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(1) For every $n\in \Bbb N$, there exist squarefree numbers $x$ and $y$ such that $x+y=n$, i.e., every natural number can be written as the sum of two squarefree numbers;
(2) For every $n\in \Bbb N$, there exist squarefree numbers $k$ and $l$ such that $k-l=n$, i.e., every natural number can be written as the difference of two squarefree numbers.

For the squarefree numbers, I only know the definition of it, and I really did not quite understand how to prove these two from the definition.

Any help, please. Thanks in advance.

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2 Answers 2

By the inclusion-exclusion principle, the number of squarefree numbers not exceeding $N$ is

$$\begin{align} \operatorname{sf}(N) &= \sum_{k = 0}^\infty \mu(k)\left\lfloor\frac{N}{k^2}\right\rfloor\\ &= N\sum_{k=0}^\infty \frac{\mu(k)}{k^2} - \sum_{k \leqslant\sqrt{N}}\mu(k)\left(\frac{N}{k^2} - \left\lfloor\frac{N}{k^2}\right\rfloor\right) - \sum_{k > \sqrt{N}} \mu(k)\frac{N}{k^2}\\ &= \frac{6}{\pi^2}N - \sum_{k \leqslant\sqrt{N}}\mu(k)\left(\frac{N}{k^2} - \left\lfloor\frac{N}{k^2}\right\rfloor\right) - \sum_{k > \sqrt{N}} \mu(k)\frac{N}{k^2}\\ &\geqslant \frac{6N}{\pi^2} - 2\sqrt{N}, \end{align}$$

where $\mu$ is the Möbius function.

For $N > 1192$, that bound implies $\operatorname{sf}(N) \geqslant \frac{11}{20}N$. For $N \leqslant 1192$, this bound can easily be verified by explicit computation.

Since the smallest (non-negative) squarefree number is $1$, to write $0$ or $1$ as the sum of two squarefree numbers, we need to allow negative squarefree numbers. If we do, we have $0 = 1 + (-1)$ and $1 = 2 + (-1)$.

For $n \geqslant 2$, the above bound implies that $n - S(n)$ and $S(n)$ intersect, where $S(n) = \{ k : 0 \leqslant k \leqslant n, k \text{ squarefree}\}$, so $n$ can be written as the sum of two non-negative squarefree numbers.

For the difference, $0 = 1 - 1$, and for positive $n$, we set $N = 20n$, and find that there are at least $10n$ squarefree numbers $n < k \leqslant N$, and at least $\left\lceil \frac{209}{20}n\right\rceil > 10n$ squarefree numbers $\leqslant 19n$, hence the sets $S(20n)$ and $n + S(19n)$ must intersect, since there are only $19n$ numbers between $n$ and $20n$, hence $n$ can be written as the difference of two squarefree numbers.

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What is your $\mu(k)$? –  Scorpio19891119 Sep 3 '13 at 13:10
    
Oh, sorry, the Möbius function, $\mu(n) = (-1)^k$ if $n$ is the product of $k$ distinct primes, and $\mu(n) = 0$ is $n$ is not squarefree. –  Daniel Fischer Sep 3 '13 at 13:11
    
Thanks for your answer. But could you please explain also that how you got the number 1192? Just by solving the inequality? Does this number have some special meaning? –  Scorpio19891119 Sep 4 '13 at 0:33
    
Just by solving the inequality. Had I chosen a lower bound of $\frac{51}{100}N$, the estimate would yield that for $N > 417$. Had I chosen $\frac35 N$, for $N > 63654$. It has no special meaning, I just picked a bound roughly half way between $\frac{6}{\pi^2}$ and $\frac12$. –  Daniel Fischer Sep 4 '13 at 10:12

Hint for the sum version: if $n$ cannot be represented as a sum of squarefree numbers, then at least half of the numbers less than $n$ must be squareful. But the limit density of squarefree numbers is $\frac{6}{\pi^2} \gt 0.5$, so already we have that there can only be a finite number of exceptions.

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Limit density does not apply. You have to know actual bounds. –  marty cohen Sep 3 '13 at 0:30
    
It's enough to show that it's almost always true, but yes we need actual bounds to show it always is. Also it is seemingly a complete argument for the difference version of the problem since we can check for differences starting from arbitrarily high up. –  Dan Brumleve Sep 3 '13 at 0:35

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