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

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    @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.2012-06-26

2 Answers 2

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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.

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    I like (+1) brute force work combined with mathematical intuition...2012-06-25
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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.

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    @AgainstASicilian: I'll elaborate more on my answer tomorrow. I have to attend to something first.2012-06-26