# Finding solutions to $(4x^2+1)(4y^2+1) = (4z^2+1)$

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 $(a^2+1)(b^2+1)=c^2+1$

• Jun 25 '12 at 14:31
• Are there general methods? Jun 25 '12 at 15:34
• You say you have looked at that earlier question. Have you looked at the Kashihara paper cited there? What did it tell you about the question you are asking? Jun 26 '12 at 5:19
• @GerryMyerson I looked through the paper but had a hard time understanding it. I couldn't find any section that simply outlined how to find solutions. It looked like more of a proof with a lot of notation I didn't understand, and I couldn't find a single page that gave me something I could work with. Jun 26 '12 at 13:05

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.

• How do you find the others? What method is this? Jun 25 '12 at 18:08
• I found the first ones by writing program code and noticing the pattern, and then doing the algebra to confirm it. But I also found other solutions which I haven't spotted a pattern for yet. Jun 25 '12 at 18:15
• I like (+1) brute force work combined with mathematical intuition... Jun 25 '12 at 21:00

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.

• 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 $0<a<b\;$? Jun 25 '12 at 20:07
• 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 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. 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. 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. Jun 25 '12 at 20:58