1.4: Pythagoras' Theorem

From here on the idea of “mental skills” tends to refer to ways of thinking rather than to doing everything in your head.

Pythagoras’ Theorem is one of the first truly surprising results in school mathematics: it is hard to see why anyone would think of “adding the squares of the two shorter sides”. Despite the apparent attribution to a named person (Pythagoras), the origin of the theorem, and its proof, are unclear. There certainly was someone called Pythagoras (around 500 BC). But the main ancient references to him were written many hundreds of years after he died, and are not very reliable. The truth is that we know very little about him, or his theorem. The proof in Problem 18 below appeared in Book I of Euclid’s thirteen books of Elements (written around 300 BC – two hundred years after Pythagoras). Much that is said (wrongly) to stem from Pythagoras is attributed in some sources to the Pythagoreans – a loose term which refers to any philosopher in what is seen as a tradition going back to Pythagoras. (This is a bit like interpreting anything called Christian in the last 2000 years as stemming directly from Christ himself.)

Clay tablets from around 1700 BC suggest that some Babylonians must have known “Pythagoras’ Theorem”; and it is hard to see how one could know the result without having some kind of justification. But we have no evidence of either a clear statement, or a proof, at that time. There are also Chinese texts that refer to Pythagoras’ Theorem (or as they call it, “Gougu”), which are thought to have originated BC – though the earliest surviving edition is from the 13th century AD. There is even an interesting little book by Frank Swetz, with the tongue-in-cheek title Was Pythagoras Chinese?.

The history may be confused, but the result – and its Euclidean proof – embodies something of the surprise and elegance of the very best mathematics. The Euclidean proof is included here partly because it is one that can, and should, be remembered (or rather, reconstructed – once one realises that there is really only one possible way to split the “square on the hypotenuse” in the required way). But, as we shall see, the result also links to exact mental calculation with surds, to trigonometry, to the familiar mnemonic “CAST”, to the idea of a “converse”, to sums of two squares, and to Pythagorean triples.

1.4.1: Pythagoras' Theorem, Trig for Special Angles, and CAST

Problem 18 (Pythagoras' Theorem)

Let \(\triangle ABC\) be a right angled triangle, with a right angle at \(C\). Draw the squares \(ACQP\), \(CBSR\), and \(BAUT\) on the three sides, external to \(\triangle ABC\). Use the resulting diagram to prove in your head that the square \(BAUT\) on \(BA\) is equal to the sum of the other two squares by:

• drawing the line through \(C\) perpendicular to \(AB\), to meet \(AB\) at \(X\) and \(UT\) at \(Y\)
• observing that \(PA\) is parallel to \(QCB\), so that \(\triangle ACP\) (half of the square \(ACQP\), with base \(AP\) and perpendicular height \(AC\)) is equal in area to\(\triangle ABP\) (with base \(AP\) and the same perpendicular height)
• noting that \(\triangle ABP\) is SAS-congruent to \(\triangle AUC\), and that \(\triangle AUC\) is equal in area to \(\triangle AUX\) (half of rectangle \(AUYX\), with base \(AU\) and height \(AX\)).
• whence \(ACQP\) is equal in area to rectangle \(AUYX\)
• similarly \(BCRS\) is equal in area to \(BTYX\).

The proof in Problem 18 is the proof to be found in Euclid’s Elements Book 1, Proposition 47. Unlike many proofs,

• it is clear what the proof depends on (namely SAS triangle congruence, and the area of a triangle), and
• it reveals exactly how the square on the hypotenuse \(AB\) divides into two summands – one equal to the square on \(AC\) and one equal to the square on \(BC\).

Problem 19

(a) Use Pythagoras’ Theorem in a square \(ABCD\) of side \(1\) to show that the diagonal \(AC\) has length \(\sqrt{2}\). Use this to work out in your head the exact values of \(\sin{45^{\circ}}, \cos{45^{\circ}}, \tan{45^{\circ}}\).

(b) In an equilateral triangle \(\triangle ABC\) with sides of length \(2\), join \(A\) to the midpoint \(M\) of the base \(BC\). Apply Pythagoras’ Theorem to find \(AM\). Hence work out in your head the exact values of \(\sin{30^{\circ}}, \cos{30^{\circ}}, \tan{30^{\circ}}, \sin{60^{\circ}}, \cos{60^{\circ}}, \tan{60^{\circ}}\).

(c) (i) On the unit circle with centre at the origin \(O\) : \((0, 0)\), mark the point \(P\) so that \(P\) lies in the first quadrant, and so that \(OP\) makes an angle \(\theta\) with the positive \(x\)-axis (measured anticlockwise from the positive \(x\)-axis). Explain why \(P\) has coordinates (\(\cos{\theta}, \sin{\theta})\).

(ii) Extend the definitions of \(\cos{\theta}\) and \(\sin{\theta}\) to apply to angles beyond the first quadrant, so that for any point \(P\) on the unit circle, where \(OP\) makes an angle \(\theta\) measured anticlockwise from the positive \(x\)-axis, the coordinates of \(P\) are \((\cos{\theta}, \sin{\theta})\). Check that the resulting functions \(\sin\) and \(\cos\) satisfy:

• \(\sin\) and \(\cos\) are both positive in the first quadrant,
• \(\sin\) is positive and \(\cos\) is negative in the second quadrant,
• \(\sin\) and \(\cos\) are both negative in the third quadrant, and
• \(\sin\) is negative and \(\cos\) is positive in the fourth quadrant.

(iii) Use (a), (b) to calculate the exact values of \(\cos{315^{\circ}}, \sin{225^{\circ}}, \tan{210^{\circ}}, \cos{120^{\circ}}, \sin{960^{\circ}}, \tan{( -135^{\circ})}\).

(d) Given a circle of radius \(1\), work out the exact area of a regular n-gon inscribed in the circle:

(i) when \(n = 3\)

(ii) when \(n = 4\)

(iii) when \(n = 6\)

(iv) when \(n = 8\)

(v) when \(n = 12\).

(e) Given a circle of radius \(1\), work out the area of a regular n-gon circumscribed around the circle:

(i) when \(n = 3\)

(ii) when \(n = 4\)

(iii) when \(n = 6\)

(iv) when \(n = 8\)

(v) when \(n = 12\).

Knowing the exact values of \(\sin, \cos\) and \(\tan\) for the special angles \(0^{\circ}, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}\) is like knowing one’s tables. In particular, it allows one to evaluate trigonometric functions mentally for related angles in all four quadrants (using the CAST mnemonic – C being in the SE of the unit circle, A in the NE quadrant, S in the NW quadrant, and T in the SW quadrant – to remind us which functions are positive in each quadrant). These special angles arise over and over again in connection with equilateral triangles, squares, regular hexagons, regular octagons, regular dodecagons, etc., where one can use what one knows to calculate exactly in geometry.

Problem 20

(a) Use Pythagoras’ Theorem to calculate the exact length of the diagonal \(AC\) in a square \(ABCD\) of side length \(2\).

(b) Let \(X\) be the centre of the square \(ABCD\) in part (a). Draw lines through \(X\) parallel to the sides of \(ABCD\) and so divide the large square into four smaller squares, each of side \(1\). Find the length of the diagonals \(AX\) and \(XC\).

(c) Compare your answers to parts (a) and (b) and your answer to Problem3(b)(i).

Pythagoras’ Theorem holds the key to calculating exact distances in the plane. To calculate distances on the Earth’s surface one needs a version of Pythagoras for “right angled triangles” on the sphere. We address this in Chapter 5.

1.4.2: Converses and Pythagoras' Theorem

Each mathematical statement of the form

“if . . . (Hypothesis H),

then . . . (Consequence C)”

has a converse statement – namely

“if C,

then H”.

If the first statement is true, there is no a priori reason to expect its converse to be true. For example, part (c) of Problem 25 below proves that

“if an integer has the form \(4k + 3\),
then it cannot be written as the sum of two squares”.

However, the converse of this statement

“if an integer cannot be written as a sum of two squares,

then it has the form \(4k + 3\)”

is false – since 6 cannot be written as the sum of two squares.

Despite this counterexample, whenever we prove a standard result, it makes sense to ask whether the converse is also true. For example,

“if \(PQRS\) is a parallelogram, then opposite angles are equal: \( \angle P = \angle R\), and \( \angle Q = \angle S\)” (see Problem 157(ii)).

However you may not have considered the truth (or otherwise) of the converse statement:

If \(ABCD\) is a quadrilateral in which opposite angles are equal \(( \angle A = \angle C\) and \( \angle B = \angle D)\), is it true that \(ABCD\) has to be a parallelogram?

The next problem invites you to prove the converse of Pythagoras’ Theorem. You should not use the Cosine Rule, since this is a generalisation of both Pythagoras’ Theorem and its converse.

Problem 21

Let \(ABC\) be a triangle. We use the standard labelling convention, whereby the side \(BC\) opposite \(A\) has length \(a\), the side \(CA\) opposite \(B\) has length \(b\), and the side \(AB\) opposite \(C\) has length \(c\).

Prove that, if \(c^{2} = a^{2} + b^{2}\), then \(\angle BCA\) is a right angle.

1.4.3: Pythagorean Triples

The simplest example of a right angled triangle with integer length sides is given by the familiar triple \(3, 4, 5\):

\(3^{2} + 4^{2} = 5^{2}\).

Any such integer triple is called a Pythagorean triple.

The classification of all Pythagorean triples is a delightful piece of elementary number theory, which is included in this chapter both because the result deserves to be memorised, and because (like Pythagoras’ Theorem itself) the proof only requires one to juggle a few simple ideas that should be part of one’s armoury.

Pythagorean triples arise in many contexts (e.g. see the text after Problem180). The classification given here shows that Pythagorean triples form a family depending on three parameters \(p, q, s\) (in which s is simply a “scaling” parameter, so the most important parameters are \(p, q\)). As a warm-up we consider two “one-parameter subfamilies” related to the triple \(3, 4, 5\).

Problem 22

Suppose \(a^{2} + b^{2} = c^{2}\) and that \(b\), \(c\) are consecutive integers.

(a) Prove that \(a\) must be odd – so we can write it as \(a = 2m + 1\) for some integer \(m\).

(b) Prove that \(c\) must be odd – so we can write it as \(c = 2n + 1\) for some integer \(n\). Find an expression for \(n\) in terms of \(m\).

Problem 22 reveals the triple \((3,4,5)\) as the first instance \((m = 1)\) of a one-parameter infinite family of triples, which continues

\((5,12,13) (m = 2), (7,24,25) (m = 3), (9,40,41) (m = 4),...,\)

whose general term is

\((2m + 1, 2m(m +1), 2m (m + 1) +1)\).

The triple \(3, 4, 5\) is also the first member of a quite different “one-parameter infinite family” of triples, which continues

\((6,8,10), (9,12,15), ....\)

Here the triples are scaled-up versions of the first triple \((3, 4, 5)\).

In general, common factors simply get in the way:

If \(a^{2} + b^{2} = c^{2}\) and \(HCF(a,b) = s\), then \(s^{2}\) divides \(a^{2} + b^{2}\), and \(a^{2} + b^{2} = c^{2}\); so \(s\) divides \(c\).

And if \(a^{2} + b^{2} = c^{2}\) and \(HCF(b,c) = s\), then \(s^{2}\) divides \(c^{2} - b^{2} = a^{2}\), so \(s\) divides \(a\).

Hence a typical Pythagorean triple has the form \((sa, sb, sc)\) for some scale factor \(s\), where \((a, b, c)\) is a triple of integers, no two of which have a common factor: any such triple is said to be primitive (that is, basic – like prime numbers). Every Pythagorean triple is an integer multiple of some primitive Pythagorean triple. The next problem invites you to find a simple formula for all primitive Pythagorean triples.

Problem 23

Let \((a, b, c)\) be a primitive Pythagorean triple.

(a) Show that \(a\) and \(b\) have opposite parity (i.e. one is odd, the other even) – so we may assume that \(a\) is odd and \(b\) is even.

(b) Show that

\(\left( \frac{b}{2} \right)^{2} = \left( \frac{c-a}{2} \right) \left( \frac{c+ a}{2} \right)\),

where

\(HCF \left( \frac{c -a}{2}, \frac{c +a }{2}\right) = 1\)

and \(\frac{c-a}{2}, \frac{c+a}{2}\) have opposite parity.

(c) Conclude that

\(\frac{c+a}{2} = p^{2}\) and \(\frac{c-a}{2} = q^{2}\)

where \(HCF(p,q) = 1\) and \(p\) and \(q\) have opposite parity, so that \(c = p^{2} + q^{2}, a = p^{2} - q^{2}, b = 2pq\).

(d) Check that any pair \(p, q\) having opposite parity and with \(HCF(p,q) = 1\) gives rise to a primitive Pythagorean triple

\(c = p^{2} + q^{2}, a = p^{2} - q^{2}, b = 2pq\)

satisfying \(a^{2} + b^{2} = c^{2}\).

Problem 24

The three integers \(a = 3, b = 4, c = 5\) in the Pythagorean triple \((3, 4, 5)\) form an arithmetic progression: that is, \(c - b = b - a\). Find all Pythagorean triples \((a, b, c)\) which form an arithmetic progression – that is, for which \(c - b = b - a\).

1.4.4: Sums of Two Squares

The classification of Pythagorean triples tells us precisely which squares can be written as the sum of two squares. We now turn to the wider question: “Which integers are equal to the sum of two squares?”

Problem 25

(a) Which of the prime numbers \( \lt 100\) can be written as the sum of two squares?

(b) Find an easy way to immediately write \((a^{2} + b^{2})(c^{2} + d^{2})\) in the form \((x^{2} + y^{2})\). (This shows that the set of integers which can be written as the sum of two squares is “closed” under multiplication.)

(c) Prove that no integer (and hence no prime number) of the form \(4k + 3\) can be written as the sum of two squares.

(d) The only even prime number can clearly be written as a sum of two squares: \(2 = 1^{2} + 1^{2}\). Euler (1707–1783) proved that every odd prime number of the form \(4k + 1\) can be written as the sum of two squares in exactly one way. Find all integers \(\lt 100\) that can be written as a sum of two squares.

(e) For which integers \(N \lt 100\) is it possible to construct a square of area \(N\), with vertices having integer coordinates

In Problem 25 parts (a) and (d) you had to decide which integers \(\lt 100\) can be written as a sum of two squares as an exercise in mental arithmetic. In part (b) the fact that this set of integers is closed under multiplication turned out to be an application of the arithmetic of norms for complex numbers. Part (e) then interpreted sums of two squares geometrically by using Pythagoras’ Theorem on the square lattice. These exercises are worth engaging in for their own sake. But it may also be of interest to know that writing an integer as a sum of two squares is a serious mathematical question – and in more than one sense.

Gauss (1777–1855), in his book Disquisitiones arithmeticae (1801) gave a complete analysis of when an integer can be represented by a ‘quadratic form’, such as \(x^{2} + y^{2}\) (as in Problem 25) or \(x^{2} - 2y^{2}\) (as in Problem 54(c) in Chapter 2).

A completely separate question (often attributed to Edward Waring (1736–1798)) concerns which integers can be expressed as a \(k\)th power, or as a sum of \(n\) such powers. If we restrict to the case \(k = 2\) (i.e. squares), then:

• When \(n = 2\), Euler (1707–1783) proved that the integers that can be written as a sum of two squares are precisely those of the form

\(m^{2} \times p_{0} \times p_{1} \times p_{2} \times … \times p_{s}\)

where \(p_{0} = 1\) or \(2\), and \(p_{1} \lt p_{2} \lt … \lt p_{s}\) are odd primes of the form \(4l + 1\)

• When \(n = 3\), Legendre (1752–1833) and Gauss proved between them that the integers which can be written as a sum of three squares are precisely those that are not of the form \(4^{m} \times (8l +7)\)
• When \(n = 4\), Lagrange (1736–1813) had previously proved that every positive integer can be written as a sum of four squares.