The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$

Question

Proof that there exists infinity positive integers triple $x^2+y^2=z^2$ that $x,y$ are consecutive integers, then exhibit five of them.

This is a question in my number theory textbook, the given hint is that
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$"
I wondered how someone come up with this idea.

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My solution is letting $x=2st, y=s^2-t^2, z=s^2+t^2$ by $s>t, \gcd(s,t)=1$.then consider two cases: $y=x+1$ and $y=x-1$
Case 1: $y=x+1$
Gives me $(s-t)^2-2t^2=1$ then I found this is the form of Pell's equation, I then found $$\begin{align}s&=5,29,169,985,5741\\t&=2,12,20,408,2378\end{align}$$then yields five triples $$(20,21,29),(696,697,985),(23660,23661,33461),(803760,803761,1136689),(27304196,27304197,38613965)$$ Case 2:$y=x-1$
Using the same method, I come up with Pell's equation $(s+t)^2-2s^2=1$, after solve that I also get five triples: $$(4,3,5),(120,119,169),(4060,4059,5741),(137904,137903,195025),(4684660,4684659,6625109)$$

I have wondered why the gaps between my solution are quite big, with my curiosity, I start using question's hint and exhibit ten of the triples:$$(3,4,5),(20,21,29),(119,120,169),(696,697,985),(4059,4060,5741),(23660,23661,33461),(137903,137904,195025),(803760,803761,1136689),(4684659,4684660,6625109),(27304196,27304197,38613965)$$ These are actually the same as using solutions alternatively from both cases. But I don't know is this true after these ten triples

Basically the problem was solved, but I would glad to see if someone provide me a procedure to come up with the statement
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$", and prove that there are no missing triplet between it.

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--After edit--

Thanks to @Dr Peter McGowan !, by the matrix
$$ \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2\\ 2 & 2 & 3 \end{bmatrix} \begin{bmatrix} x\\x+1\\z \end{bmatrix} = \begin{bmatrix} 3x+2z+2\\3x+2z+1\\4x+3z+2 \end{bmatrix}$$ gives me the hinted statement.

Answer 1

You are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5) $ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a "vector" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1) $, we may calculate the next triple as follows:

$$\begin {align} x_2&=x_1+2y_1+2z_1 \\ y_2&=2x_1+y_1+2z_1 \\ z_2&=2x_1+2y_1+3z_1 \end {align} $$

The hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are:

$(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely.

The sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation.

Answer 2

$x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$.

Let $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\mathbb Z[\sqrt 2]$ of norm $-1$.

It is clear that $\omega=1+\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\omega$.

Thus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by

$$\begin{align} u_{k+1}+z_{k+1}\sqrt 2&=(u_k+z_k\sqrt 2)\omega^2 \\ &=(u_k+z_k\sqrt 2)(3+2\sqrt 2) \\ &=(3u_k+4z_k)+(2u_k+3z_k)\sqrt 2 \end{align}$$ So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$.

Now let $u_{k+1}=2x_{k+1}+1$.

Then $$x_{k+1}=\frac{u_{k+1}-1}{2}=\frac{(3u_k+4z_k)-1}{2}=\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed.

Answer 3

By the nature of the formulas below, we can see that there are infinite Pythagorean triples where $A-B=\pm1$.

We can generate these triples sequentially with a formula that starts with a seed of $A_0,B_0,C_0=0,0,1$: $$A_{n}=3A_{n-1}+2C_{n-1}+1\quad B_{n}=3A_{n-1}+2C_{n-1}+2\quad C_{n}=4A_{n-1}+3C_{n-1}+2$$

or we can generate the $(m,n)$s for Euclid's formula using Pell numbers to find $x^{th}$ triple directly:

$$m_x=\frac{(1+\sqrt{2})^{x+1}-(1-\sqrt{2})^{x+1}}{2\sqrt{2}}\quad n_x=\frac{(1+\sqrt{2})^{x}-(1-\sqrt{2})^{x}}{2\sqrt{2}}$$

$$A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$ $$f(2,1)=(3,4,5)$$ $$f(5,2)=(21,20,29)$$ $$f(12,5)=(119,120,169)$$ $$f(29,12)=(697,696,985)$$ $$f(29,12)=(4059,4060,5741)$$ $$f(169,70)=(23661,23660,33461)$$ $$f(408,169)=(137903,137904,195025)$$ $$f(985,408)=(803761,803760,1136689)$$ $$f(2378,985)=(4684659,4684660,6625109)$$ $$f(5741,2378)=(27304197,27304196,38613965)$$ $$f(13860,5741)=(159140519,159140520,225058681)$$ $$f(33461,13860)=(927538921,927538920,1311738121)$$ $$f(80782,33461)=(5406093003,5406093004,7645370045)$$ $$f(195025,80782)=(31509019101,31509019100,44560482149)$$ $$f(470832,195025)=(183648021599,183648021600,259717522849)$$ $$f(1136689,470832)=(1070379110497,1070379110496,1513744654945)$$ $$f(2744210,1136689)=(6238626641379,6238626641380,8822750406821)$$ $$f(6625109,2744210)=(36361380737781,36361380737780,51422757785981)$$ $$f(15994428,6625109)=(211929657785303,211929657785304,299713796309065)$$