Four integers such that the sum of every pair is a power of two

Question

Numberphile Youtube channel just made a video telling us about an open question regarding natural numbers and the powers of two.

In a nutshell, it states that there are no 4 integers whose sum of each two pairs is a power of two.

Let's start with a simpler example with 3 such numbers:

$-1,\ 3\ $ and $5$.

$$-1 + 3 = 2,\ {-1} + 5 = 4 = 2^{2},\ 3+5=8=2^{3}$$ While it's easy to find examples with 3 numbers, finding examples with 4 numbers is not so easy.

In the video, Neil Sloane states that it's an open question, and the best solution known so far has 4 pairs. I thought it looks like a fun problem to play around with so I've written it in formal math:

Let $x,\ y,\ z,\ q \in \mathbb{Z}$ for the sake of nontriviality assume they are different. And right $a,\ b,\ c,\ d,\ e,\ f \in \mathbb{N}$ some given constants.

Well, we have the following equation system:

$$ x+y=2^a\ [1] $$ $$ x+z=2^b\ [2] $$ $$ x+q=2^c\ [3] $$ $$ y+z=2^d\ [4] $$ $$ y+q=2^e\ [5]$$ $$ z+q=2^f\ [6] $$

After solving the first 4 equations we get:

$$x = 2^{a-1} + 2^{b-1} - 2^{d-1}\ [7]$$ $$y = 2^{a-1} + 2^{d-1} - 2^{b-1}\ [8]$$ $$z = 2^{b-1} + 2^{d-1} - 2^{a-1}\ [9]$$ $$q = 2^c - 2^{a-1} - 2^{b-1} + 2^{d-1}\ [10]$$

All nice and dandy. Let's continue with the last 2 equations. Replace y, z, and q with their other forms and get the following equations:

$$2^c+2^d-2^b = 2^e\ [11]$$ $$2^c+2^d-2^a = 2^f\ [12]$$

I conjectured that these equations have no nontrivial natural solutions. So I asked it here on StackExchange and concluded it has none, just as suspected.

This means that the equations $[11]$ and $[12]$ have no valid solutions, and thus the (problem or) "square" with 4 integers cannot be completed.

Q.E.D.

But it seems too easy to be a complete and correct proof. I made some mistakes and it's not valid. I had fun playing around with it. I am curious to know where the mistakes are and why it's not valid.

Note that this proof can be generalized to any number of variables. The only exception where it doesn't work is for 2 variables ($x+y=2^a$ it works since there are infinitely many solutions), and for 3 variables, where the number of equations is equal to the number of variables.

Answer 1

It seems correct, however if you want to strip out unnecessary parts you can just evaluate $[4]-[1]+[3]$ to get $z+q=2^d-2^a+2^c$. Since $z+q=2^f$, you get $2^d+2^c=2^a+2^f$ and as you already noticed, then either $d=a$ or $d=f$. That means $y+z=2^d=2^a=x+y$ and $z=x$ (impossible), or $y+z=2^d=2^f=z+q$ and $y=q$ (again impossible).

This proves it is impossible to have all six pairs to be a powers of two. But, if you look close enough, you might notice we haven't used all equations (for example $[5]$ wasn't used at any point), so this is actually also proof for impossibility of five pairs.

Anyway as also pointed out in comments, after the video has been recorded there was a progress on the problem (including solution of the four numbers case) (https://www.numberphile.com/stop-press). It's probably not surprise that the proof uses the same idea as above.

Answer 2

The two final equations can be deduced directly by the compatibility conditions of linear systems $$\det\begin{vmatrix}1&1&0&0&2^a\\1&0&1&0&2^b\\1&0&0&1&2^c\\0&1&1&0&2^d\\1&1&0&1&2^e\end{vmatrix}=2(2^b-2^c-2^d+2^e)=0$$ and similarly with $$\det\begin{vmatrix}1&1&0&0&2^a\\1&0&1&0&2^b\\1&0&0&1&2^c\\0&1&1&0&2^d\\0&0&1&1&2^f\end{vmatrix}=2(2^a-2^c-2^d+2^f)=0$$ from which $$2^c+2^d=2^b+2^e=2^a+2^f$$ Taking for example,$$2^b+2^e=2^a+2^f$$ this is possible only when the minimum exponents in both sides are equal because if not we would have by division the false equality $odd=even$ but this implies the equality of the other exponents.

We are done.