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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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Related: On the equation $(a^2+1)(b^2+1)=c^2+1$. –  Martin Sleziak Jun 25 '12 at 14:31
    
Are there general methods? –  MyNameIsKhan 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? –  Gerry Myerson 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. –  MyNameIsKhan Jun 26 '12 at 13:05

2 Answers 2

up vote 4 down vote accepted

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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How do you find the others? What method is this? –  MyNameIsKhan 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. –  Peter Phipps Jun 25 '12 at 18:15
    
I like (+1) brute force work combined with mathematical intuition... –  draks ... 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}, with {a, b, c, d} chosen such that one term on the RHS is equal to unity. An easier (but not complete) solution to (3) is,

$$\tag{5}(p^2q+p-q)^2+(2pq+1)^2=(p^2q+p+q)^2+1$$

Equating the LHS of (2) and (5), and solving for {x, y} should not be too difficult.

EDITED LATER:

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

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

Equate the LHS of (2) and (6) and we find that,

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

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

To solve this form,

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

one simply chooses,

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

for any {u, v}. Of course, since you want integer n, then all you have to do is solve the Pell equation,

$$u^2-av^2 = \pm 1$$

In this case, since a = 3, then solutions to $u^2-3v^2 = 1$ are {2,1}, {7,4}, {26,15}, etc. I trust you can take it from here.

P.S. It is quite easy to find other solutions similar to (6), and appropriate ones would need other Pell equations. For example, try,

$$\tag{7}(70n+6)^2+(70n^2+12n-17)^2=(70n^2+12n+18)^2+1$$

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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
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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 a*c-b*d. 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

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