How to prove the existence of odd numbers $a$ and $b$ in $2m\equiv a^{20}+b^{11}\pmod{2^n}$ 
Show that for any natural numbers $m,n$, there exist odd numbers $a,b$ such that
  $$2m\equiv a^{20}+b^{11}\pmod{2^n}$$

Thank you everyone.

We proceed by induction on $n$.
When $n=1$, we have $1^{20}+1^{11} \equiv 0 \equiv 2m \pmod{2^1}$.
Suppose that the statement holds for $n=k$, so $\exists a, b \in \mathbb{Z}$, $a, b$ odd, such that $2m \equiv a^{20}+b^{11} \pmod{2^k}$. Thus $a^{20}+b^{11} \equiv 2m, 2m+2^k \pmod{2^{k+1}}$. If $a^{20}+b^{11} \equiv 2m \pmod{2^{k+1}}$, we are done. Otherwise we have $a^{20}+b^{11} \equiv 2m+2^k \pmod{2^{k+1}}$. Then $a^{20}+(b+2^k)^{11} \equiv a^{20}+b^{11}+11b^{10}2^k \equiv (a^{20}+b^{11})+2^k \equiv 2m \pmod{2^{k+1}}$.  We are thus done by induction.
Therefore for any $m, n \in \mathbb{Z}^+$, there exist odd $a, b \in \mathbb{Z}$ such that $2m \equiv a^{20}+b^{11} \pmod{2^n}$.
