equiv 0 \mod(5^n)\end{cases} $$
and it's easy to check that thesis is true for the first two cases, just by simple operations. But how to check the last two?

(3)If $x≡0\pmod{2^n}=a2^n$  for some integer $a$
$3x^2-2x^3=3(a2^n)^2-2(a2^n)^3$ is clearly divisible by $2^{2n}$ 
Here $x≡1\pmod{5^n}=(b5^n+1)$ for some integer $b$
$3x^2-2x^3=3(b5^n+1)^2-2(b5^n+1)^3=(b5^n+1)^2(3-2(b5^n+1))$
$≡(1+2b5^n+b^25^{2n})(1-2b5^n)≡(1+2b5^n)(1-2b5^n)=1-4b^25^{2n}≡1\pmod{5^{2n}}$
(4)If $x≡0\pmod{5^n}=c5^n$  for some integer $c$
$3x^2-2x^3=3(c5^n)^2-2(c5^n)^3$ is clearly divisible by $5^{2n}$ 
Here $x≡1\pmod{2^n}=(d2^n+1)$ for some integer $d$
$3x^2-2x^3=3(d2^n+1)^2-2(d2^n+1)^3=(d2^n+1)^2(3-2(d2^n+1))$
$≡(1+2d2^n)(1-2d2^n)=1-4b^22^{2n}≡1\pmod{2^{2n}}$
Alternatively, if $x=ma+b$  where $0 ≤b<m$ So,$x≡b\pmod {m}≡b\pmod {m^2}$,
$3x^2-2x^3=3(ma+b)^2-2(ma+b)^3≡3b^2-2b^3-6mab(b-1)\pmod {m^2}$
If  $3b^2-2b^3-6mab(1-b)≡b\pmod {m^2}$
$m^2\mid 2b^3-3b^2+b+6mab(b-1)$ 
$m^2\mid b(b-1)(2b-1+6ma)$ 
Clearly two of the three solutions are $b=0,1\pmod {m^2}$
$\implies$ 
if $x=ma,m^2\mid(3x^2-2x^3)$
if $x=ma+1≡1\pmod {m}, 3x^2-2x^3≡1\pmod {m^2}$
