Solving Pell's equation $x^2-5y^2=\pm4$ using elementary methods.

Question

Solve Pell's equation $x^2-5y^2=\pm4$.

This equation arises when I tried to prove that the units of $\mathbb{Z}[\varphi]$, where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio, are of the form $\pm\varphi^n$. I found that $x+\varphi y$ is a unit iff $x^2+xy-y^2=\pm1$, i.e. $(2x+y)^2-5y^2=\pm4$. Yet I am unable to solve this equation. I saw here a solution using algebraic number theory, but I am interested in how to solve this equation using elementary methods, without using results from algebraic number theory. Thanks in advance!

I put the full diagram at http://math.stackexchange.com/questions/512621/finding-all-solutions-of-the-pell-type-equation-x2-5y2-4/512649#512649 I typed in the $-4$ and you can read off the $+4$ — Will Jagy, Jun 03 '17 at 03:23
It seems that your solution uses too many results with which I'm not familiar... Is there any solution using, say, continued fractions? — Yuxiao Xie, Jun 03 '17 at 03:31
If you want an "elementory" method, maybe continued fractions? Calculate your partial quotients, $a_n$ and the corresponding values of $p_n$ and $q_n$(I can't remember if these quantities had names) and then compute $p_n^2-Nq_n^2$. Eventually the $p_n^2-Nq_n^2$ becomes periodic so you just want to see if there are any values of $p_n$ and $q_n$ that gives you $\pm 1,\pm 4$ — daruma, Jun 03 '17 at 03:33
In general there are no solutions or there are infinitely many solutions. — daruma, Jun 03 '17 at 03:36
@daruma Could you please elaborate? I'm unsure what to do... — Yuxiao Xie, Jun 03 '17 at 03:58

score 4 · Accepted Answer · answered Jun 03 '17 at 04:36

4

Let's take a solution $(x,y)$ of $x^2-5y^2=\pm4$. Assume $x>0$ and $y>0$. Clearly also $x$ and $y$ have the same parity. Define $$x'=\frac{5y-x}2,\qquad y'=\frac{x-y}2.$$ Then $x'$ and $y'$ are integers, and $$x'^2-5y'^2=\frac{(5y-x)^2-5(x-y)^2}4=\frac{20y^2-4x^2}4=\pm4.$$ Therefore $(x',y')$ is also a solution. I claim that $y'\ge0$ and $x'>0$.

If $y'<0$ then $x<y$ and $x^2-5y^2<-4x^2<-4$, which is false. So $y'\ge0$. If $x'\le0$ then $x\ge5y$ and $x^2-5y^2\ge20y^2>4$ which is false. So $y'\ge0$.

I claim that as long as $y\ge2$, then $y'<y$. Otherwise, $x\ge3y$ and $\pm 4=x^2-5y^2\ge4y^2$. This is only possible if $y=1$.

So, iterating the operation $(x,y)\mapsto(x',y')$ eventually reduces $(x,y)$ to a solution $(X,Y)$ with $X>0$ and $Y\in\{0,1\}$. Therefore to $(X,Y)=(2,0)$, $(1,1)$ or $(3,1)$. All of these reduce down to $(2,0)$.

Therefore we can start with $(x_0,y_0)=(2,0)$ and reversing the operation generate all positive solutions. The iterative process is $$(x_{n+1},y_{n+1})=\left(\frac{x+5y}2,\frac{x+y}2\right).$$

answered Jun 03 '17 at 04:36

Angina Seng

158,341

Wow, that was brilliant! Is this a method that could be used to solve other Pell equations or is it just designed for this specific problem? – Yuxiao Xie Jun 03 '17 at 04:42
@YuxiaoXie It is just a translation of the algebraic number theory method into elementary language, knowing that the fundamental unit is $\frac12(1+sqrt5)$. – Angina Seng Jun 03 '17 at 04:43
How did you get $(5y-x)/2$ and $(x-y)/2$ then? Through algebraic number theory? – Yuxiao Xie Jun 03 '17 at 04:50
@YuxiaoXie Just dividing $x+y\sqrt5$ by $\frac12(1+\sqrt5)$. – Angina Seng Jun 03 '17 at 04:51
Though this may lead to another question, I feel compelled to ask, is it possible to generalize this to other Pell equations of the form $x^2-dy^2=k$ and how? – Yuxiao Xie Jun 03 '17 at 04:55
@YuxiaoXie Yes, find the fundamental unit! – Angina Seng Jun 03 '17 at 04:56
Awesome approach. – BAI Jun 04 '17 at 03:01

Will Jagy · Answer 2 · 2017-06-04T00:57:16.420

A little better format. Given any solution to $u^2 - 5 v^2 = \pm 4$ with large positive $u,v,$ a smaller solution can be constructed using $$ (u,v) \mapsto (9 u - 20 v, \; -4 u + 9 v). $$ As a result, we get a finite number of "seed" solutions, those being $u,v,\geq 0$ with either $9u - 20 v < 0$ or $-4u + 9v < 0.$ The set of such seeds is guaranteed finite; with large positive values and $u^2 - 5 v^2 = \pm 4,$ we have $u/v \approx \sqrt 5 \approx 2.236,$ arbitrarily close. The seed pairs have either $u/v < 2.22222$ or $u/v > 2.250.$

Put it together, all solutions are given by $u$ a Lucas number and $v$ Fibonacci.

======================================

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental_plus_minus
  automorphism matrix:  
    9   20
    4   9
   backwards:  
    9   -20
    -4   9

  9^2 - 5 4^2 = 1

 u^2 - 5 v^2 = +- 4

Sat Jun  3 11:46:39 PDT 2017

u:  2  v:  0  SEED   KEEP +- 
u:  1  v:  1  SEED   BACK ONE STEP  -11 ,  5
u:  3  v:  1  SEED   KEEP +- 
u:  4  v:  2  SEED   BACK ONE STEP  -4 ,  2
u:  7  v:  3  SEED   BACK ONE STEP  3 ,  -1
u:  11  v:  5  SEED   BACK ONE STEP  -1 ,  1
u:  18  v:  8
u:  29  v:  13
u:  47  v:  21
u:  76  v:  34
u:  123  v:  55
u:  199  v:  89
u:  322  v:  144
u:  521  v:  233
u:  843  v:  377
u:  1364  v:  610
u:  2207  v:  987
u:  3571  v:  1597
u:  5778  v:  2584
u:  9349  v:  4181
u:  15127  v:  6765
u:  24476  v:  10946
u:  39603  v:  17711
u:  64079  v:  28657
u:  103682  v:  46368
u:  167761  v:  75025
u:  271443  v:  121393
u:  439204  v:  196418
u:  710647  v:  317811
u:  1149851  v:  514229
u:  1860498  v:  832040
u:  3010349  v:  1346269
u:  4870847  v:  2178309
u:  7881196  v:  3524578
u:  12752043  v:  5702887

Sat Jun  3 11:46:59 PDT 2017

 u^2 - 5 v^2 = 4

jagy@phobeusjunior:~$

=======================================

Same problem with $13$ rather than $5.$ Instead of $u_{n+2} = u_{n+1} + u_n,$ this time we get $u_{n+2} = 3 u_{n+1} + u_n.$

=====================================

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental_plus_minus
  automorphism matrix:  
    649   2340
    180   649
   backwards:  
    649   -2340
    -180   649

  649^2 - 13 180^2 = 1

 u^2 - 13 v^2 = +- 4

Sat Jun  3 17:46:55 PDT 2017

u:  2  v:  0  SEED   KEEP +- 
u:  3  v:  1  SEED   BACK ONE STEP  -393 ,  109
u:  11  v:  3  SEED   KEEP +- 
u:  36  v:  10  SEED   BACK ONE STEP  -36 ,  10
u:  119  v:  33  SEED   BACK ONE STEP  11 ,  -3
u:  393  v:  109  SEED   BACK ONE STEP  -3 ,  1
u:  1298  v:  360
u:  4287  v:  1189
u:  14159  v:  3927
u:  46764  v:  12970
u:  154451  v:  42837
u:  510117  v:  141481
u:  1684802  v:  467280
u:  5564523  v:  1543321
u:  18378371  v:  5097243

Sat Jun  3 17:47:25 PDT 2017

 u^2 - 13 v^2 = +- 4

jagy@phobeusjunior:~$

=====================================

This time 29

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental_plus_minus
  automorphism matrix:  
    9801   52780
    1820   9801
   backwards:  
    9801   -52780
    -1820   9801

  9801^2 - 29 1820^2 = 1

 u^2 - 29 v^2 = +- 4

Sat Jun  3 17:54:38 PDT 2017

u:  2  v:  0  SEED   KEEP +- 
u:  5  v:  1  SEED   BACK ONE STEP  -3775 ,  701
u:  27  v:  5  SEED   KEEP +- 
u:  140  v:  26  SEED   BACK ONE STEP  -140 ,  26
u:  727  v:  135  SEED   BACK ONE STEP  27 ,  -5
u:  3775  v:  701  SEED   BACK ONE STEP  -5 ,  1
u:  19602  v:  3640
u:  101785  v:  18901
u:  528527  v:  98145
u:  2744420  v:  509626
u:  14250627  v:  2646275

Sat Jun  3 17:55:38 PDT 2017

 u^2 - 29 v^2 = +- 4

jagy@phobeusjunior:~$

==============================

@YuxiaoXie revised the answer. Comes out Fibonacci and Lucas; Also, about as elementary as is possible. — Will Jagy, Jun 03 '17 at 19:47

score 0 · Answer 3 · answered Jun 04 '17 at 07:41

There is a good story of solutions in Wikipedia. For the solution of Pell's

like equation x^2−Dy^2=±C^2. You can always use following simple algorithm:

A:

x^2−5y^2 = 4

suppose x = y + a then we have:

y^2+2ay+a^2 -5y^2= 4 or 4y^2-2ay-a^2+4=0

y= [a±(5a^2-16)^0.5]/4

with a= 10 we get y = 8 and x = 10 + 8 = 18

I guess there is more solutions in N for this equation.

with a=0 we get complex solutions x = y = ±i and with any a other than these

two we may get real solutions but it must be checked in the equation.

B:

For equation x^2−5y^2 = -4 assuming again x = y+a we get:

4y^2 - 2ay -a^2-4=0

y = [a±(5a^2+16)^0.5]/4

with a=2 we get y =2 and y=-1 ; if y=-1 then x=±1 these are solutions in R. There may be infinitely many solutions in R .

With a=6 we get y=5 and x = 6+5=11. More solutions in N is probable.

with a=0 we get y = x = ±1

Generally there is an algorithm that says there can be infinitely many Pell's like equations in which triples x, D and y are functions of a parameter like m.Now I solve equation x^2−5y^2 = 4 with this algorithm:

Suppose x=m^2 +2 we have:

m^2(m^2+4) = 5y^2

if y^2=m^2 then y = ±m and also:

m^2 +4 = 5 ; m=±1 which results x = 3

and triples (x, D, y)=(m^2+2, 5m, ±m) for m=-1 and

(x, D, y) = (m^2+2, -5m, ±m) for m=1.

Now we can make more Pell's like equations; for example with m =2 we have:

D = 5 . 2 =10 that gives x^2 -10y^2 = 16 which it's solutions are x=2^2 +2 =6

and y=±2.

score 0 · Answer 4 · answered Jan 09 '23 at 22:21

For $x^2-5y^2=-4$ we have a base solution at $(x,y) = (1,1)$ and the unit solution for $x^2-5y^2=\pm1$ is $(x,y) = (2,1)$. So you can generate infinitely many solutions with: \begin{equation*} x^2-5y^2 = (1-\sqrt{5})(1+\sqrt{5})(2-\sqrt{5})^n(2+\sqrt{5})^n = (-4)(-1)^n \end{equation*} With $n=1$ that gives $(1+\sqrt{5})(2-\sqrt{5}) = -3 +\sqrt{5}, (1-\sqrt{5})(2+\sqrt{5}) = -3 -\sqrt{5}$ and we thus have $3^2-5(1^2) = 4$. We also have another base solution for $x^2-5y^2=4$ with $(x,y) = (\pm2, 0)$ which generates another family of solutions: \begin{equation*} x^2-5y^2 = 4(2-\sqrt{5})^n(2+\sqrt{5})^n = (4)(-1)^n \end{equation*}

Solving Pell's equation $x^2-5y^2=\pm4$ using elementary methods.

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