# Simplest diophantine equation with open solvability

What is the simplest diophantine equation for which we (collectively) don't know whether it has any solutions? I'm aware of many simple ones where we don't know (whether we know) all the solutions, but all of these that I know have some solution.

Yes, I know that "simplest" is subjective. I'd be satisfied if it could be typeset in one line in a LaTeX document. Also, it would be nice if it could be easily memorized (no seven-digit numbers with nonobvious patterns:) though that's secondary.

• $(x+1)(y+1)=2^{2^{1000}}+1$ [wanting solutions in positive integers only]. Mar 8 '20 at 6:35
• $x^5+y^5=z^5+w^5$ [positive integers, $\{\,x,y\,\}\ne\{\,z,w\,\}$]. Mar 8 '20 at 11:51

Determining which integers $$n$$ are a sum of three cubes is a very famous open problem:

$$a^3 + b^3 + c^3 = n, \quad a,b,c \in \mathbb{Z}.$$ Conjecturally, $$n$$ is a sum of three cubes iff $$n \not \equiv 4,5 \bmod 9$$.

Note that this is really a family of Diophantine equations, rather than a single Diophantine equation.

• The smallest open case now is $n=114$: en.wikipedia.org/wiki/Sums_of_three_cubes#Computational_results Mar 8 '20 at 10:56
• This one, together with @MattF.'s comment, fits most what I wanted. "Perfect cuboid" is nice over positive integers, but encoding positive integers themselves in Z really complicates the equation. Thank you very much!
– Veky
Mar 10 '20 at 3:58

The perfect cuboid problem: We do not know if there is a common integer solution to $$a^2+b^2+c^2=d^2$$

$$a^2+b^2=e^2$$ $$a^2+c^2=f^2$$ $$b^2+c^2=g^2$$ with $$a,b,c \ge 1$$.

The last condition can also be replaced with $$abc=1+h^2+i^2+j^2+k^2$$.

The equations can also be reduced from four or five to one, taking $$\sum(LHS-RHS)^2=0$$ as a single equation in 7 or 11 variables for which we don’t know any solutions.

It's more complicated than the other answers by MattF and DanielLoughran, but the Erdős–Straus conjecture states that for every integer $$n \ge 2$$, there exist positive integers $$x, y, z$$ such that

$$\frac4n = \frac1x + \frac1y + \frac1z$$

Again, this is a family of Diophantine equations, related to Egyptian fractions, and computer research has verified the conjecture holds for all $$n$$ up to a rather large number, but whether it holds for all $$n$$ ($$\ge 2$$) is an open problem.

• I'm sure that the smallest $n$ for which no solution is known is more than a 7-digit number. Mar 8 '20 at 21:10
• @GerryMyerson In fact, it more than a 16-digit number.. Mar 11 '20 at 11:59