6
$\begingroup$

Is there some (huge) positive integer $M$ with the following property: for any $z>M$, there exist positive integers $x, y_{1}, y_{2},..., y_{z}$ such that $x^x$ $=$ $y_{1}^{y_{1}}$+ $y_{2}^{y_{2}}$+ ... +$y_{z}^{y_{z}}$ ?

[Please remark that the $y$'s are $\geq$ $1$ and need not to be necessarily distinct.]

As a [rather naive] way to attack this problem [which may (perhaps) be related to some works of Robinson, Matiasevich, M. Davis, and Chao-Ko], I'm thinking about lots of $1$'s, lots of $2$'s, and lots of $(x-1)$'s. Also, let us observe that, if $z$ has this property and $y_{i}$ $=$ $2$ for some $i$, then $z+3$ has the same property, too...

$\endgroup$

1 Answer 1

2
$\begingroup$

Have you tried looking at the density of possible $z$'s?

I think the answer might be "no". Here's my heuristics (hoping it's not bogus): the smallest $z$ we can achieve using $x$ is at least $\frac{x^x}{(x-1)^{(x-1)}}\approx ex$, the second smallest would be at least $\frac{x^x-(ex-1)(x-1)^{(x-1)}}{(x-2)^{(x-2)}}\approx e^2x$ and so on.

Let $N$ be a very large integer. we look at the interval $[1,N]$ and see how many such $z$'s are in it. From the above argument there are at most $\approx \frac{N}{e}+\frac{N}{e^2}+\cdots=\frac{N}{e-1}< N$ solutions. This contradicts your conjecture that there are asymptotically $\approx N$ solutions (in terms of values of $z$)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.