Subfield: Number Theory


Question: Prove that there is a constant $c>0$ and infinitely many positive integers $n$ with the following property: there are infinitely many positive integers that cannot be expressed as the sum of fewer than $c n \log (n)$ pairwise coprime $n^{\text {th }}$ powers.

Solution:
Suppose, for an integer $n$, that we can find another integer $N$ satisfying the following property:

$$
n\text{ is divisible by }\varphi\left(p^{e}\right)\text{ for every prime power }p^{e} \text{exactly dividing }N.
\tag{1}
$$

This property ensures that all $n^{\text {th }}$ powers are congruent to 0 or 1 modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\text {th }}$ powers is congruent to $m$ or $m-1$ modulo $p^{e}$, since at most one of the $n^{\text {th }}$ powers is divisible by $p$. Thus, if $k$ denotes the number of distinct prime factors of $N$, we find by the Chinese Remainder Theorem at most $2^{k} m$ residue classes modulo $N$ which are sums of at most $m$ pairwise coprime $n^{\text {th }}$ powers. In particular, if $N>2^{k} m$ then there are infinitely many positive integers not expressible as a sum of at most $m$ pairwise coprime $n^{\text {th }}$ powers.

It thus suffices to prove that there are arbitrarily large pairs $(n, N)$ of integers satisfying $(1)$ such that

$$
N>c \cdot 2^{k} n \log (n)
$$

for some positive constant $c$.

We construct such pairs as follows. Fix a positive integer $t$ and choose (distinct) prime numbers $p \mid 2^{2^{t-1}}+1$ and $q \mid 2^{2^{t}}+1$; we set $N=p q$. It is well-known that $2^{t} \mid p-1$ and $2^{t+1} \mid q-1$, hence

$$
n=\frac{(p-1)(q-1)}{2^{t}}
$$

is an integer and the pair $(n, N)$ satisfies $(1)$.

Estimating the size of $N$ and $n$ is now straightforward. We have

$$
\log _{2}(n) \leqslant 2^{t-1}+2^{t}-t<2^{t+1}<2 \cdot \frac{N}{n}
$$

which rearranges to

$$
N>\frac{1}{8} \cdot 2^{2} n \log _{2}(n)
$$

and so we are done if we choose $c<\frac{1}{8 \log (2)} \approx 0.18$.
we seek arbitrarily large pairs of integers $n$ and $N$ satisfying ( $\dagger$ ) such that $N>c 2^{k} n \log (n)$.

This time, to construct such pairs, we fix an integer $x \geqslant 4$, set $N$ to be the lowest common multiple of $1,2, \ldots, 2 x$, and set $n$ to be twice the lowest common multiple of $1,2, \ldots, x$. The pair $(n, N)$ does indeed satisfy the condition, since if $p^{e}$ is a prime power divisor of $N$ then $\frac{\varphi\left(p^{e}\right)}{2} \leqslant x$ is a factor of $\frac{n}{2}=\operatorname{lcm}_{r \leqslant x}(r)$.

Now $2 N / n$ is the product of all primes having a power lying in the interval $(x, 2 x]$, and hence $2 N / n>x^{\pi(2 x)-\pi(x)}$. Thus for sufficiently large $x$ we have

$$
\log \left(\frac{2 N}{2^{\pi(2 x)} n}\right)>(\pi(2 x)-\pi(x)) \log (x)-\log (2) \pi(2 x) \sim x
$$

using the Prime Number Theorem $\pi(t) \sim t / \log (t)$.

On the other hand, $n$ is a product of at most $\pi(x)$ prime powers less than or equal to $x$, and so we have the upper bound

$$
\log (n) \leqslant \pi(x) \log (x) \sim x
$$

again by the Prime Number Theorem. Combined with the above inequality, we find that for any $\epsilon>0$, the inequality

$$
\log \left(\frac{N}{2^{\pi(2 x)} n}\right)>(1-\epsilon) \log (n)
$$

holds for sufficiently large $x$. Rearranging this shows that

$$
N>2^{\pi(2 x)} n^{2-\epsilon}>2^{\pi(2 x)} n \log (n)
$$

for all sufficiently large $x$ and we are done.