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I have a polynomial of the second degree $a\cdot n^2 + b \cdot n + c$ and I need to find out natural numbers $n$, such that $\sqrt{a\cdot n^2 + b \cdot n + c}$ is also a natural number.

After thinking about this problem, my idea was to rewrite it into $a\cdot n^2 + b \cdot n + c = x^2$ and rewrite it to look like a Pell's equation:

$$(2an + b)^2 - 4ax^2= b^2 - 4 ac$$ which kind of resembles $x^2 - n y ^2 = 1$, but not really close enough to for me to solve it.

Then I tried to find some similarities in solutions for particular cases. For example when I took the equation $\sqrt{3\cdot n^2 - 2 \cdot n - 1}$ and wrote the program, the couple of first values were: 1, 5, 65, 901, 12545 (with no visible pattern for me).


So how should I solve this problem?

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  • $\begingroup$ Do you really want to have a general solution that works for all values of $a,b,c$? If you are eventually only interested in some special cases, it would be easier to tackle those. $\endgroup$ – Joonas Ilmavirta Jul 15 '15 at 7:51
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    $\begingroup$ It's difficult to take your question seriously with that as your profile picture. $\endgroup$ – user230734 Jul 15 '15 at 7:53
  • $\begingroup$ @JoonasIlmavirta if the general solution is too hard, you can show me how to come up with a solution for $3n^2 - 2n - 1$ $\endgroup$ – Salvador Dali Jul 15 '15 at 8:09
  • $\begingroup$ If $c=k^2$ it is easy to rewrite in another Pell equation. In another case decided the same view through the Pell equation. $\endgroup$ – individ Jul 15 '15 at 8:41
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If you wish to have an infinite number of integer solutions to,

$$an^2+bn+c = d^2\tag1$$

but not necessarily all of them, then one way is, yes, to solve a Pell equation. First, as pointed out in the other answer, you need a initial solution. Second, if you limit it to only integers, then $a$ must not be a square.

Given an initial $n,d$ to $(1)$, then an infinite more can be found as,

$$ax^2+bx+c = (-d+py)^2$$

where,

$$x = n+qy$$

$$y = 2dp+(b+2an)q$$

and $p,q$ solve the Pell equation $p^2-aq^2 = 1$.

Example:

$$3n^2-2n-1 = d^2$$

with initial $n,d = 5,8$, you get,

$$3x^2-2x-1 = (-8+py)^2$$

where,

$$x = 5+qy$$

$$y = 16p+28q$$

and $p,q$ is any solution to $p^2-3q^2=1$.

Note: This is an easy method to get an infinite number of solutions, but not all of them.

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  • $\begingroup$ Any idea how can I find initial solution? $\endgroup$ – Salvador Dali Jul 15 '15 at 18:00
  • $\begingroup$ @SalvadorDali: You can use mod arguments like in this answer. Or, since it is the smallest , then a simple program can quickly establish if it has a solution within a bound, say $|n|<1000$. If there is, the method above then guarantees there are an infinite more. $\endgroup$ – Tito Piezas III Jul 16 '15 at 2:49
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With slight change of notation, you are looking for, given $a,b,c\in \mathbf{Z}$, points with integer co-ordinates on the curve $ax^2-y^2+bx+c=0$.

This is a conic. If you know one solution exists, then we can find infinitely many through rational parametrization. This is well-known (for example can be found in Silverman-Tate). Call this one solution point $Q_0$

Now fix a line $L$ whose equations has integer/rational coefficients. For a variable point $P\in L$ with RATONAL co-ordinates, write the equation of the line connecting $P$ with $Q_0$; as the intersection of a conic with line has 2 points you get the other point $P'$. This will be a rational point.(Convince yourself). So you get solutions you want.

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