<s>[INST] $a^4+b^4+c^4+d^4 \neq 2^{2011}$ Prove (elementary, meaning no high level theorems used) that there can not exist 4 prime numbers a,b,c,d $\geq$ 7 such that
\begin{equation}a^4+b^4+c^4+d^4=2^{2011}\end{equation}
I tried the following:
The last digit l(d) of the fourth power of a prime number is 1 or 5.
The last digit of $2^{2011}$ is 8.
But there exists a combination that holds true: 
\begin{equation}l(a)=1,l(b)=1,l(c)=1,l(d)=5\end{equation} 
\begin{equation}1+1+1+5=8\end{equation}
 [/INST] If $n$ is odd, $n^4\equiv 1\pmod{16}$, hence if $a,b,c,d$ are prime numbers and at least one of them is odd,
$$ a^4+b^4+c^4+d^4 \in \{1,2,3,4\}\pmod{16}.$$
However, $2^{2011}\equiv 0\pmod{16}$ and $2^4+2^4+2^4+2^4$ is way smaller than $2^{2011}$.
</s>