Last three digits of $\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$ Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$
I don't know how to continue my work:
$$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}+3^{4032}}=\sqrt{2^{4032}+2^{2017}\cdot 3^{2016}+3^{4032}}=?$$
 A: HINT:
This can be factorised like this
$$(2^{2016})^2+2\cdot2^{2016}\cdot3^{2016}+({3^{2016}})^{2}=(2^{2016}+3^{2016})^2$$
So the question is to find last three digits of
$$2^{2016}+3^{2016}$$
A: The presence of the squares $4$ and $9$ here is a big hint to help you find that the expression $E := \sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}} = \sqrt{(2^{2016}+3^{2016})^2} = 2^{2016}+3^{2016}$.
Then finding hte last three digits of this expression is a matter of finding the remainders $2^{2016}\bmod 1000$ and $3^{2016}\bmod 1000$ and then adding.
The Carmichael function $\lambda$ gives us the maximum cyclic order of exponentiation and here $\lambda(1000) = {\rm lcm}(\lambda(2^3), \lambda(5^3)) = {\rm lcm}(2, 100) = 100 $. Any smaller cycle will divide this value. We could also use Euler's totient $\phi(1000)=400$ to similar effect. So this gives us $a^{2016} \equiv a^{16} \bmod 1000$ (noting $16\ge 3$, the greatest prime exponent in $1000$) and we can calculate this easily enough for both cases:
$ 2^{16}\equiv 1024\cdot 64 \equiv 24.64 \equiv 1536 \equiv 536 \bmod 1000$
$ 3^{16} \equiv 81^4 \equiv 6561^2 \equiv 561^2 \equiv 314721 \equiv 721 \bmod 1000$ 
And then we can add to get $E\equiv 257 \bmod 1000$.
A: If you follow the hint from @user35508, lets make a congruence.
$$2^{2016}+3^{2016} \text{ mód } 1000$$
Thanks to the Euler's totient function, we know that $a^{\phi(b)}\equiv 1 \text{ mod } b$ if mcd(a,b)=1. The factorization of $1000 = (5*2)^3$ so its trivial that $mcd(2,3,1000)=1$.
$$\phi(1000)=\phi((5*2)^3)=\phi(5^3)\cdot\phi(2^3)=(5^3-5^2)\cdot(2^3-2^2)=400$$
Finally, lets divide $2016/400 = 5\cdot 400 + 16$ and apply it to the first formula.
$$2^{5*400+16}+3^{5*400+16}\text{ mod } 1000 = (2^5)^{400}\cdot{2^{16}}+(3^5)^{400}\cdot{3^{16}}\text{ mod } 1000 = 2^{16}+3^{16}\text{ mod } 1000$$
The remainder is $43112257\text{ mod } 1000 = 257$ that is also the three last digits.
