Ancient Greek mathematics had whole numbers, that show up in counting and in
the ratios of lengths of strings on a lyre whose sounds go nicely together, so
naturally started out doing what it could with those. In geometry it was
clearly possible to repeat a length any whole number of times to obtain a length
that was a multiple of the one you started with; and, given that, it turns out
to be quite easy to likewise subdivide a length into any (positive) whole number
of equal pieces. By combining these, one can get a vast diversity of lengths
from a given length; indeed, given any two distinct lengths it's posssible to
scale and evenly subdivide any other length to obtain one whose length lies
between those two distinct lengths, no matter how close together they may be.
That, at least, means that scaling and subdividing one given length can get
you arbitrarily close to any other. Two lengths are said to
be rationally commensurate if scaling and subdividing one
can deliver the other. A question naturally arises: are all lengths rationally
commensurate ? Does getting arbitrarily close mean that you can
necessarily get exact ?
As we'll shortly see, Pythagoras'
theorem shows that some triangles have sides whose lengths are not
rationally commensurate. That then opens up the question of which right-angle
triangles' sides do have rationally commensurate lengths, which I'll
explore in detail.
An awkward example
Let's start by looking (as the greeks did)
at an isosceles right-angle triangle
– that is, one whose two perpendicular sides are of equal length. The
squares on those two equal sides are equal and Pythagoras then tells us that the
square on the hypotenuse has area equal to their sum, hence to twice either of
them.
Let S be the length of either perpendicular side of our isosceles
right-angle triangle and consider any length L rationally commensurate with it.
Being rationally commensurate means that subdividing S into some whole number q
of equal parts and then repeating one of those parts some whole number p of
times will give us L. If the length of each of those parts is U, we have S =
q.U and L = p.U; whence the area of a square on L is p.p.U.U while the area of a
square on S is q.q.U.U and twice this is 2.q.q.U.U. For the hypotenuse to be
rationally commensurate with the perpendicular sides, we'd need p.p.U.U to be
equal to 2.q.q.U.U for some whole numbers p and q; since U.U is non-zero we can
divide each by it to conclude that we'd need p.p to equal 2.q.q – which is
purely a statement about the arithmetic of whole numbers. Thus we can reduce
our geometric question, of whether the hypotenuse can be rationally commensurate
with the two perpendicular sides, to: can doubling the square of one positive
whole number give us the square of another ?
We could prove that can't happen
by reasoning from
the uniqueness of prime factorisation,
but let's take a more primitive approach. Every positive whole number is either
odd or even; if it is even, it is twice some strictly smaller positive whole
number, half the one we started with. As long as we get an even result we can
keep halving, getting a strictly smaller positive whole number each time; and
there are only finitely many positive whole numbers less than the one we started
with, so we must eventually run out of factors of two, after finitely many
halvings. Since we only stop on an odd number, we thus express our original
number as some odd number times power(i, 2) for some natural i, known as the
multiplicity of 2 as a factor of our original number. For an odd number, i is
zero, for an even number that's not a multiple of four it's one, for multiples
of four but not eight it's two, and so on. The result of multiplying an odd
number by an odd number is an odd number, so the result of squaring our original
number is an odd square times the square of power(i, 2), which is power(2.i, 2);
thus 2 has even multiplicity as a factor of any perfect square. Since doubling
a perfect square thus results in a number of which 2 has an odd multiplicity,
and no odd number is even, the result of doubling a perfect square is never a
perfect square.
Taking this purely arithmetic result back to our isoscelese right-angle
triangle, we find that no length rationally commensurate with that of one of its
perpendicular sides is a side of a square with twice the area of the square on
one of those perpendicular sides. Since the square on the hypotenuse does have
area twice that of the square on either perpendicular side, the hypotenuse of an
isosceles right-angle triangle is not rationally commensurate with that
triangle's perpendicular sides.
In
particular, there exist lengths that are not rationally commensurate.
The rest of this page is devoted to the study of right-angle triangles
– and their higher-dimensional peers, the simplices – whose
sides are rationally commensurate. That shall, in
due course, let us see how close to isosceles a right-angle triangle can
be.
Pythagorean triangles
I'll describe a right-angle triangle as pythagorean
precisely if its sides' lengths are rationally commensurate. (One may likewise
pose the corresponding question for non-right
angles.) A pythagorean triple is a set of three whole
numbers in which the squares of two sum to the square of the third: these can be
construed as the lengths of the sides of a pythagorean triangle. Where several
pythagorean triples have the same largest member (hypotenuse), their lower
members identify whole-number co-ordinates for
distinct points on a circle with that
largest member as radius.
Rationally commensurate
If one length is rationally commensurate with another, duplicating the first
one number of times and the second another number of times will produce lines of
equal length, if we chose the numbers correctly; equivalently, sub-dividing each
into suitably many equal pieces will yield a unit for which each is obtained by
duplicating the unit. Thus being rationally commensurate is a symmetric
relationship. If lengths K and L are rationally commensurate and L is also
rationally commensurate with M, then there are some whole numbers k and m for
which dividing L into k equal parts gives a length that, when duplicated enough,
yields K; while dividing L into m equal parts gives a unit of which M is a
multiple. So dividing L into k.m equal parts gives a length of which both K and
M are multiples; thus K and M are also rationally commensurate and rationally
commensurate is transitive. Since it is fatuously reflexive (every length
is rationally commensurate with itself), rationally
commensurate is thus an equivalence relation.
To classify pythagorean triangles, it suffices to consider the case where
our right-angle triangle's lengths are all whole multiples of some given unit;
this reduces our description of the triangle to three whole numbers. The
question thus reduces to finding cases where one perfect square (that is, the
square of a whole number) is equal to the sum of two others. We've shown,
above, that those two must be distinct.
Scaling a pythagorean triangle necessarily produces a pythagorean triangle;
just scale the unit used for the original to get a unit that describes the
scaled triangle's sides using the same numbers, whose squares haven't changed so
still satisfy Pythagoras's equation. Sub-dividing our unit likewise just
increases the numbers we use to describe the lengths of the edges, all by some
common factor i, a whole number; if n.n +m.m = k.k then, i.n.i.n +i.m.i.m =
i.k.i.k, so any whole-number scaling of a solution to our problem is, trivially,
another. So we can scale up our three whole numbers by a common factor to turn
one solution into another, albeit one that's not interestingly different.
Given that we can do this, it's sufficient to classify the solutions that
have no common factor, since all others can be obtained from these by such
scalings. If two of the sides share a common factor i then their squares have
i.i as a common factor; hence the sum or difference of thsese squares, which is
the square of the other side, also has i.i as a factor; hence the other side has
i as a factor. Thus any common factor of two of the sides is in fact a common
factor of all three, so we only need to consider cases where no two sides have
any common factor; the three sides are mutually comprime.
Odd and even
In particular, at most one of the sides has two as a factor so at least two
sides have odd lengths. Any odd number is 2.n +1 for some whole number n; its
square is then (2.n).(2.n) +2.(2.n) +1 = 4.n.n +4.n +1 = 4.n.(n+1) +1. Of any
two adjacent whole numbers, such as n and n+1, one is odd and the other is even;
so n.(n+1) is even and 4.n.(n+1) is a multiple of eight. Thus the square of any
odd number is one more than a multiple of eight; so the two odd sides have
squares whose difference is a multiple of eight and whose sum is two more than a
multiple of eight; one of these is the square of the other side, which thus
isn't one more than a multiple of eight, so that other side isn't odd.
Given that the square of an odd number is one more than a multiple of eight,
the square of twice an odd number is four more than a multiple of (thirty-two,
hence also of) eight. All other even numbers are multiples of four so have
squares that are multiples of 16, hence in particular of eight. Thus no whole
number's square is two more than a multiple of eight, so the sum of two odd
squares is never a perfect square. So our pythagorean equation n.n +m.m = k.k,
in which two terms are one more than multiples of eight, must have these two
terms on opposite sides of the equation with the remaining term equal to a
multiple of eight; i.e. the hypotenuse is odd, as is one of the other two sides,
and the remaining side's square is a multiple of eight.
One square is a multiple of eight: thus two has multiplicity at least three
as a factor of that square. As two has even
multiplicity as a factor of every perfect square, this at least three
must in fact be at least four, the smallest even number that's at least three.
This makes the even square a multiple of 16 and the side it's a square of a
multiple of four. So our hypotenuse is odd, as is one of the other sides, and
the remaining side is a multiple of four.
Classification
Suppose, then, that m is the multiple of four, with k and n
odd; we have n.n = k.k −m.m = (k +m).(k −m) with k, n, m coprime.
As k and m are coprime and m is even, k ±m are also coprime (had k and m
both been odd, k ±m would have shared two as a common factor); and their
product is n.n, a perfect square. With {x, y} = {k ±m} in either order,
any prime factor p of x is a prime factor of x.y = n.n hence also of n, hence
has even multiplicity as a factor of n.n = x.y; but p is not a factor of y,
hence p in fact has even multiplicity as a factor of x; thus the prime
factorisation of x only has even multiplicities for its prime factors and x is a
perfect square. Thus k ±m are mutually coprime odd perfect squares.
In particular, the two odd numbers whose squares are k ±m have
product n; and we can obtain k and m from them as the mean and half-difference
of their squares. Thus, for any coprime pythagorean triangle, there are
naturals u, v with v > u for which k +m = 4.v.(v+1) +1, k −m = 4.u.(u
+1) +1 so:
k = 2.(v.(v +1) +(u +1).u) +1,
m = 2.(v.(v +1) −(u +1).u) and
n = (2.v +1).(2.u +1).
Furthermore, for any naturals u, v, we can compute k, m and n as above;
they might not be coprime, but they do always satisfy n.n = (k +m).(k −m)
= k.k −m.m hence k.k = n.n +m.m, so they do describe a pythagorean
triangle, even if it's not coprime. (For example, [v,u] = [4,1] yields
[45,36,27] which just scales the familiar [5,4,3] from [v,u] = [1,0] by a factor
of 9.) Every pythagorean triangle is a whole multiple of some triangle
generated in this way. Note that the hypotenuse, k, is always one more than a
multiple of four (since both u.(u+1) and v.(v+1) are even; which also ensures m
is a multiple of four).
If your browser supports SVG (and CSS floats), you should see the first 58
distinct hypotenuses (with short side
horizontal and intermediate side vertical) depicted at right (possibly
above); these are all the examples with 0 ≤ u < v < 12.
One consequence of this classification is that the set of pairs of rationals
whose sum is 1, i.e. the set of points on the unit circle of the two-dimensional
rational plane, is countable – it is a union of two images (using the
perpendicular sides above in each order) of the set of pairs of naturals [u, v],
via the combinations of them used to make the two perpendicular sides above.
These give us (countably many) angles whose Sin and Cos are rational.
(Technically, these are all in the first quadrant (from zero through a quarter
turn); we can add and subtract them to obtain angles in the full range, still
with rational Sin and Cos – but this in fact necessarily just gives the
same set of angles as adding multiples of a quarter turn to the ones obtained
directly, since each implies a pythagorean triangle, albeit potentially in other
quadrants.) While this set of angles is (though I have not shown this)
dense in the turn, it leaves plenty of gaps in the unit circle of the rational
plane; e.g. the line {[r, r]: r is a positive rational} runs from [0,0] inside
the unit circle to outside that circle but has no intersection with the
circle.
Euclid's formulæ
The constraint on our parameterisation, v > u, can be re-written by
taking v = u +1 +w for some natural w; so v.(v +1) = (u +w +1).(u +w +2) = u.u
+2.u.w +w.w +3.u +3.w +2. We then get
k = 2.(v.(v +1) +(u +1).u) +1
= 4.u.u +4.u.w +2.w.w +8.u +6.w +5
= (2.u).(2.u) +2.(2.u).w +2.w.w +4.(2.u) +6.w +5
= (2.u +w +2).(2.u +w +2) +(w +1).(w +1),
m = 2.(v.(v +1) −(u +1).u)
= 2.(2.u.w +w.w +2.u +3.w +2)
= 2.(2.u.(w +1) +w.(w +1) +2.(w +1))
= 2.(2.u +w +2).(w +1) and
n = (2.v +1).(2.u +1)
= (2.u +2.w +3).(2.u +1)
which is rather messier, but applies for any pair u, w of
naturals, u and w. In particular, we see that the hypotenuse, k, of a coprime
pythagorean triple is itself a sum of two squares – and, indeed, of an odd
square and an even square.
If we write p = 2.u +w +2 = v +u +1, q = w +1 = v −u, this last form
of the formulæ becomes k = p.p +q.q, m = 2.p.q, n = p.p −q.q. This
formulation is known as Euclid's formulæ and works for any p and q with p
> q. It only gives coprime results if p and q are coprime, with odd
difference; that gives u = (p −q −1)/2, v = u +q.
3blue1brown
has a video explaining
this and depicting the result. This includes the delightful observation
that any pair of whole numbers p, q can be represented as the complex number p
+q.i (with i.i + 1 = 0) whose magnitude is √(p.p +q.q); it square, p.p
−q.q +2.i.p.q, thus has magnitude p.p +q.q, which is necessarily a whole
number.
In Euclid's formula, we have p = v +u +1 = w +2.u +2 and q = v −u = w
+1; these imply p +q = 2.w +2.u +3 and p −q = 2.u +1, still with k = p.p
+q.q, m = 2.p.q and n = p.p −q.q.
Mathologer points out
that, from a triangle described by p −q, q, p, p +q in this way, there are
matching triangles that can be obtained from
b −a = p −q, a = p, b = 2.p −q, b +a = 3.p −q
b −a = p +q, a = p, b = 2.p +q, b +q = 3.p +q
b −a = p +q, a = q, b = p +2.q, b +a = p +3.q
with b > a > 0 whenever p > q > 0 and a coprime to b
whenever p is coprime to q. Going in the opposite direction (recovering p, q
from a, b):
p −q = b −a, q = 2.a −b, p = a, p +q = 3.a −b
p −q = 3.a −b, q = b −2.a, p = a, p +q = b −a
p −q = b −3.a, q = a, p = b −2.a, p +q = b −a
in which we always get the same four values, possibly shuffled and with
some signs flipped – just what we need to recover p, q positive; and, for
any positive a, b with a < b, there's a unique choice of which of these three
reversals gets both p and q positive with p > q. If 2.a > b, we use the
first; for 2.a < b < 3.a the second, otherwise the last. In each case, p
is coprime to q precisely if b is coprime to a and a > 1. However,
b > a > 0 with a coprime to b allows the possibility a = 1, coprime to
everything, notably 2 and 3, the boundary values at which we would change
between which of the three to use. For a = 1, b = 3 (implying k = 10, m = 6, n
= 8, a scaled version of 5, 4, 3) we get q = 1 = p (implying k = 2, m = 2, n =
0); for a = 1, b = 2 (implying the well-known k = 5, m = 4, n = 3) we get q = 0,
p = 1 (implying k = 1, m = 0, n = 1). In each case, stepping back from (a form
of) the 5, 4, 3 triangle gets us the degenerate right-angled triangle
with a zero side perpendicular to a side equal to the hypotenuse, with q in {p,
0} failing to satisfy the requirement p > q > 0.
The fun thing is that these steps between generators for pythagorean triples
suffice to generate all distinct triples, starting from the p = 2, q = 1
generator of the simplest of them all. Using the same forward step at each
iteration will either take you towards one side being tiny compared to the other
two (which differ in length by 1 on one side or 2 on the other), or towards
ever-more-nearly isoscelese triangles (with ever longer perpendicular sides
differing in length by 1). As a bonus, if q and p are consecutive members
of the Fibonacci sequence, then k is in that
sequence.
Other moduli
Every coprime triple is [k, m, n] given, for some naturals u, w,
by:
k = (2.u +w +2).(2.u +w +2) +(w +1).(w +1)
m = 2.(2.u +w +2).(w +1)
n = (2.u +2.w +3).(2.u +1)
Our choice of parmeterisation by natural u, w ensures k and n are odd
while m is a multiple of four; so they don't share a factor of two. Let's
consider what their remainders on division by other numbers may be. In
particular, let's see how this constrains u, w for the triple to be coprime. As
before, I'll also use p = 2.u +w +2 and q = w +1 where they simplify what I need
to write.
If n is a multiple of any odd prime z, whence p = i.z ±q for some odd
i, we have k = i.i.z.z ±2.i.z.q +2.q.q and m = 2.i.z.q ±2.i.q,
whence p = ±q modulo z implies k = ±m modulo z.z, with the same
sign.
For the largest member of a coprime triple to be a multiple of any prime z
(necessarily odd, since the largest member of a coprime triple is never even),
there must be non-zero perfect squares modulo z that are each other's negation,
m.m = −n.n. As z is prime, the multiplication of non-zero values modulo
z is a group, so m has a
multiplicative inverse i, yielding n.n.i.i = −m.i.m.i = −1, so the
arithmetic modulo z has a square root of −1. This
happens precisely when z is one
more than a multiple of four. So only primes one more than a multiple of four
can appear as factors of the largest member of a coprime triple.
Modulo 3 or 9
The square of a multiple of three is a multiple of 9; any other number's
square is one more than a multiple of three. Adding two of the latter gives two
more than a multiple of three, which can't be a perfect square, so one of the
two shorter sides must be a multiple of three; for a coprime triple, this means
the other two are not. In particular, no coprime triple's largest member is a
multiple of 3. Modulo 9, their squares must be equal; these shall be 1, 4 or 7
= −2, the squares of ±1, ±2 and ±4,
respectively.
If p is a multiple of 3, so is m (making it a multiple of 12) and, modulo 9,
k and −n are both q.q; for a coprime triplet, q is not a multiple of 3 so
k = −n is one of 1, 4 or 7 modulo nine.
If q is a multiple of 3, so again is m (making it a multiple of 12) and,
modulo 9, k and n are both p.p; for a coprime triplet, p is not a multiple of 3
so k = n is one of 1, 4 or 7 = −2.
The remaining case, with neither p nor q a multiple of 3, makes n a multiple
of 3, hence p = ±q modulo 3 and k = ±m modulo 9, with the same
sign, and k is one of 2, 5 = −4 and 8 = −1 modulo 9.
Modulo 5 or 25
Modulo 5, (: power(2) |5) = [0, 1, −1, −1, 1] and the potential
values for [k.k, m.m, n.n] are ±[0, 1, −1], ±[1, 1, 0] and
±[1, 0, 1]; so at least one member of any pythagorean triple is a
multiple of five. If the multiple of 5 is the biggest, one of the smaller
members is ±1 and the other is ±2 modulo 5. If the multiple of
five isn't the biggest, then either the other two are both in {±1} or
both are in {±2} modulo 5.
When n is a multiple of 5, so p = ±q modulo 5 and k = ±m
modulo 25, a coprime triple won't have both p and q multiples of 5, so neither
is; and we get m = 2.p.q = ±2.q.q modulo 5 and both k and m are in
{±2} modulo 5, so [k.k, m.m, n.n] is [−1, 0, −1], without the
± initially considered.
When m is a multiple of 5 (hence also of 20) we have either p or q a
multiple of 5, whence k = ±n modulo 25 (not just modulo 5).
When k is a multiple of 5, so p.p = −q.q, a coprime triple has one of
p, q in {±1} and the other in {±2}, modulo 5. Thus m is in
{±1} and n is in {±2}, modulo 5. So [k.k, m.m, n.n] is in fact
[0, 1, −1] modulo 5, without the ± that initially seemed
applicable.
Higher moduli
From the known case of [5, 4, 3] we know that there is no number above 5
that necessarily has a multiple in every pythagorean triple.
The squares, modulo z, of numbers whose double exceeds z are equal to those
of their differences from z, so we only actually need to look at the first half
of (: power(2) |z) modulo z (including the mid-point, when there is one). This
tells us the values squares can have, constraining what the squares of our
triple can be, if they're to add up. For each perfect square x, modulo z, we'll
get x = 0 +x as an option for our triple of squares; and of course we can swap
the two squares being added, unless our modulus is even, in which case just one
of them is even and is m.m, while n.n is the odd one. When the modulus has
factors, we can discard triples of their multiples, as they're not
coprime.
Modulo 6, (: power(2) |4) = [0, 1, −2, ±3] allowing 1 = 3
+(−2), so any coprime triple with no multiple of 6 has k = ±1
modulo 6, n an odd multiple of 3 and m = ±4 modulo 12. (Note that we
discard −2 = 3 +1 because it doesn't let m be even.)
Modulo 7, (: power(2) |4) = [0, 1, −3, 2] allowing 1 +1 = 2, 2 +2 =
−3 and (−3) +(−3) = 1, These give m = ±n modulo 7 with
[k, n] as [±1, ±2], [±2, ±3] or [±3,
±1] in all the coprime triplets that lack a multiple of 7; and no coprime
triplet's largest member is a multiple of 7. In particular, the sum of the two
smaller members of a pythagorean triple is always a multiple of 7.
Modulo 10, (: power(2) |6) = [0, 1, 4, −1, −4, ±5],
with m = 10.even, 10.even ±4 or 10.odd ±2, with (modulo ten) k =
±n for the first, [k, n] in {[5, ±3], [±1, 5]} for the
second and [k, n] in {[5, ±1], [±3, 5]} for the last. At first
sight, we also get x +(−x) = 0 cases, but the x = ±4 and x =
±5 cases have common factors, and the x = ±1 case fails becaue we
know m is even.
Modulo 11, (: power(2) |6) = [0, 1, 4, −2, 5, 3], allowing 1 +3 =
4, 1 +4 = 5, 5 +(−2) = 3, 3 +(−2) = 1, 4 +5 = −2 as the sum of
squares; there are several ways for a pythagorean triple to not have a multiple
of 11. Furthermore, no coprime triple has a multiple of 11 as its largest
member.
Modulo 12, (: power(2) |7) = [0, 1, 4, −3, 4, 1, 0] allowing 4
+(−3) = 1, in which m (which can't be either ±2 or ±6, as
they're not multiples of 4) must be the ±4 whose square is 4, n is an odd
multiple of 3 and k is in {±1, ±5}. The duplicated entries (only
1 actually matters, as the duplicated 4 and 0 would give the triple a common
factor) also make the m = 0 case more interesting: each of k, n in {±1,
±5}, chosen independently. However, all of this simply restates what we
saw modulo 6.
Modulo 13, (: power(2) |7) = [0, 1, 4, −4, 3, −1, −3],
allowing 1 +3 = 4, 4 +(−1) = 3, 4 +(−3) = 1 and the negation of each
of these, along with x +(−x) = 0, for each x in the list, in addition to
the usual 0 +x = x; and, unlike the z = 10 case, the x +(−x) = 0 cases
aren't ruled out. So k can be a multiple of 13 when either [m, n] or [n, m] is
one of [±1, ±5], [±2, ±3] or [±4, ±6];
in particular, the [−1, +5] case is seen in the [13, 12, 5] triple.
There are also plenty of cases with no multiple of 13 in the triple, along with
the usual cases where one of n, m is a multiple of 13 and the other is ±k
modulo 13.
Modulo 14, (: power(2) |8) = [0, 1, 4, −5, 2, −3, −6,
±7] allows, for the squares of coprime triples (ruling out 2 +2 = 4 and 4
+4 = −6 as all even; but also requiring m.m even among the values added),
1 +(−6) = −5, 7 +(−6) = 1, 4 +(−3) = 1, 4 +7 = −3,
2 +(−5) = −3. Once again, there are many options.
Modulo 15, (: power(2) |8) = [0, 1, 4, −6, 1, −5, 6, 4],
allowing 1 +(−6) = −5, 6 +(−5) = 1, 6 +4 = −5,
(−5) +(−6) = 4. Again, several other candidates (including 6
+(−6) = 0) are ruled out by common factors of 5 or 3, despite which we're
left with ample choice, amplified by the repetition of 1 and 4 in the
list.
Modulo 16, (: power(2) |9) = [0, 1, 4, −7, 0, −7, 4, 1, 0],
with only the usual x +0 = x options, replaying what we already know from
smaller powers of 2.
Modulo 17, (: power(2) |9) = [0, 1, 4, −8, −1, 8, 2,
−2, −4], allowing x +(−x) = 0, x +x = 2.x and 2.x +(−x)
= x solutions, notably including 2×8 = −1 variants, for each power
of 2 as ±x, in addition to the usual x +0 = x ones. A coprime triple
thus can have a multiple of 17 as its largest member, e.g. [17, 15, 8].
Modulo 18, (: power(2) |10) = [0, 1, 4, ±9, −2, 7, 0,
−5, −8, ±9], allowing coprime triples with 9 +4 = −5, 9
+(−8) = 1, with the usual x +0 = x made more flexible by 0 being the
square of every multiple of 6, not just the multiples of 18.
Modulo 19, (: power(2) |10) = [0, 1, 4, 9, −3, 6, −2,
−8, 7, 5], allowing 1 +4 = 5, 1 +(−3) = −2, 1 +5 = 6, 1 +6 =
7, 4 +(−3) = 1, 4 +7 = −8, 4 +5 = 9, 9 +(−3) = 6, 9
+(−2) = 7, 9 +(−8) = 1, 9 +7 = −3, 7 +(−3) = 4, 6
+(−2) = 4, 6 +(−8) = −2, 6 +5 = −8, 7 +(−2) = 5, 5
+(−8) = −3. Once again, we have many ways for a triplet to include
no multiple of 19; however, no coprime triplet has a multiple of 19 as its
largest member.
Every pythagorean triple includes at least one multiple of each of 3, 4
and 5, the smaller two of them add up to a multiple of 7; and the largest member
(hypotenuse) of any coprime pythagorean triple has, as prime factors, only those
that are one more than multiples of four. Each prime less than 313 that is one
more than a multiple of 4 does show up in the diagram above as a hypotenuse; and
25×25 = 625 = 313 +312 = 313×313 −312×312 so we have
[313, 312, 25] as a pythagorean triple, generated by (some pair of) p = 13, q =
12 = v, u = 0, w = 11, making it the first omission due to limiting the diagram
to 0 ≤ u < v < 12.
Inscribed circle
I learn,
via Catriona
Shearer and Ben Orlin, that the inscribed circle of a pythagorean triangle
has a radius that's a whole multiple of the same unit as the triangle's sides
are whole multiples of. We now have the general formula for the pythagorean
triangle's sides, at least when they're coprime, so what does that give us as
the radius of the incircle ?
Let the radius be r; use our two perpendicular sides of the triangle as
co-ordinate axes, so the origin is the right angle corner, with [0, m] and [n,
0] as the other two corners, distance k apart. The centre of the circle is at
[r, r]; the radius that meets the hypotenuse does so at [x, y] = [r, r] +r.[m,
n]/k = r.[k +m, k +n]/k; the one-form m.x +n.y is constant on the hypotenuse,
equal to m.n, so, using Euclid's form of the
formulæ,
n.m
= r.(k.m +m.m +k.n +n.n)/k
= r.(m +n +k), so
r = n.m / (m +n +k)
= 2.p.q.(p.p −q.q) / (2.p.q +p.p −q.q +p.p +q.q)
= 2.p.q.(p.p −q.q) / (2.p.q +2.p.p)
= q.(p +q).(p −q) / (q +p)
= q.(p −q)
or, in the other two representations, r = (v −u).(2.u +1) = (w
+1).(2.u +1). Sure enough, the radius of the incircle of a pythagorean triangle
is a whole number multiple of the highest common factor of its sides.
Furthermore, for every positive whole number, there is at least one pythagorean triangle
whose incircle has that whole number, times the unit of length that all sides of
the triangle are multiples of, as its radius.
Rewriting r = p.q −q.q in terms of the side lengths, we get r = m/2
−(k −n)/2 = (m +n −k)/2; writing s = (m +n +k)/2 for the
semi-perimeter of the triangle, this is r = s −k, conforming
to the cotangent rule, with the
half-angle opposite k being an eighth of a turn, whose Tan is 1. Note, also,
that the area of the triangle is
m.n/2 = r.(m +n +k)/2, the radius of the incircle times the semi-perimeter,
Approximately Isosceles
Let's pause to consider which triangles are closest to isosceles, with a
difference of just 1 between the lengths of the two perpendicular sides.
Dividing the perpendicular sides by the hypotenuse of such triangles provides
good approximations to √2. These triangles correspond to the triples
generated by u, v for which
2.(v.(v +1) −(u +1).u) = (2.v +1).(2.u +1) ±1
The ± gives us two cases to consider:
2.(v.(v +1) −(u +1).u) = (2.v +1).(2.u +1) +1
v.v +v −u.u −u = 2.u.v +v +u +1
v.v = 2.u.v +u.u +2.u +1
(v −u).(v −u) = (u +1).(u +1) +u.u
2.(v.(v +1) −(u +1).u) = (2.v +1).(2.u +1) −1
v.v = 2.u.v +u.u +2.u
(u +v +1).(u +v +1) = (v +1).(v +1) +v.v
In each case, solutions to this problem generate new solutions, by
using the shorter perpendicular edge as u or v and the hypotenuse as either v
−u or v +u +1. The second case also has the square of v −u equal to
2.u.(u +1), requiring one of u and u +1 (since they're coprime) to be an odd
perfect square while the other is twice a perfect square. This is relatively
rare, with only four such u < 1000 and two more <
10,000:
v
u
k
m
n
3
1
29
20
21
20
8
985
696
697
119
49
33461
23660
23661
696
288
1136689
803760
803761
4059
1681
38613965
27304196
27304197
23660
9800
1311738121
927538920
927538921
The first case, where our ± is +, has the square of u +v +1
equal to 2.v.(v +1), likewise requiring one of v and v +1 to be an odd perfect
square and the other to be twice a perfect square.
v
u
k
m
n
1
0
5
4
3
8
3
169
120
119
49
20
5741
4060
4059
288
119
195025
137904
137903
1681
696
6625109
4684660
4684659
9800
4059
225058681
159140520
159140519
Taking the two cases together, we see that we can start with the
degenerate 0, 1, 1 triad, which is trivially a case of two adjacent numbers
whose squares add up to a perfect square, use its short side as u in the first
table and v in the second then, from each of the pythagorean triangles these
generate, use the short side likewise to produce more rows of both tables and so
on. (The initial degenerate case can be thought of as the result of u = 0 = v
from the even more degenerate all sides zero triangle, which is neither coprime
nor generated by our formulæ; and, of course, it violates the u < v
constraint.)
Right-angle simplices
A simplex is the generalisation of a two dimensional triangle to other
dimensions; the triangle is the two-simplex, the tetrahedron is the
three-simplex, a simple line segment is the one-simplex and a point is the
zero-simplex. For any natural n, {mappings ({positives}:f:1+n): sum(f) = 1}
provides a canonical n-simplex. However, I'm here more interested in a
less symmetric n-simplex – namely, one in which n of the edges are (of
non-zero length and) mutually perpendicular.
Simplices
One important feature of a simplex is that each vertex is connected to every
other vertex by an edge (and, in fact, for every set of i vertices, there is an
i-simplex face of the simplex which has these vertices as its corners).
Given a set of edges, a path among those edges is a list (:f|1+i) of
vertices among which, for each j in i, f(j) and f(1+j) are connected by one of
the edges in our set; f(0) and f(i) are described as the ends of the path
and the path is described as a path between these two vertices, or from f(0) to
f(i). A set of edges is connected if, between any two vertices of edges in the
set, there is some path within the set.
A path is closed if its ends are the same vertex; i.e. f(0) is
f(i). This is easy to achieve by traversing some sequence of edges and then
simply reversing the sequence of edges to come back to where you started; but
this is the boring case. If you can get back to where you started without doing
that, then your path can be reduced to one which never re-uses an edge; a path
is described as a closed loop if it is closed and uses each edge at most
once. Among a set of mutually perpendicular edges there is, necessarily, no
closed loop. A path among some edges, among which there is no closed loop, can
always be reduced to a path that uses each edge at most once.
A set of i edges (for positive natural i), with as few vertices as possible
subject to containing no closed loop, always has i+1 vertices and is connected:
when i is 1, this is trivial. Thereafter, each added edge must add at least one
vertex – because the prior edges are connected, so connecting two of their
vertices would form a loop. If all vertices are reached by prior edges, it is
no longer possible to add an edge without creating a loop; otherwise, connecting
any vertex of an existing edge to any vertex not reached by existing edges does
indeed add only the one vertex it must, to the set of edges, without adding a
closed loop. If we have n edges of an n-simplex and no closed loop, this
requires us to use all 1+n vertices that our n-simplex has; and (since this
means our set of edges uses as few vertices as possible, while having no closed
loop) ensures that they are all connected. So any n mutually perpendicular
edges of an n-simplex are necessarily all connected and reach all n +1 of its
vertices.
Hypotenuses
Every vertex of our n-simplex is an end of at least one of our mutually
perpendicular edges; and the mutually perpendicular edges are all connected, so
there is some path among these edges from any given vertex to any other. As
there are no closed loops, this path need only use each edge at most once; and
there is only one such path, for each given pair of vertices. Thus, for every
edge of our n-simplex, there is such a path connecting its end-points; if this
path traverses i edges (once each), I'll refer to the edge whose end-points it
connects as an i-hypotenuse of our n-simplex. Each of our perpendicular edges
is a 1-hypotenuse (which is boring).
If the n mutually perpendicular edges form a path, that uses each edge
exactly once, then we have an n-hypotenuse and, for each positive i in n, each
n−i sub-path of length i provide us with an i-hypotenuse. However, if
three (or more) of the perpendicular edges meet in one vertex, no path among the
perpendicular edges can use each exactly once; at most two of the edges at any
given vertex can appear in any such path, as we have no closed loops.
In two dimensions, we had two perpendicular sides to a triangle and they
couldn't avoid forming a connected path, so life was simple: we only had one
other side and it was a 2-hypotenuse. In higher dimensions, we have more things
to consider:
We can restrict attention to the case with a single n-hypotenuse,
We can consider all the i-hypotenuses, for each i in n, or
We can restrict attention to the case where all the perpendicular edges
meet in one vertex; all other edges are 2-hypotenuses.
This leads to various possible analogues of pythagorean that we
could investigate for n-simplices with n perpendicular sides. Describe an edge
as tidy, for present purposes, precisely if its length is rationally
commensurate with those of the perpendicular sides. No edge can be tidy unless
all the mutually perpendicular edges have rationally commensurate lengths, of
course. We could study n-simplices with n perpendicular edges and all edges
tidy. A little less generally, we could restrict attention to the case where
the perpendicular edges form a path. As a simplified case of this, we could
ignore all edges except the n-hypotenuse; this amounts to looking for lists
({naturals}:|n) whose sum of squares is a perfect square.
Incrementalism
Refining the simplest notion slightly, we can ask for lists of naturals
where each initial sub-list has a perfect square as its sum of squares; thus,
for example, [3, 4, 12, 84, 132] has 3×3 +4×4 = 5×5; adding
12×12 to that we get 13×13; adding 84×84 to that we get
85×85; and adding 132×132 to that we get 157×157. Each of [3,
4], [3, 4, 12], [3, 4, 12, 84] is an initial prefix of the list and has a
perfect square as its sum of squares. This means we can have an n-simplex with
these as the lengths of successive perpendicular edges, such that each
i-hypotenuse associated with a path starting from the start of the first edge is
tidy. However, notice that 4×4+12×12 = 160 isn't a perfect square,
so at least one of the other other 2-hypotenuses isn't tidy.
From the above, it should be clear that we can build such incrementally
pythagorean lists quite straightforwardly; indeed, in the example above, [3, 4,
12, 84] follow a pattern that can be continued indefinitely. As 3 is odd, its
square is the sum of adjacent naturals, 4 and 5, hence equal to the difference
of their squares, so [3, 4] has length 5. Then 5×5 is odd and likewise
12+13, so [5, 12] has length 13, whose odd square is 84+85, so [13, 84] has
length 85; whose square is 7225 = 3612 +3613, so [85, 3612] has length 3613,
which is (inevitably) odd again, so we can keep doing this for ever. Such a
list grows rapidly, approximating (: 2.power(power(2, 1+i), x/2) ←i :) for
some x, but can clearly be done for any odd start-value: for example, [7, 24,
312, 48984]. None the less, as illustrated above, they're not the only game in
town; we could use 132 after 84 instead of 3612.
Even so, all cases of this kind can be built using the two-dimensional
pythagorean triangles, incrementally: the sum of squares of the first n entries
in such a list is the square of some natural, p, so any pythagorean triangle of
which p is a perpendicular side can be used to obtain the other perpendicular
side as n-th entry in the list, making the hypotenuse take the place of p for
the thus-extended list.
Suppose, then, that m is the multiple of four, with k and n
odd; we have n.n = k.k −m.m = (k +m).(k −m) with k, n, m coprime.
As k and m are coprime and m is even, k ±m are also coprime (had k and m
both been odd, k ±m would have shared two as a common factor); and their
product is n.n, a perfect square. With {x, y} = {k ±m} in either order,
any prime factor p of x is a prime factor of x.y = n.n hence also of n, hence
has even multiplicity as a factor of n.n = x.y; but p is not a factor of y,
hence p in fact has even multiplicity as a factor of x; thus the prime
factorisation of x only has even multiplicities for its prime factors and x is a
perfect square. Thus k ±m are mutually coprime odd perfect squares. 
