Math Forum / Mathematics / Undergraduate Math / May 2009

    Insults? What insults? Over the many years in Usenet you have
only written crap. That's not an insult: It's a fact.

    Also, you have been unable to learn from your many mistakes -
another fact.

    I would pity you if you were not so stupid.

§8. On The Equation x2-Dy2=z2

We shall now develop a general theory by means of which the solutions of the
equation

may be found. Naturally D is assumed to be an integer. Without loss of
generality it may be taken positive; for if it were negative the equation
might be written in the form s2 - (- D)y2 = x2, where - D is positive. If D
is the square of an integer, say, D = d2, the equation may be written

x2 = z2+(dy)2,

so that the theory becomes essentially that of the Pythagorean equation
x2+y2 = z2. Accordingly, we shall suppose that D is not a square.

By suitably specializing Eq. (2) of the preceding section we readily obtain
the following two-parameter solution of (1) :

x = m2+Dn2, y = 2mn, z = m2-Dn2.

But there is no ready means for determining whether this is the general
solution. Consequently we shall approach from another direction the problem
of finding the solution of (1).

We shall first show that Eq. (1) possesses a non-trivial solution for which
z - i; that is, we shall prove the existence of a solution of the equation

x2-Dy2 = i. (2)

different from the trivial solutions x = ±i, y=o.

For this purpose we shall first show that integers u, v exist such that the
absolute value * of the (positive or negative) real quantity u-vvD is less
than i/v and also less than any pre- assigned positive constant e. (By VD we
mean the positive square root of D.) Let / be an integer such that te> i.
Now give to v successively the integral values from o to t and in each case
choose for u the least integral value greater than

* By the absolute value of A is meant A itself when A is positive and - A
when A is negative. We denote it by 'A'.

In each case the quantity u-v^/D lies between o and i and in no two cases
are its values equal. If we divide the interval from o to i into t
subintervals, each of length i/t, then two_pf the above values of u-vVD, say
u'-v'VD and u" - v"VD, must lie in the same interval. Then the expression

(u'-u")-(v'-v")VD

is different from zero, is of the form u-vVD and has an absolute value less
than i/t and hence less than e. That this absolute value is less than that
of i/(v'-v") follows from the fact that the difference of v' and v" is not
greater than t. This completes the proof of the above statement concerning
the existence of u, v with the assigned properties.

From the existence of one such set of integers u, v it follows readily that
there is an infinite number of such sets. For, let u, v be one such set. Let
ei be a positive constant less than u-i)VL>|. Then integers ui, vi can be
determined such that ui-viVD is in absolute value less than i/vi, and also
less than «i. It is then less than e. Thus we have a second set ui, vi
satisfying the original conditions. Then, letting e2 be a positive constant
less than ui - viV7)l, we may proceed as before to find a third set u2, v2
with the required properties. It is obvious that this process may be
continued indefinitely and that we are thus led to an infinite number of
sets of integers u, v such that u-v\/D is in absolute value less than ? and
also less than the absolute value of i/v.

Now let u and v be a pair of integers determined as above. Then we have

2vV7)l<

v

Hence

r

i v

so that

I + 2VD.

+ |*'V5|},

Since u2-Dv2l is less than i + 2VZ) for every one of the infinite number of
sets u, v in consideration, and since its value is always integral, it
follows . that an integer / exists such that

u2-Dv2 = l

for an infinite number of sets of values u, -u. It is then obvious that
there is an infinite number of these pairs Mi, »r, u2, V2; Ms, vz; ., such
that ut - Uj and vt - vj are both divisible by / for every i and j. Let u',
v'; u", v" be two pairs belonging to this last infinite subset and chosen so
that M"^±m' and z/'^dbo'. It is obvious that this choice is possible. From
the equations

u'2-Dv'2 = l, u"2-Dv"2=l,

we have (by Formula (2) of § 7) :

(u'u"-Dv'v")2-D(u'v"-u"v')2=l2.

Here we take u' =m, u" = p, v' = n, v" = -q, D= -b, in applying the formula
referred to. Setting

u'u" -Dv'v" u'v"-u"v' f .

*- - -, y=- - - , (3)

we have

1. (4)

It remains to show that the values of x and y in (3) are integers. On
account of (4) it is obviously sufficient to show that y is an integer. That
y is an integer follows at once from the equations .u'=u"+1Jil, v" = v'+vl,
by multiplication member by member. We show further that y^o. If we suppose
that y = o, we have

u'v"-u"v' = o, u'u"-Dv'v" = ±l.

These equations are satisfied only if u" = ±u', v" = ±z/, relations which
are contrary to the hypothesis concerning u', u", v', v".

We have thus established the fact that Eq. (2) has at least one integral
solution which is not trivial. Since we may associate with any solution x, y
of (2) the other solutions - x, y; -x, -y; x, -y; it is clear that there is
at least one solution of (2) in which x and y are positive.

Let xi, yi, and X2, y2 be any solutions of (2), whether the same or
different. Then we have

i = (xi2-Dyi2)(x22-Dy22) = (xix2+Dyiy2)2-D(xiy2+x2yi)2,

so that Xix2+Dyiy2 and Xiy2+x2yi afford a solution of (2). Hence from the
solution x, y, whose existence has already been proved, we have a second
solution x2-{-Dy2, 2xy. It is easy to show that this process may be
continued and that it will lead to an infinite number of solutions of (2).
But this problem is a special case of one to be treated presently; and hence
will not be further pursued now.

In order to come upon the more general problem let us seek solutions of Eq.
(1) in which z shall have the positive value <r; that is, let us seek
solutions of the equation

x2-Dy2 = a2. (5)

If x=xi, y = yi is a positive solution of Eq. (2) then it is clear that x -
axi, y = ayi is a positive solution of (5). Hence from what precedes we have
at least two positive solutions of (5).

Now let # = /i, y=ui; x = t2, y = u2 be any two solutions of Eq. (5) and
write

t2+u2VD t+uV~D , .

= , -, (6)

where t and u are rational numbers. Then

(7)

<r From (6) we have

ti-uiVD t2-uoV~D t- ,-,

= . (Q)

a ff a

Multiplying Eqs. (6) and (8) member by member and making use of the
relations

ti2-Dui2 = a2, t22-Du22 = <r2, (9)

we have

p-Du2 = o2. (10)

Hence x = t, y = u afford a rational solution of (5), / and u having the
values given in (7).

We shall now point out two cases in which this solution is integral.

Suppose that a2 is a factor of D. Then from (9) it follows that o- is a
factor of both ti and h and hence from (7) that u is an integer. Then from
(10) it follows that t is an integer.

Suppose that *

4D = a2 mod 40^;

that is, that <r2 is a remainder obtained on dividing 4D by 4<r2. Then a is
evidently an even number. Write er=2p. Then we have D=p2mod4p2. Hence D is
divisible by p2. Then from (9) it follows that both ti and t2 are divisible
by p, since <r = 2p. Put

D=dp2, ti = 8ip, t2 = 82p.

Then d is odd. Moreover, the following relations exist, as we see from (9)
and (7) :

From Eqs. (u) we see that 6i and Mi are both odd or both even, and also that
62 and uz are both odd or both even. Then from (12) it follows that u is an
integer and hence from (10) that t is an integer.

We are now in position to prove readily the following theorem :

Let D be any positive non-square integer and let a be any positive integer
such that D=o mod a2 or 4D=a2 mod 4<r2. Let x = h and y = uibe the least
positive integral solution of the equation

x2-Dy2 = a2. (5bis)

Then all the positive integral solutions f of this equation are contained in
the set

x = tn, y = un, « = i, 2, 3, . . .,

Did you READ any of this JSH ?  If so try the books, if not, "trolls die
horrible deaths"

Cite a reference where someone has lied about this.

Marcus.

On May 15, 6:08 pm, juandiego <sttscitr...@tesco.net> wrote:

The real ones have.  I said so myself, noting that it was probably
known to Fermat and Euler.

The result is trivial.

<snip math>

You lie all the time, you're projecting again.

You always lie, it is easy for you.

You have no idea how to do Mathematics, any of it.

This is you in spades;
http://en.wikipedia.org/wiki/Psychological_projection

You always make mistakes in your post, cause you like to get *screwed* like
stupid bitch whore.

Work an example.

I'll work one with what I show:

2^2 - 5*1^2 = -1, so x = 2*2^2 + 1 = 9, and 9^2 - 5*4^2 = 1.

I've repeatedly said it is trivial.

Now then, work an example, with what you gave.  I dare you.

James Harris

Agreed.  I was pointing out why the method you give here is inferior
to the continued fraction method, which does work for cases like D =
7.  Given that a superior method exists it is not surprising that the
inferior method gets less mention in the textbooks.

Agreed.

You are obviously reading this thread.  IIRC there are three
references to the literature in this thread from 630 CE to 2009 CE.
References to this method are less common than references to the
continued fraction method but they do exist.

rossum