# 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 $2^5$?

• Did Duncan Moore notice the divisibility by at least $2^5$ in a paper or elsewhere, and then state he could not establish it? If so maybe this question is a bit advanced for this site, and would be better on a research site like Overflow. – coffeemath Dec 19 '14 at 12:26
• @coffeemath: He pointed that out to me in an email back in 2009 and suggested it might be due to a parameterization like the one for $a^4+b^4+c^4 = d^4$ by Elkies. I think it is a general property of $(2)$, but there's just three solutions to test it on. – Tito Piezas III Dec 19 '14 at 15:58

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 $105+23=2^7$.

• Thanks. Two quick questions: a) Does this analysis also apply to the general $x_1^5+x_2^5+\dots+x_n^5 = 0$ with exactly two odd terms? I tested it on various examples and it seems to be true in general. b) What about the cases when there are an odd number of odd terms? – Tito Piezas III Dec 22 '14 at 18:55
• For a) do you require a sum only divisible by $2^5$ or some higher power of $2$? b) [An odd number of odd terms] + [some even terms] = [odd] ≠ 0 :) – punctured dusk Dec 22 '14 at 19:31
• @barto: Re: a) Only by $2^5$. b) Oops. You're right. :) – Tito Piezas III Dec 22 '14 at 22:36
• @AdamBailey: That's a good generalization! For exactly four odd terms, perhaps it was just a fluke that two pairs were divisible by $2^5$. Unless another solution appears with the same property... – Tito Piezas III Dec 22 '14 at 22:43