Number theory on multiples of squares

Question

If we take the numbers $ 48 $, $ 49 $ and $ 50 $, we can see that they are all consecutive integers and multiples of squares ($48$ is multiple of $2^2$, $49$ is multiple of $7^2$ and $50$ is multiple of $5^2$).

Is it possible to prove that there are infinitely many triples of consecutive integers which satisfy this condition, and is there a quadruple of consecutive integers which satisfy it? I would define $n$, $n+1$ and $n+2$, and by trying to square them I got $n^2$, $n^2+2n+1$ and $n^2+4n+2$. What can I do to prove the statement?

https://mathoverflow.net/questions/149194/consecutive-non-squarefree-integers/149198#149198 — TravorLZH, Sep 05 '23 at 13:45
Does this answer your question? Consecutive non square free numbers And for a more general statement, see For any positive integers $m$ and $n$, there are $n$ consecutive positive integers divisible by $a^m$ for some integer $a$. — Anne Bauval, Sep 05 '23 at 13:49
If $n$ is a solution to the negative Pell equation $n^2-2m^2=-1$, then $n^2+1$ is twice a square, $n^2$ is a square, and $n^2-1$ is a multiple of $2^2=4$ — J. W. Tanner, Sep 05 '23 at 13:53

score 1 · Answer 1 · answered Sep 05 '23 at 13:43

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If you add $2^2\cdot 7^2\cdot 5^2=4900$ to your solution, you get another $4948,4949,4950$ that satisfy the same property (with the same squares in fact). You can do this as many times as you like, so there are infinitely many solutions.

Say we want $x$ such that

\begin{align*} x&\equiv 0 \bmod 2^2\\ x+1&\equiv 0 \bmod 7^2\\ x+2 &\equiv 0\bmod 5^2 \end{align*}

then the Chinese remainder theorem gives us infinitely many solutions. The same is true if we choose four square moduli and four consecutive integers etc.

answered Sep 05 '23 at 13:43

Joshua Tilley

7,070

Just to add to this, note that $9$ doesn't divide $4900$; so you'll easily be able to extend to four consecutive integers (and so on). Of course, some of these solutions with four will trivially extend to a solution with five consecutive integers; but you can extend this as much as you want. – Chris Lewis Sep 05 '23 at 13:48
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Better link to a duplicate than edit another answer, Joshua. As for an even more general statement, @ChrisLewis, see For any positive integers $m$ and $n$, there are $n$ consecutive positive integers divisible by $a^m$ for some integer $a$. – Anne Bauval Sep 05 '23 at 13:51

Number theory on multiples of squares

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