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Consider the diophantine

$$ x^2 + t(n) \space y^2 = n $$

Where $x,y> 0$, $ n>1$ and $t(n) > -1$.

For a given $n$ we want to find the smallest possible integer $t(n)$.

Clearly when $n$ is a square then $t(n) = 0 $. When $ n = a^2 \space b $ then $ t(n)$ is at most $t(b)$. If $ n $ is a prime 1 mod 4 then $t(n) = 1 $. Also $ t(n) < n$ is trivial as $1^2+(n-1)1^2=n$

Clearly, primes of type $s \mod d$ with prime $d$ and $s > 0$ are related. And trivially Pell’s equation, and hence also continued fractions are related.

I have seen a lot of related things but I have not seen $t(n)$ discussed.

What is known about $t(n) $ ??

The sequence starts like

$$ t(2) = 1, t(3) = 2 , t(4) = 0 , t(5) = 1 , t(6) = 2 , t(7) = 3 , t(8) = 1 , t(9) = 0 , ... $$

I could not find it on the OEIS.

mick
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    Related: I've written a little bit about the number of solutions to $x^2+ay^2=n$. https://math.stackexchange.com/questions/2834085/how-many-integer-pairs-satisfy-the-ellipse-x2ay2-r/2836230#2836230 – Mason Jul 08 '18 at 19:24
  • Can you write out the first few $t(n)$s? – Mason Jul 08 '18 at 21:05
  • $ t(2) = 1, t(3) = 2 , t(4) = 0 , t(5) = 1 , t(6) = 2 , t(7) = 3 , t(8) = 1 , t(9) = 0 , ... $ – mick Jul 08 '18 at 21:31
  • I considered using theta functions but it did not work out. – mick Jul 08 '18 at 22:49
  • We do not have a Pell-type equation here because such an equation must have the form $x^2-dy^2=n$ with a positive integer $d$ – Peter Jul 09 '18 at 05:34
  • If $n$ is not a perfect square, than there is a unique positive integer $k$ with $k^2<n<(k+1)^2$ giving $0<n-k^2<2k+1$ , which gives the bound $t(n)<2k+1<2\sqrt{n}+1$ – Peter Jul 09 '18 at 06:21

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