Find integers solutions for which bivariate polynomial with bi-quadratic form: $4x^2y^2-4xy^2+1$, becomes a square number

Question

Could you help me to find all integer $x$ and $y$ for which the bivariate polynomial:

$$4x^2y^2-4xy^2+1$$

is a square number, i.e., it can be expressed as $z^2$ for some integer $z$? From the above, one can see that $z$ is an odd number.

I heuristically found a solution $x=2,y=6$ for which it becomes $17^2$. However, I am seeking a systemic way to find all the solutions, or a specialized way that can solve my specific problem.

Note that the above polynomial is bi-quadratic, that is, by fixing each variable, it becomes a univariate quadratic in the other one.

Answer 1

It is closed form if you know how to deal with the traditional Pell equation; a useful topic that is part of that story is (simple) continued fractions. With your letters, restrict $x \geq 2.$ Then give names $$W = (2x-1)^2 - 1 \; , \; \; \; k = 4x - 2$$ you are asking for positive $y$ in $$ z^2 - W y^2 = 1 $$ Worth learning how this works; the $y$ values are $$ y: \; \; 0, 1, k, k^2 - 1, k^3 -2k, .... $$ which gives a simple linear recurrence when we index the $y's,$ $$ y_{j+2} = k y_{j+1} - y_j $$ The $z$ values have the same recurrence $ z_{j+2} = k z_{j+1} - z_j $ However, the starting values are a little different, so we get $$ z: \; \; 1, \frac{k}{2}, \frac{k^2}{2} - 1, \frac{k^3}{2} - 1, .... $$

The relation between the $y$ values and the $z$ values is analogous to that between the Fibonacci and Lucas numbers.

You can also write a Binet type formula by introducing a quantity $\lambda$ that satisfies $\lambda^2 - k \lambda + 1 = 0,$ or $$ \lambda = \frac{k + \sqrt{k^2-4}}{2} $$

Oh, I thought negative $x$ would need work, but $x$ and $1-x$ get the same sequence of $y$ values.

I did a raw search, your $2 \leq x \leq 5$ and $y \geq 0$ up to fairly large.

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Sat 13 Jan 2024 09:20:21 AM PST

  x: 2  y: 0  z:  1  
  x: 2  y: 1  z:  3  
  x: 2  y: 6  z:  17  
  x: 2  y: 35  z:  99  
  x: 2  y: 204  z:  577  
  x: 2  y: 1189  z:  3363  
  x: 2  y: 6930  z:  19601  
  x: 2  y: 40391  z:  114243  
  x: 2  y: 235416  z:  665857  

  x: 3  y: 0  z:  1  
  x: 3  y: 1  z:  5  
  x: 3  y: 10  z:  49  
  x: 3  y: 99  z:  485  
  x: 3  y: 980  z:  4801  
  x: 3  y: 9701  z:  47525  
  x: 3  y: 96030  z:  470449  
  x: 3  y: 950599  z:  4656965  

  x: 4  y: 0  z:  1  
  x: 4  y: 1  z:  7  
  x: 4  y: 14  z:  97  
  x: 4  y: 195  z:  1351  
  x: 4  y: 2716  z:  18817  
  x: 4  y: 37829  z:  262087  
  x: 4  y: 526890  z:  3650401  
  x: 4  y: 7338631  z:  50843527  

  x: 5  y: 0  z:  1  
  x: 5  y: 1  z:  9  
  x: 5  y: 18  z:  161  
  x: 5  y: 323  z:  2889  
  x: 5  y: 5796  z:  51841  
  x: 5  y: 104005  z:  930249  
  x: 5  y: 1866294  z:  16692641  
  x: 5  y: 33489287  z:  299537289  

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Answer 2

$$4x^2y^2-4xy^2=a^2-1$$ $$4y^2 \cdot(x^2-x) = a^2-1$$

As $a$ is prime, you can mostly write it as $a=2b+1$, or:

$$4y^2 \cdot(x^2-x) = 4b^2+4b$$

Or: $$y^2 \cdot [x \cdot (x-1)] = b \cdot (b+1)$$

So, set $y=1$ and $x=b+1$ and should get something.

E.g. for $a=17$:$ b=8$, set $y=1$ and $x=9$.

(This is not the general answer, but it gives a good idea.)