Representation of integers by $A = B^{\frac{5}{4}} + C^{\frac{5}{4}}$

Question

Let $n>-1$ be an integer and $f(n)$ be the integer part of $n^{\frac{5}{4}}$.

And to avoid confusion define $f(0)=0,f(1)=1$ and all $f(n)$ as the integer part of the positive real values for $n^{\frac{5}{4}}$. ( no complex or negative branches for the power )

Conjecture :

Every positive integer can be expressed as

$$ A = f(B) + f(C)$$

For any given positive integer $A$, and some positive integers $B,C$.

How to prove this ?

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Some details or background ...

I have no formal method for the value $\frac{5}{4}$.

A generalization is ofcourse $f(q,n)$ be the integer part of $n^{q}$ for $1<q<2$. This is similar to Waring problem and Eulers son noticed the connection to

$$v(k) = 2^k + **(3/2)^k** - 2.$$

where the ** ** means integer part.

which you can see explained on wiki.

See : https://en.wikipedia.org/wiki/Waring%27s_problem

the similar function

$$ w(k) = 2^k + (3/2)^k - 2 $$

has as solution

$$ w(k) = 2 $$

$$k = 1.2333...$$

what is close to $1.25 = \frac{5}{4}$.

How large $a$ can be so that we can represent

$$ A = f(q,B) + f(q,C)$$

is a natural question.

I am not conjecturing q can be at most $1.25$ although it might be so. ( I think $q = \frac{4}{3}= 1.333...$ fails and $q = 1/2$ fails so it seems reasonable )

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Answer 1

I don't usually post empirical results as an answer, but I find it difficult to see how to make analytical progress.

Claim. (Additive basis of order 2) Every positive integer can be written in the form $\lfloor m^q\rfloor+\lfloor n^q\rfloor$ for some positive integers $m\le n$.

The difficulty lies in the difference that you request "every positive integer" instead of "every sufficiently large positive integer". The latter phrase is easier as we can make (sometimes quite crude) number-theoretic bounds to arrive at the result. As noted in the comments, several publications have resolved the claim for various ranges of $q$.

In the "sufficiently large" case, these include:

• $q\in(1,3/2)$: Konyagin, S.V., 2002. An additive problem with fractional powers.

• $q\in(1,11/10)$ when one of $m,n$ is prime: Yu, G., 2020. On a binary additive problem involving fractional powers. Journal of Number Theory, 208, pp.101-119.

Your post has two questions:

1. Does the claim hold when $q=5/4$?
2. What is $\sup q$ such that the claim holds?

For empirical results, I will look at the smallest positive integer that cannot be written in said form with $n$ given.

1. The R code to compute the smallest positive integer is as follows.

floorfunc <- function(q,a){
  p <- array(NA,c(a,a))
  for(m in 1:a){for(n in 1:a){p[m,n] <- floor(m^q)+floor(n^q)}}
  return(sort(setdiff(2:p[a,a],unique(as.vector(p))))[1])}

The horizontal axis in the plot is $n/10$ and the vertical axis is the smallest positive integer. The growth rate is slowly exponential, and with Konyagin's result, it makes sense to believe the claim holds.

2. I believe $$\sup q=1.26966\cdots$$ (note this does not appear in OEIS). The following function calculates the smallest non-representable positive integer as $q$ increments from $a+b/100$ to $a+b$ in increments of $1/100$, when $n=200$.

floorarray <- function(a,b){
  r <- array(NA,100)
  for(k in 1:s){r[k] <- floorfunc(a+b*k/100,200)}
  return(r)}

When $a=b=1$, we observe a very apparent drop in the $27$th element; that is, $q=1.27$. This means that $31$ cannot be written as $\lfloor m^{1.27}\rfloor+\lfloor n^{1.27}\rfloor$.

 [1]   NA   NA   NA  493   NA   NA   NA  609  639  675  661  705  743  757
[15]  790  835  874  876  967 1032 1055 1115 1097 1254 1330 1387   31 1544
[29]   30   17   15   13   13   11   11   11    9    9    9    9    7    7
[43]    7    7    7    7    5    5    5    5    5    5    5    5    5    5
[57]    5    5    3    3    3    3    3    3    3    3    3    3    3    3
[71]    3    3    3    3    3    3    3    3    3    3    3    3    3    3
[85]    3    3    3    3    3    3    3    3    3    3    3    3    3    3
[99]    3    3

With this knowledge, narrowing the interval $(a+b/100,a+b)$ repeatedly leads to the interval $(1.26966+0.00004/100,1.26967)$, in which we obtain

 [1] 1375 1375 1375 1375 1375 1375 1375 1375 1375 1375 1375   31   31   31
[15]   31   31   31   31   31   31   31   31   31   31   31   31   31   31
[29]   31   31   31   31   31   31   31   31   31   31   31   31   31   31
[43]   31   31   31   31   31   31   31   31   31   31   31   31   31   31
[57]   31   31   31   31   31   31   31   31   31   31   31   31   31   31
[71]   31   31   31   31   31   31   31   31   31   31   31   31   31   31
[85]   31   31   31   31   31   31   31   31   31   31   31   31   31   31
[99]   31   31