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Is there a sequence of 5 consecutive positive integers such that none are square free? A number is square free if there is no prime number p such that $p^2 \mid n$

What I've tried doing so far is to show that for every sequence of 5 integers there will be a number that is not square free but I dont think that this helps to prove the initial statement.

andrew749
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4 Answers4

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Using the Chinese Remainder Theorem you can find $X$ such that each of

  • $X \equiv 0 \mod 2^2$,
  • $X- 1 \equiv 0 \mod 3^2$,
  • $X- 2 \equiv 0 \mod 5^2$,
  • $X- 3 \equiv 0 \mod 7^2$,
  • $X- 4 \equiv 0 \mod 11^2$

holds.

Of course this can be generalized for any desired number of consecutive non-squarfree numbers.

quid
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    Great answer. In this particular case, though, if $x$ is a multiple of $4$, $x-4$ is already non-squarefree. – Mike Mar 15 '15 at 20:30
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    Good point. The method is certainly not optimized in this regard. – quid Mar 15 '15 at 20:33
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How about $$ 844, 845, 846, 847, 848? $$ Here, $$2^2 \mid 844, \quad 13^2 \mid 845, \quad 3^2 \mid 846, \quad 11^2 \mid 847, \quad 4^2 \mid 848.$$

This is the smallest example.

More generally, the smallest number that starts a consecutive run of $n$ non-squarefree integers is enumerated by the sequence A045882, starting (with $n=2$) $$ 8, 48, 242, 844, 22020, \ldots $$

Travis Willse
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Yuval Filmus
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    Side-note: Put some brackets around each pair, to make it more obvious to the reader (the divides operator has the appearance of a delimiter, where in fact the commas are the delimiters here). – barak manos Mar 15 '15 at 20:49
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Yes, there is, more than one of those, in fact. If $n$ is odd, then $n^2 - 1$ is a multiple of $4$, as is $n^2 + 3$. If $n^2 + 1$ just happens to be twice the square of an odd number then you just need $n^2 + 2$ to have a repeated prime factor.

For example, $n = 41$. Then $1680$ and $1684$ are both multiples of $4$ and $1682 = 2 \times 29^2$. And we're in luck as $1683 = 3^2 \times 11 \times 17$.

If you need exactly $5$ consecutive non-squarefree numbers, then this run of $1680$ to $1684$ works for you, because $1679 = 23 \times 73$ and $1685 = 5 \times 337$. Not that we should be concerned about $6$ consecutive non-squarefree numbers, because the first such run of those isn't until $22020$.

James47
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Some good answers on here already. I just want to mention the Möbius function $\mu(n) = \delta_{\omega(n)}^{\Omega(n)} (-1)^{\omega(n)}$. Basically this says that if $n$ has any repeated prime factors, then $\mu(n) = 0$, otherwise $\mu(n) = -1$ or $1$ depending on a fact not relevant for this particular application. (Note that $\mu(1) = 1$, this is a consequence of $1$'s non-primality, not a reason for it.) Some computer algebra systems implement this as MoebiusMu or moebiusmu.

Your question then becomes one of finding $n$ such that $\mu(k) = 0$ for $n \leq k \leq n + 4$. Or $n$ such that $$\sum_{k = n}^{n + 4} |\mu(k)| = 0.$$ There are infinitely many such runs and a few different ways of finding them.

If you want to find the first $n$, then I suppose you have to test numbers one by one. But if you're more interested in the factorization of $n$ having certain properties, then you can use the Chinese remainder theorem. Keep in mind that since $4 = 2^2$, if you choose to make $n$ a multiple of $4$, then you automatically get $n + 4$ non-squarefree without having to work for it. But this means that $n + 2$ must be divisible by an odd square. You also need this for $n + 1$ and $n + 3$, of course. One possible system of congruences you can use is:

  • $n \equiv 0 \pmod 4$
  • $n \equiv 8 \pmod 9$
  • $n \equiv 23 \pmod{25}$
  • $n \equiv 46 \pmod {49}$

One solution is $n = 29348$. This gives us:

  • $29348 = 4 \times 7337$
  • $29349 = 9 \times 3261$
  • $29350 = 25 \times 1174$
  • $29351 = 49 \times 599$
  • $29352 = 4 \times 7338$

If for whatever reason you only want one multiple of $4$ in the run, that can be found, too. Try $n = 4225074$ for instance.