38 It is required in the present case to determine the
values which are to be adopted for , in order that the
formula may become a real square; and,
consequently, that a rational root of it may be assigned.
Now, the letters , , and , represent given numbers; and
the determination of the unknown quantity depends chiefly
on the nature of these numbers; there being many cases in
which the solution becomes impossible. But even when it is
possible, we must content ourselves at first with being able to
assign rational values for the letter , without requiring those
values also to be integer numbers; as this latter condition
produces researches altogether peculiar.
39 We suppose here that the formula extends no farther
than the second power of ; the higher dimensions require
different methods, which will be explained in their proper
places.
We shall observe first, that if the second power were not
in the formula, and were = 0, the problem would be attended with no difficulty;
for if were the given formula, and it were required to determine ,
so that might be a square, we should only have to make ,
whence we should immediately obtain . Now,
whatever number we substitute here for , the value of would always
be such, that would be a square, and consequently,
would be a rational quantity.
40 We shall therefore begin with the formula ;
that is to say, we are to find such values of , that,
by adding unity to their squares, the sums may likewise be squares;
and as it is evident that those values of cannot be integers,
we must be satisfied with finding the fractions which express
them.
41 If we supposed , since must be a square, we should have
, and ;
so that in order to find we should have to seek numbers
for , whose squares, diminished by unity, would also leave
squares; and, consequently, we should be led to a question as
difficult as the former, without advancing a single step.
It is certain, however, that there are real fractions, which,
being substituted for , will make a square; of
which we may be satisfied from the following cases:
- If , we have ; and consequently .
- becomes a square likewise, if , which gives .
- If we make , we obtain , the square root of which is .
But it is required to show how to find these values of ,
and even all possible numbers of this kind.
42 There are two methods of doing this. The first requires us to make
;
from which supposition we have , where the square
destroys itself; so that we may express without a radical sign.
For, cancelling on both sides of the equation, we obtain ; whence we find
; a quantity in which we may substitute for any number
whatever less than unity.
Let us therefore suppose , and we have
and if we multiply both terms of this fraction by , we shall find .
43 In order, therefore, that may become a square,
we may take for and all possible integer numbers, and
in this manner find an infinite number of values for .
Also, if we make, in general, , we find,
by squaring,
or, by putting in the numerator,
a fraction which is really a square, and gives
We shall exhibit, according to this solution, some of the
least values of .
If = |
2, |
3, |
3, |
4, |
4, |
5, |
5, |
5, |
5, |
and = |
1, |
1, |
2, |
1, |
3, |
1, |
2, |
3, |
4, |
We have = |
ΒΎ, |
β΄ββ, |
β΅βββ, |
ΒΉβ΅ββ, |
β·βββ, |
ΒΉΒ²ββ
, |
Β²ΒΉβββ, |
βΈβββ
, |
βΉβββ. |
44 We have, therefore, in general,
and, if we multiply this equation by , we find
so that we know, in a general manner, two squares, whose
sum gives a new square. This remark will lead to the
solution of the following question:
To find two square numbers, whose sum is likewise a
square number.
We must have ; we have therefore only to make
, and , then we shall have
.
Farther, as , we may
also resolve the following question:
To find two squares, whose difference may also be a square
number.
Here, since , we have only to suppose
, and , and we obtain .
We might also make , and ,
from which we should find .
45 We spoke of two methods of giving the form of a
square to the formula . The other is as follows:
If we suppose , we shall have
subtracting 1 from both sides, we have
This equation may be divided by , so that we have
or , whence we find . Having found
this value of , we have
which is the square of . Now, as we obtain from that, the
equation
we shall have, as before,
that is, the same two squares, whose sum is also a square.
46 Tlie case which we have just analysed furnishes two
methods of transforming the general formula
into a square. The first of these methods applies to all
the cases in which is a square; and the second to those in
which is a square. We shall consider both these suppositions.
First, let us suppose that is a square, or that the given
formula is . Since this must be a square,
we shall make
and shall thus have
in which the terms containing destroy each other, so that
If we multiply by , we obtain
hence we find ; and,
substituting this value for , we shall have
47 As we have got a fraction for , namely,
, let us make , then
, and ; so that the formula
is a square; and as it contains a square, though it be multiplied by the square , it follows,
that the formula is also a square, if we suppose ,
and . Hence it is evident, that an infinite
number of answers, in integer numbers, may result from
this expression, because the values of the letters and are
arbitrary.
48 The second case which we have to consider, is that in
which , or the first term, is a square. Let there be proposed, for example,
the formula , which it
is required to make a square. Here, let us suppose
and we shall have
in which equation the terms destroying each other, we may
divide the remaining terms by , so that we obtain
or
or
or, lastly,
If we now substitute this value instead of , we have
and making , we may, in the same manner as before,
transform the expression , into a square,
by making , and .
49 Here we have chiefly to distinguish the case in which
, that is to say, in which it is required to make a
square of the formula , for we have only to
suppose , from which we have
the equation ; which, divided
by , and multiplied by , gives ;
and, consequently,
If we seek, for example, all the triangular numbers
that are at the same time squares, it will be necessary that
, which is the form of triangular numbers, must be
a square; and, consequently, must also be a
square. Let us, therefore, suppose
to be that square, and we shall have , and
; in
which value we may substitute, instead of and , all possible
numbers;
but we shall generally find a fraction for ,
though sometimes we may obtain an integer number. For
example, if , and , we find , the triangular
number of which, or 36, is also a square.
We may also make , and ; in this case,
, the triangle of which, 1225, is at the same time
the triangle of +49, and the square of 35. We should
have obtained the same result by making and ;
for, in that case, we should also have found .
In the same manner, if and , we obtain
, its triangular number is
which is a square, whose root is 12 Β· 7 = 204.
50 We may remark, with regard to this last case, that
we have been able to transform the formula into a
square from its having a known factor, ; this observation
leads to other cases, in which the formula
may likewise become a square, even when neither nor
are squares.
These cases occur when may be resolved
into two factors; and this happens when
is a square: to prove which, we may remark, that the factors depend
always on the roots of an equation; and that, therefore, we must
suppose . This being laid down, we have
, or
whence we find
or
and, it is evident, that if be a square, this quantity
becomes rational.
Therefore let ; then the roots will be
, that is to say, ;
and, consequently, the divisors of the formula are
, and . If we multiply these factors together, we are
brought to the same formula again, except that it is divided
by ; for the product ;
and since , we have
which being multiplied by , gives . We have, therefore, only to multiply
one of the factors by , and we obtain the formula in question expressed by the product
and it is evident that this solution must be applicable whenever is a square.
51 From this results the third case, in which the formula
may be transformed into a square; which we
shall add to the other two.
52 This case, as we have already observed, takes place,
when the formula may be represented by a product, such
as . Now, in order to make a square of this
quantity, let us suppose its root, or
and we shall then have
and, dividing this equation by , we have
or
and, consequently, .
To illustrate this, let the following questions be proposed.
Question 1. To find all the numbers, , such, that if 2
be subtracted from twice their square, the remainder may be
a square.
Since is the quantity which is to be a square, we must observe,
that this quantity is expressed by the factors, .
If, therefore, we suppose its root , we have
dividing by , and multiplying by , we obtain
and .
If, therefore, we make , and , we find ,
and .
If and , we have . Now, as is
only found in the second power, it is indifferent whether we
take , or ; either supposition equally
gives .
53 Question 2. Let the formula
be proposed to be transformed into a square. Here, we have
, , and , in which neither nor
is a square. If, therefore, we try whether becomes
a square, we obtian 25; so that we are sure the formula
may be represented by two factors; and those factors are
If is their root, we have
which becomes , whence we find
Now, m order that the numerator of this fraction may become positive,
must be greater than ; and, consequently, less than
:
that is to say, must be less than Β³ββ. With regard
to the denominator, if it must be positive, it is evident that
must exceed ; and, consequently, must be greater
than β
. If, therefore, we would have the positive values of , we must
assume such numbers for and , that may be less
than Β³ββ, and yet greater than β
.
For example, let , and ; we shall then have
, which is less than Β³ββ, and evidently
greater than β
, whence .
54 This third case leads us to consider also a fourth,
which occurs whenever the formula may be
resolved into two such parts, that the first is a square, and
the second the product of two factors: that is to say, in this
case, the formula must be represented by a quantity of the
form , in which the letters , , and express
quantities of the form . It is evidcut that the rule for this
case will be to make ;
for we shall thus obtain
in which the terms vanish; after which we may divide by ,
so that we find
or , and equation from which is easily determined.
This, therefore, is the fourth
case in which our formula may be transformed into a square;
the application of which is easy, and we shall illustrate it by
a few examples.
55 Question 3. Required a number, , such, that double
its square, shall exceed some other square by unity; that is,
if we subtract unity from this double square, the remainder
may be a square.
For instance, the case applies to the number 5, whose
square 25, taken twice, gives the number 50, which is
greater by 1 than the square 49.
According to this enunciation, must be a square;
and as we have, by the formula, , , and ,
it is evident that neither nor is a square; and farther,
that the given quantity cannot be resolved into two factors,
since which is not a square; so that none of
the first three cases will apply. But, according to the fourth,
this formula may be represented by
If, therelore, we suppose its root ,
we shall have
This equation, after having expunged, the terms , and
divided the other terms by , gives
whence we find ; and, since in our
formula, , the square alone is found, it is indifferent
whether we take positive or negative values for . We may at first
even write , instead of , in order to have
If we make , and , we find , and ; or if
we make , and , we find , and ;
lastly, if we suppose , and , we find , or , and .
56 Question 4. To find numbers whose squares doubled
and increased by 2, may likewise be squares.
Such a number, for instance, is 7, since the double of
its square is 98, and if we add 2 to it, we have the square
100.
We must, therefore, have a square; and as
, , and , so that neither nor ,
nor , the last being = -16, are squares, we must, therefore,
have recourse to the fourth rule.
Let us suppose the first part to be 4, then the second will be
, which presents
the quantity proposed in the form
Now, let be its root, and we shall have
the equation
in which the squares 4, are destroyed; so that after having
divided the other terms by , we have
and consequently,
If, in this value, we make , and , we find
, and . But if , and , we
have , and .
57 It frequently happens, also, when none of the first
three rules applies, that we are still able to resolve the
formula into such parts as the fourth rule requires, though
not so readily as in the foregoing examples.
Thus, if the question comprises the formula
, the resolution we speak of is possible, but the
method of performing it does not readily occur to the mind.
It requires us to suppose the first part to be
or , so that the other may be :
and we perceive that this part has two factors, because
, = 1, is a square. The two factors
therefore are ; so that the formula
becomes , which we may
now resolve by the fourth rule.
But, as we have observed, it cannot be said that this
analysis is easily found; and, on this account,
we shall explain a general method for discovering, beforehand, whether
the resolution of any such formula be possible or not; for
there is an infinite number of them which cannot be resolved at all:
such, for instance, as the formula ,
which can in no case whatever become a square. On the
other hand, it is sufficient to know a single case, in which a
formula is possible, to enable us to find all its answers; and
this we shall explain at some length.
58 From what has been said, it may be observed, that all
the advantage that can be expected on these occasions, is
to determine, or suppose, any case in which such a formula
as , may be transformed into a square ; and
the method which naturally occurs for this, is to substitute
small numbers successively for , until we meet with a case
which gives a square.
Now, as may be a fraction, let us begin with substituting
for the general fraction ; and, if the formula
which results from it, be a square, it
will be so also after having been multiplied by ; so that
it only remains to try to find such integer values for and , as will
make the formula a square; and it is evident, that after this,
the supposition of cannot fail to give the formula
equal to a square.
But if, whatever we do, we cannot arrive at any satisfactory case,
we have every reason to suppose that it is altogether
impossible to transform the formula into a square; which, as
we have already said, very frequently happens.
59 We shall now show, on the other hand, that when one
satisfactory case is determined, it will be easy to find all the
other cases which likewise give a square; and it will be perceived, at the same time, that the number of those solutions
is always infinitely great.
Let us first consider the formula , in which ,
, and . This evidently becomes a square, if we
suppose ; let us therefore make , and, by
substitution, we shall havc , and our
formula becomes , in which the first term is
a square; so that we shall suppose, conformably to the second
rule, the square root of the new formula to be
, and we shall thus obtain the equation
in which we may expunge 9 from both sides, and divide by :
which being done, we shall have ; whence
; and consequently
, in which we may substitute any values we please
for and .
If we make , and , we have ; or,
since the second power of stands alone, , wherefore
.
If , and , we have , or .
But if , and , we have ; which gives
, the square of 45.
If , and , we shall then have, in the same
manner, , or .
But, by making , and , we find ;
so that .
60 Let us now examine the formula ,
which becomes a square by the supposition of .
Here, if we make , our formula will be changed into
this:
the square root of which we shall suppose to be ;
by which means we shall have
or ; whence we deduce
; and, lastly, .
If , and , we have , and consequently .
But if and , we find , and .
61 Let us now consider the formula,
, in which we must begin
with the supposition of .
Having substituted and multiplied , we obtain
, which must be a square.
Let us therefore try to adopt some small numbers as the values
of and .
If , and , the formula will become = 35
If and , the formula will become = 71
If , and , the formula will become = 11
If , and , the formula will become = 121.
Now, 121 being a square, it is proof that the value of answers
the required condition; let us therefore suppose
, and we shall have, by substituting this value
in the formula,
or .
Therefore let the root be represented by , and we
shall have , or
whence
Suppose, for example, , and ; we shall then
find , and the formula becomes
If , and , we find ; if
, and , we have , and the formula
62 But frequently it is only lost labor to endeavour to
find a case, in which the proposed formula may become a
square. We have already said that is one of those
unmanageable formulae; and, by giving it, according to this
rule, the form , we shall perceive that, whatever
values we give to and , this quantity never becomes a
square number. As the formulae of this kind are very
numerous, it will be worth while to fix on some characters,
by which their impossibility may be perceived, in order that
we may be often saved the trouble of useless trials; which
shall form the subject of the following chapter.
Editions
- Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
- Leonhard Euler. VollstΓ€ndige Anleitung zur Algebra. Mit den ZusΓ€tzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.