It is known that given a solution to,
$$a^4+b^4+c^4 = d^4\tag1$$
then either $-c+d,\;c+d$ is always divisible by $2^{10}$. For example,
$$95800^4+414560^4+217519^4=422481^4$$
then $217519+422481=2^{10}\cdot5^4$.
Duncan Moore noticed, but could not prove, a similar congruence for,
$$a^5+b^5+c^5+d^5+e^5=0\tag2$$
There are only three known primitive solutions $a,b,c,d,e$, namely,
$$27,\; 84,\; 110,\; 133,\; -144$$
$$220,\; -5027,\; -6237,\; -14068,\; 14132$$
$$55,\; 3183,\; 28969,\; 85282,\; -85359$$
And we have,
$$27 + 133 = \color{blue}{2^5}\cdot5$$
$$-5027 -6237 = -\color{blue}{2^{10}}\cdot 11,\quad \text{and}\quad -14068 + 14132 = \color{blue}{2^6}$$
$$55 + 28969 = \color{blue}{2^5}\cdot907,\quad \text{and}\quad 3183 + (- 85359) = -\color{blue}{2^8}\cdot327$$
Question: Is it true that solutions to $(2)$ always have a pair of addends such that $a+b$ is divisible by $2^5$?