Pythagorean triples
One
of the numerous formulations of the theorem of Pythagoras says that if
a and b are the cathets of a right-angled triangle and
c is the hypotenuse, we have
a2 + b2 = c2.
The inverse is also valid:
If sides a, b and c of a triangle verify the relation
a2 + b2 =c2, then the triangle is right-angled,
a and b are the cathets and c the hypotenuse.
The demonstration is very simple. Let us draw a right-angled triangle
with the cathets a and b, and let d be its hypotenuse.
For the theorem of Pythagoras we have d2 = a2 +
b2, while supposing that a2 +
b2 = c2. It follows that d2 =
c2, hence d = c, so that the two triangles
have the the same three sides, and therefore they are equals.
But the second was a right-angled triangle with the cathets
a and b, and therefore the same applies to the second.
The previous result gives us a very simple method to draw right-angled triangles
without having to measure the angles. In fact it is enough to find three
numbers a, b and
c, that verify the relation a2 + b2
= c2; the triangle with sides a, b and c
will obviously be right-angled.
An example is the triangle with sides 3, 4 and 5; since
32+42=9+16=25=52,
the triangle with these sides is right-angled.
This triangle is often found in the most ancient books.
Other right-angled triangles are those with sides 5, 12 and 13,
or 8, 15 and 17.
We note that the sides of all these triangles are integers.
In the case of the theorem of Pythagoras this is not necessary; it is enough that
the relation a2 + b2 = c2 is
verified, like for example in the triangle having the cathets equal 1
and the hypotenuse equal the root of 2.
On the other hand the triangles with integer sides are more interesting,
in part because their sides must be chosen with care -
it is not possible, for example, to
choose one at random and then find the other two.
If three numbers a, b and c verify the relation
a2+b2=c2 it is said that they form a
Pythagorean triple. For example 3, 4 and 5 are a Pythagorean triple,
but not 1, 1 and root of 2, because this last one is not integer.
All Pythagorean triples are described by the formula
a =
m2 - n2 | b = 2mn | c = m2 + n2 | (1) |
where
m and n are two integers, with m>n.
That the numbers a, b and c form a Pythagorean triple, is
easily verifiable. In fact we have
a2 = (m2 - n2)2 =
m4 + n4 - 2m2 n2 and
b2 = (2mn)2 = 4 m2
n2
and hence
a2 + b2 = m4 + n4 -
2m2 n2 + 4 m2 n2 =
m4 + n4 + 2m2 n2 =
(m2 + n2)2 = c2.
It is more difficult to demonstrate that the formula (1) gives all the
possible Pythagorean triples. Here is how it can be done.
We start by observing that if a, b and c
form a Pythagorean triple, the same applies for ha, hb and
hc.
We can therefore limit ourselves to consider triples with
a and b prime among them; all the others will be obtained
multiplying a, b and c by the same number.
We now show that a and b must be one even and one odd,
and that consequently c must be odd.
That
a and b are not both even depends on the fact that
they are prime numbers.
To show that they cannot both be odd is a little more delicate.
If a and
b were odd, so would be a2 and
b2 , such that c2 , sum
of two odd numbers, would be even, and hence c would be even.
On the other hand, if
a and b are odd, it must be that
a = 2k+1 and b=2h+1, from which a2 =
(2k+1)2 = 4k2 +4k+1, b2
=4h2 +4h+1 and summing we get c2
=a2 +b2 = 4(k2 +k+h2 +h)
+ 2.
From this formula follows that dividing c2
by 4 we get the quotient k2 +k+h2 +h
and the remainder 2. In particular, c2 is not divisible by
4, and this is absurd since c
is even.
Finally, if a, b and c form a Pythagorean triple,
the two numbers a and b must be one even and one odd
(for example b even and a
odd), and consequently c must be odd.
In the
relation a2 +b2 = c2
take a2 as second member ; we have:
b2 = c2 - a2 = (c + a)(c –
a). Since a and c are odd, c+a and
c–a are even. If we set b=2s, c+a=2x and
c–a=2y, we'll have s2=xy. Also x and
y are prime among them; in fact if they had a common factor
q, also a = x – y would be divisible by
q, and the same applies to b2 , and thus by
b, in contradiction with the hypothesis that
a and b were prime among themselves. Since the product
xy is a square, x and y are also squares
: x=m2 and y=n2 . We'll finally have:
a=x–y=m2-n2;
c=x+y=m2+n2 and
b2=4xy=4m2n2 such that b =
2mn.
The formula (1) is thus demonstrated. Giving later to
m and ndifferent values, prime again, and one even and odd
the other, we find all the possible Pythagorean triples.
Note that when n = m –
1, we also have that b = c – 1, that can be easily demonstrated by the reader.
From the consideration of the Pythagorean triples,
Pierre Fermat (1601-1665) took his cue to try to find triples of integers
all different from zero, that would verify the relation
x3 +
y3 = z3 or, more in general,
xn + yn = zn.
At the margin of a copy of the
Arithmetica of Diofanto, a Greek author who lived around the III
century A.D., at the point where the generation of Pythagorean triples were explained
Fermat wrote:
Cubem autem in duos cubos, aut quadratoquadratum in duos
quadratoquadratos, et generaliter nullam in infinitum ultra
quadratum potestatem in duos ejusdem nominis fas est dividere:
cujus rei demonstrationem mirabilem sane detexi. Hanc marginis
exiguitas non caparet.
that is
It is not, instead, possible to divide a cube into two cubes.
a square-square into two square-squares, and in general no power bigger than
two in two powers of the same order. I found a beautiful demonstration of this
that I cannot write about because of lack of space.
This result, that was calledThe last theorem of Fermat, has stimulated
the research of many of the greatest mathematicians of the last three centuries,
and was only completely demonstrated in 1994, by Andrew Wiles.
