Congruence properties of $a^5+b^5+c^5+d^5+e^5=0$?

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 $2^5$?

It is helpful to consider separately the cases with exactly two and exactly four odd terms (these are the only possibilities since a solution with zero odd terms would not be primitive, while a sum with an odd number of odd terms could not equal zero).

The case with exactly four odd terms looks difficult, but that with exactly two odd terms (which includes two of the three known solutions) is straightforward. Without loss of generality suppose $a, b$ in $(2)$ are odd and $c, d, e$ even. Then:

$$2^5\mid -(c^5+d^5+e^5)=(a^5 + b^5) = (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)$$

The final set of brackets contains five odd terms and is therefore odd. Hence:

$$2^5 \mid a+b$$

This result generalises to any number of terms, provided that exactly two terms are odd, and also to any odd power $n$, provided that we consider divisibility by $2^n$. An example for $n=7$ is:

$$194^7 + 150^7+105^7 +23^7 -192^7-152^7-132^7-38^7=0$$

where $105+23=2^7$.