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What can we say about growth of smallest gap $g(a)$ which is the smallest $|x-y|$ where $0\leq x,y\leq\Big\lfloor\frac a2\Big\rfloor$ and $\sqrt{x(a-x)},\sqrt{y(a-y)}\in\Bbb Z$?

Is $g(a)=1\iff a=b^2+1$ (corresponding to $x=0$ and $a-x=b^2+1$ or $x=1$ and $a-x=b^2$)?

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  • $\begingroup$ This question is very similar to the question on gaps between sums of two squares, which has been studied extensively. See the recent arXiv preprint arxiv.org/pdf/1712.07243.pdf, and see also my response to your previous question mathoverflow.net/questions/289387/… to make the connection. $\endgroup$
    – GH from MO
    Commented Dec 27, 2017 at 21:15
  • $\begingroup$ @GHfromMO If $t=b^2+1$ then at $x=0$ and $x=1$ we have perfect squares and so is $g(b^2+1)=1$. Conversely if $g(a)=1$ then does $a=b^2+1$ hold? By this I mean can there be other $x$ and $x+1$ for some $t$ such that both $x(t-x)$ and $(x+1)(t-(x+1))$ are perfect squares? I think is impossible. $\endgroup$
    – Turbo
    Commented Dec 28, 2017 at 7:33

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Is $g(a)=1\iff a=b^2+1$ (corresponding to $x=0$ and $a-x=b^2+1$ or $x=1$ and $a-x=b^2$)

No. It may happen, that, say, $x,a-x$ are perfect squares and $x+1$, $a-x-1$ are twice perfect squares. These Pell type equations have infinitely many solutions, the smallest is $x=1,a-x=9$.

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  • $\begingroup$ So the behavior is not nice... but at least does $x$ have to be odd? Unlikely I think but a counter example would be illustrative. $\endgroup$
    – Turbo
    Commented Dec 28, 2017 at 8:57
  • $\begingroup$ Eg: $x=64$, $a-x=m^2\cdot 13\cdot 5+1=t^2$ and $x+1=65$, $a-x-1 =m^2\cdot 13\cdot 5$? May be some $m$ and $t$ works? $\endgroup$
    – Turbo
    Commented Dec 28, 2017 at 9:06
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    $\begingroup$ no, it may happen that $x$ is even and $x,a-x$ are perfect squares; $x+1,a-x-1$ are 5 times perfect squares. Say, $x=4$, $a-x=81$. $\endgroup$
    – Fedor Petrov
    Commented Dec 28, 2017 at 10:20

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