Pythagoras' Theorem

One of the most fundamental truths of Euclidean geometry – and, indeed, of the geometry of the real world – describes a relationship among the sides of a right-angled triangle. It has been known since antiquity. It was known in ancient Indian cultures and (probably subsequently) in Mesopotamian cultures. More recently yet, the ancient Greek school of though known as the Pythagoreans (notable for also sharing, with some Indian cultures, both vegetarianism and a belief in re-incarnation) attributed it to their (possibly mythical) founder, Pythagoras. It asserts that

The sum of the areas of the squares on the two orthogonal sides of a right-angle triangle is the area of the square on the third side.

The third side, opposite the right angle, is called the hypotenuse of the right-angle triangle. (The addition of an extra term in the cosine of the angle between two sides generalises this rule to the case of non-right angles.)

Proof

This can be proved in various ways.

Tesselation Proof

This is my favourite proof: it only requires one diagram and, thanks to its tesselation property, it would make a deeply satisfying floor-tile. The next two proofs aren't as tidy, though they do relate tidily to algebraic expressions (here given using h as the length of the hypotenuse, the other two sides being of lengths a and c). All three proofs are proofs by suitably general illustration: the construction is shown for a particular case (here a 3:4:5 triangle) but its feasibility for any other right-angle triangle is intuitively obvious from the way the construction works for the chosen example. For the above proof, I have also constructed Scalable Vector Graphics variants which (provided your browser supports SVG images) demonstrate that the same picture fits together regardless of the interior angles of the triangle.

Minimal pseudo-Algebraic Proof

Construct a square on the hypotenuse with the triangle inside the square. Make three further copies of the triangle, each rotated from the original about the square's centre so as to make its hypotenuse be a side of the square. These do not over-lap, though they abut one another. They enclose a (blue) central square whose side is the difference between the two perpendicular sides of the triangle.

Cut out two adjacent copies (magenta and green) of the triangle: translate each to the other side of the square, attaching its hypotenuse to that of the copy originally opposite to it. The resulting figure has one concave corner where the central square on the difference meets the longer perpendicular edge of one of the (blue) un-moved copies of the triangle: continue the edge of the central square through this corner, extending it until it meets the far edge of the figure (dashed line). The resulting line cuts the figure into two squares, whose sides are the two perpendicular sides of the triangle.

i.e. h.h = 2.a.c + (a−c).(a−c) = a.a + c.c

Details: The two non-right angles of the triangle sum to a right angle. Thus, at each corner of the hypotenuse-square, two copies of the triangle abut each other, rather than over-lapping (or leaving a gap), along their respective non-hypotenuse edges into the corner; for one its short edge, for the other its long edge. The latter over-runs the former by the difference to form one side of the small square enclosed by the four triangles.

Traditional pseudo-Algebraic Proof

Construct a square on the hypotenuse with the triangle outside the square. Make three further copies of the triangle, each rotated from the original about the square's centre so as to make its hypotenuse be a side of the square. The square is thus surrounded by copies of the triangle to form a big square whose side is equal to the sum of the perpendicular sides of the triangle.

Chose two adjacent copies of the triangle: they have a vertex in common, at one end of the hypotenuse of each. Rotate each through a right angle, about the other end of its hypotenuse, in the sense which moves it into the original square on the hypotenuse. Their final positions do not overlap, though they abut along an edge: extend that edge back towards the edge of the big square on which the moved triangles' shared vertex originally lay. The resulting line cuts the portion of the big square not now covered by the triangles into two squares, whose sides are the perpendicular sides of the triangle. The close similarity between this proof and the previous is no accident.

i.e. 2.a.c + h.h = (a+c).(a+c) = 2.a.c + a.a + c.c

The first of this pair of diagrams is, if I remember correctly, the figure which has been found on a shard of pottery from one of the ancient civilisations of India (far predating the earliest possible date for a Greek Pythagoras), with the word for Aha! in Sanskrit written under it. The ancient Indians clearly took the second diagram for granted. Thanks to Piers Bursill-Hall for teaching me about, and to Marc Pienaar for reminding me of, this proof ;^)

Integer Pythagorean triangles

The ancient greeks were particularly interested in one crucial implication of Pythagoras' theorem: the existence of lengths which are not rationally commensurate. Given a line, one can make arbitrarily long lines by repeating it and arbitrarily short lines by sub-dividing it into equal parts; combining these operations one may make a line of length arbitrarily close to any length one cares to nominate – but not all lengths can be arrived at from a given one by duplication and equal sub-division. Lines of lengths one can obtain from a given line by such scaling and shrinking are said to be rationally commensurate with the given line. The simple case of a right angle triangle with equal non-hypotenuse sides has a hypotenuse which is not rationally commensurate with the other two sides; the hypotenuse is √2 times as long as the others.

The question then arises: which right-angle triangles do have rationally commensurate sides ? (One may likewise pose the corresponding question for non-right angles.) 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; and it suffices to consider the case where our right-angle triangle's lengths are all whole multiples of some given unit. The question thus reduces to finding cases where the sum of two perfect squares (that is, squares of whole numbers) is itself a perfect square.

If we use a fraction of our unit length, of which each side of the triangle is a multiple, as substitute unit, then the same triangle will yield another pair of perfect squares whose sum is a perfect square; indeed, if n.n +m.m = k.k then, for any natural i, 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 solution. Conversely: if the two integers whose squares we're adding to obtain that of a third share a common factor, then the third must also share the common factor. Indeed, if one of the first two shares a common factor with the third, we can deduce (by re-writing n.n +m.m = k.k as k.k −m.m = n.n) that the other of the first two also shares that factor. It thus suffices to solve our problem for the case where no two sides have a common factor: the three sides's lengths are mutually coprime.

If we do our arithmetic module 8 (that is, considering any two numbers to be equivalent if the difference between them is a multiple of 8) then all odd numbers have square 1 (because (2.i+1).(2.i+1) = 4.(i+1).i +1 and one of i, i+1 must be even, so 4.(i+1).i must be a multiple of 8) and all even numbers have square 0 or 4 (i.e. a multiple of 4). Thus our sum of two squares equal to a third is necessarily 0 + 0 = 0, 4 + 0 = 4, 4 + 4 = 0 or 0 + 1 = 1. All but the last leave all three sides with a common factor of 2, and we're only looking at the case where there is no common factor; so we can infer that any mutually coprime solution must have a hypotenuse of odd length, with one of the other two sides being even and the other odd. Furthermore, the even side's square must be 0 modulo 8, implying that the even side must be a multiple of 4.

So let the hypotenuse be k times the largest unit of which all three sides are whole multiples, with the even side being 4.n times that unit and the odd side being m units long. Then k.k = 16.n.n +m.m and (k −4.n).(k +4.n) = m.m. Since k is odd, k ±4.n are both odd, so do not share 2 as a common factor; if their highest common factor is i, then i must also be a factor of both their sum, 2.k, and their difference, 8.n; since i is odd it must then also be a factor of k and of n, implying that it is a repeated factor of m.m and hence a factor of m; since our sides are coprime, we can infer that i = 1 and so that k ±4.n are coprime. Thus each factor of m.m is a factor of either k +4.n or k −4.m, but not of both; since m.m = (k −4.n).(k +4.n) is a perfect square, we may thus infer that each of k ±4.n is an odd perfect square.

Since, as noted above, all odd squares are equal to 1 modulo 8, pairs of odd perfect squares always differ by a multiple of 8, the 8.n difference between k ±4.n above; so take k −4.n = (2.u +1).(2.u +1) = 4.(u+1).u +1, k +4.n = 4.(v+1).v +1, whence k =2.(u.(u+1) +(v+1).v) +1, 2.n = v.(v+1) −u.(u+1) and m = (2.u+1).(2.v+1). For any naturals u, v we have

m.m +16.n.n

= (2.u+1).(2.v+1).(2.u+1).(2.v+1) + 4.(v.(v+1) −(u+1).u).(v.(v+1) −(u+1).u)

= (4.(u+1).u +1).(4.(v+1).v +1) +4.(v.v.(v+1).(v+1) +u.u.(u+1).(u+1) −2.u.(u+1).(v+1).v)

= 4.v.v.(v+1).(v+1) +4.u.u.(u+1).(u+1) +8.u.(u+1).(v+1).v +4.u.(u+1) +4.v.(v+1) +1

= 4.(u.(u+1) +(v+1).v).(u.(u+1) +(v+1).v) +4.(u.(u+1) +(v+1).v) +1

= (2.(u.(u+1) +(v+1).v) +1).(2.(u.(u+1) +(v+1).v) +1)

= k.k

Thus any u and v yield a valid integer pythagorean triangle (though not necessarily a coprime one – [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); and every such triangle is a whole multiple of some such. Thus every pythagorean triangle with rationally commensurate sides has sides equal to the result of duplicating some unit length

• 2.(v.(v+1) +(u+1).u) +1 (for the hypotenuse),
• 2.(v.(v+1) −(u+1).u) and
• (2.u+1).(2.v+1)

times, for some (not necessarily unique) naturals u and v. Note that the hypotenuse is always one more than a multiple of four (since both u.(u+1) and v.(v+1) are even). 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 left; these are all the examples with 0 ≤ u < v < 12.

Written by Eddy.