6

Consider the following equation with integral, nonzero $x,y,z$

$$(4x^2+1)(4y^2+1) = (4z^2+1)$$

What are some general strategies to find solutions to this Diophantine?

If it helps, this can also be rewritten as $z^2 = x^2(4y^2+1) + y^2$

I've already looked at On the equation $(a^2+1)(b^2+1)=c^2+1$

Community
  • 1
MyNameIsKhan
  • 908

2 Answers2

7

Let $a$ be a positive integer.
Then

\begin{align} (4a^2+1)(4((2a)^2)^2+1) &= 256a^6 + 64a^2 + 4a^2 + 1 \\ & = 4(64a^6 + 16a^4 + a^2) + 1 \\ &= 4(a^2(8a^2+1)^2)+1 \\ &= 4((8a^2+1)a)^2+1 \end{align}

so $(a, (2a)^2, (8a^2+1)a)$ is always a solution.

There are others as well.

Peter Phipps
  • 3,065
5

Here is one general approach. Since the product of the sum of two squares is itself the sum of two squares, then,

$$\tag{1}(4x^2+1)(4y^2+1) = 4z^2+1$$

is equivalent to,

$$\tag{2}(2x+2y)^2+(4xy-1)^2 = 4z^2+1$$

The complete solution to the form,

$$\tag{3}x_1^2+x_2^2 = y_1^2+y_2^2$$

is given by the identity,

$$\tag{4}(ac+bd)^2 + (bc-ad)^2 = (ac-bd)^2+(bc+ad)^2$$

One can then equate the terms of (2) and (4), solve for {x, y, z}, with {a, b, c, d} chosen such that one term on the RHS is equal to unity.

EDITED MUCH LATER:

In response to your questions, let's have a simpler solution to (3) as,

$$\tag{5}(6n+2)^2+(6n^2+4n-1)^2=(6n^2+4n+2)^2+1$$

Equate the terms of (2) and (5) and we find that,

$$x = \frac{1}{2}\big(1+3n-\sqrt{3n^2+2n+1}\big)$$

$$y = \frac{1}{2}\big(1+3n+\sqrt{3n^2+2n+1}\big)$$

$$z = (6n^2+4n+2)/2$$

To get rid of the $\sqrt{N}$ and solve the form,

$$an^2+bn+c^2 = \square$$

one simply chooses,

$$n = \frac{-2cuv+bv^2}{u^2-av^2}$$

for arbitrary {u, v}. Of course, since you want integer n, you have to solve the denominator as the Pell equation $u^2-av^2 = \pm 1$.

In summary, and after simplification, an infinite number of integer solutions to,

$$(4x^2+1)(4y^2+1) = 4z^2+1$$

is given by the rather simple,

$$x = (u-3v)(u-v)$$

$$y = 2uv$$

$$z = (u^2-2uv+3v^2)^2$$

where,

$$u^2-3v^2=1$$

P.S. It is quite easy to find other solutions similar to (5), and appropriate ones would need other Pell equations.

Tito Piezas III
  • 54,565
  • Is it right, that one could exclude cases, where $4z^2+1$ is prime due to Thue's Lemma: A prime $p=4k+1;$ has a unique representation $p=a^2+b^2$ with $0<a<b;$? – draks ... Jun 25 '12 at 20:07
  • 1
    To be honest I don't understand this answer. Are you saying (bc+ad)^2 = 1? I don't know where the p's and q's are coming from – MyNameIsKhan Jun 25 '12 at 20:13
  • @AgainstASicilian, I think he's saying that $ac-bd$, like $5\cdot 1 - 2\cdot 1$. Then you get the other terms in $(4)$, resp. $(3)$ and so forth. +1 nice answer. – draks ... Jun 25 '12 at 20:44
  • @draks I understand that ac-bd is ac-bd. I don't understand though what he means about "unity" nor do I know what he means about solving for x and y through the LHS. If I equate the LHS of (2) and (5) in Wolfram, it's a mess. – MyNameIsKhan Jun 25 '12 at 20:49
  • @AgainstASicilian 1. one term on the RHS is equal to unity: Choose $ac-bd=1$. 2. Set $p^2q+p-q=2x+2y$ and asked Wolfram again and even get some examples. – draks ... Jun 25 '12 at 20:58
  • @AgainstASicilian BTW, Tito has a very cool blog: A Collection of Algebraic Identities. Worth a look... – draks ... Jun 25 '12 at 21:14
  • Even after changing everything to p's and q's, the formulas still involve square roots and all sorts of things that don't really simplify the problem any or allow for direct calculations of valid integral solutions – MyNameIsKhan Jun 25 '12 at 22:25
  • @AgainstASicilian: I'll elaborate more on my answer tomorrow. I have to attend to something first. – Tito Piezas III Jun 26 '12 at 02:35