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1 PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient Babylonian tablet listing 15 Pythagorean triples. It is called Plimpton 3 (George Arthur Plimpton donated it to Columbia University). More information about it can be found at [1]. Plimpton 3 Some Pythagorean triples are scalar multiples of other triples: (6, 8, 10) is twice (3, 4, 5). We call a triple (a, b, c) primitive when the three integers have no common factor. For any triple (a, b, c), if d is the greatest common divisor of all three terms then (a/d, b/d, c/d) is a primitive triple and the original triple is a scalar multiple of this, so finding all Pythagorean triples is basically the same as finding all primitive Pythagorean triples. Our goal is to describe the primitive Pythagorean triples. We will be using different characterizations of primitive triples, as described in the following lemma. Lemma 1.1. For a Pythagorean triple (a, b, c), the following properties are equivalent: 1

2 KEITH CONRAD (1) a, b, and c have no common factor, i.e., the triple is primitive, () a, b, and c are pairwise relatively prime, (3) two of a, b, and c are relatively prime. Proof. To show (1) implies (), we prove the contrapositive. If () fails then two of a, b, and c have a common prime factor. Suppose a prime p divides a and b. Then c = a + b is divisible by p, and since p c also p c, so p is a common factor of a, b, and c. A similar argument works using a common prime factor of a and c or of b and c. Thus (1) fails. It is easy to see that () implies (3) and (3) implies (1). Theorem 1.. If (a, b, c) is a primitive Pythagorean triple then one of a or b is even and the other is odd. Taking b to be even, a = k l, b = kl, c = k + l for integers k and l with k > l > 0, (k, l) = 1, and k l mod. Conversely, for such integers k and l the above formulas yield a primitive Pythagorean triple. k l a b c Table 1. Examples of Primitive Triples Table 1 lists examples of the values of k and l for some primitive triples. Notice in particular that the triple (8, 15, 17) is recorded in the table as (15, 8, 17) because of the convention in Theorem 1. that the middle term of the triple is even. Since k l mod, one of k and l is even and the other is odd. The first two rows in Table 1 shows that either k or l can be even; it depends on the particular triple (a, b, c). How it depends on the triple will be seen at the end of Section 3. Let s first check that the formula in Theorem 1. always yields primitive Pythagorean triples. For all k and l in Z, the formula (k l ) + (kl) = (k + l ) is true, so (k l, kl, k + l ) is a Pythagorean triple when k > l > 0. When (k, l) = 1 and k l mod, let s check (k l, kl, k + l ) is a primitive triple. This follows (by Lemma 1.1) from showing k l and k + l are relatively prime. If d is a (positive) common divisor of k l and k + l then d divides their sum and difference, which are k and l. Since k + l and k l are odd, d is odd, so d k and d l. Since k and l are relatively prime, so are k and l, so d = 1. If we relax any of the conditions on k and l in Theorem 1. we won t get a primitive Pythagorean triple. For instance, taking k = 3 and l = 1 produces the triple (8, 6, 10). The condition k l mod is violated here. Next we will show, in two ways, that a primitive Pythagorean triple arises in the form given by Theorem 1.: using unique factorization in Z (an algebraic proof) and using intersections of lines and circles (a geometric proof). Then we will put the formula to work and see some generalizations.

3 PYTHAGOREAN TRIPLES 3. Proof of Theorem 1. by algebra To show that one of a and b is odd and the other is even, suppose a and b are both odd. Then a b 1 mod 4, so c = a + b mod 4. However, mod 4 is not a square. Thus a or b is even. If both a and b are even the triple (a, b, c) is not primitive, a contradiction. That shows one of a or b is odd and the other is even, so c = a + b is odd. Taking b to be the even choice among a and b, rewrite the equation a + b = c as b = c a = (c + a)(c a). Both a and c are odd, so c + a and c a are even. Dividing by 4, (.1) ( b ) = c + a c a. The two integers on the right, (c+a)/ and (c a)/, are relatively prime: if d is a common divisor then it divides their sum and difference, which are c and a, and those are relatively prime, so d = 1. Since c > a > 0, both factors on the right side of (.1) are positive. Since they are relatively prime and their product is a square, unique factorization in Z tells us that (.) c + a = k, c a for some positive integers k and l. Since, as we noted already, (c + a)/ and (c a)/ are relatively prime, also k and l are relatively prime. Adding and subtracting the equations in (.), so = l c = k + l, a = k l, ( ) b = k l = b = 4k l = b = kl, the last step holding because b, k, and l are positive. It remains to check that k l mod. Since k and l are relatively prime, they are not both even. If they were both odd then k + l, kl, and k l would all be even, contradicting primitivity of (a, b, c). 3. Proof of Theorem 1. by geometry Pythagorean triples are connected to points on the unit circle: if a + b = c then (a/c) + (b/c) = 1. So we get a rational point (a/c, b/c) on the unit circle x + y = 1. Draw the line through the points ( 1, 0) and (a/c, b/c), as in the figure below. Its slope is m = b/c 1 + a/c = b a + c and the line passes through ( 1, 0), so the equation of the line is y = m(x + 1).

4 4 KEITH CONRAD y (a/c, b/c) ( 1, 0) x Substituting the equation for the line into the equation for the unit circle gives us an equation whose roots are the x-coordinates of the two points on both the line and the circle: so 1 = x + y = x + (m(x + 1)) = (1 + m )x + m x + m, (3.1) 0 = (1 + m )x + m x + m 1. Since ( 1, 0) and (a/c, b/c) lie on both the line and the circle, the two roots of (3.1) are 1 and a/c. The sum of the roots is m /(1 + m ), so 1 + a c = m 1 + m. Thus a c = 1 m 1 + m = 1 m 1 + m. Since (a/c, b/c) is on the line y = m(x + 1), b ( a ) ( ) 1 m c = m c + 1 = m 1 + m + 1 = m 1 + m. Since m is the slope of a line connecting ( 1, 0) to a point on the unit circle in the first quadrant, 0 < m < 1. Write m = b/(a + c) in reduced form as m = l/k with positive relatively prime integers k and l. (Why don t we write m = k/l? Because the choice we made will match the formulas as written in Theorem 1..) Since m < 1, k > l. The formulas for a/c and b/c in terms of m become (3.) and (3.3) a c = 1 (l/k) 1 + (l/k) = k l k + l b c = (l/k) 1 + (l/k) = kl k + l. By definition, k and l are relatively prime. Let s check that k l mod. If k and l have the same parity then they must both be odd (they are not both even since they are

5 PYTHAGOREAN TRIPLES 5 relatively prime). Then the fraction kl/(k + l ) has an even numerator and denominator, so it simplifies further to kl k + l = kl (k + l )/. Here the numerator is odd, so the reduced form representation b/c must have an odd numerator. But b is even by hypothesis, so we have a contradiction. From the end of Section 1, the three numbers k l, kl, and k +l form a Pythagorean triple and the first and third numbers are relatively prime, so the triple is primitive by Lemma 1.1. Therefore the fractions on the right in (3.) and (3.3) are in reduced form. The fractions a/c and b/c are also in reduced form since the triple (a, b, c) is primitive, so the (positive) numerators and denominators match: a = k l, b = kl, c = k + l. This concludes the geometric proof of Theorem 1.. Remark 3.1. While this derivation was motivated by geometry, the calculations were pure algebra and we could formulate them without references to a picture. That is, if we have numbers x and y with x + y = 1 and x 1, we can (without motivation) define the number m to fit the condition y = m(x + 1). Then substituting this into x + y = 1 tells us x + m (x + 1) = 1, and expanding the square gives 0 = (1 + m )x + m x + (m 1) = (1 + m )(x + 1) ( ) x + m m, so from x 1 we have x = (1 m )/(1 + m ) and y = m(x + 1) = m/(1 + m ). One nice consequence of our geometric derivation of the formula for primitive Pythagorean triples (a, b, c) is that it tells us which of k or l will be even. The formula m = b a + c = l k says that l/k is the reduced form fraction for b/(a + c). We know either l or k is even. We will have k even when a+c is divisible by a higher power of than b is, and l is even when b is divisible by a higher power of than a + c is. For instance, in the triple (3, 4, 5), a + c = 8 and b = 4, so k is even. (Here k = and l = 1.) The triple (5, 1, 13) has a + c = 18 = 9 and b = 1 = 3, so l is even. (Here k = 3 and l =.) 4. Applications Using the parametric formula for primitive Pythagorean triples, we can address questions concerning relations among the sides of a primitive right triangle. The most well-known Pythagorean triples, (3, 4, 5), and (5, 1, 13), have consecutive terms. We ask: what are all the Pythagorean triples (a, b, c) with a pair of consecutive terms (either a and b or b and c)? Any such triple is automatically primitive since consecutive integers are relatively prime. First we consider the case when a and b (the two legs) are consecutive. For a and b to differ by 1 means (k l ) kl = ±1.

6 6 KEITH CONRAD This can be rewritten as (k l) l = ±1, where k l is positive and odd and l is positive. Conversely, if we have a solution to x y = ±1 in positive integers x and y, then necessarily x is odd and (x, y) = 1. Let k = x + y and l = y, so k > l > 0, (k, l) = 1, and k l mod. Thus (k l, kl, k + l ) is a primitive triple, so finding Pythagorean triples whose legs differ by 1 is the same as finding positive integer solutions to the Pell equation x y = ±1. Some examples are in Table. x y k l a b c Table. Consecutive Legs Even if two legs in a primitive triple don t differ by 1, the formula for their difference is still (k l) l, so the possible differences between legs in a primitive triple are precisely the odd values of x y for positive integers x and y. Not every odd number can arise in this way, e.g., the equation x y = 5 has no integral solution (why?), so no primitive Pythagorean triple has its legs differing by 5. Important results in number theory have developed from the study of values of x dy, where d is a fixed nonsquare integer. The special case d = shows up in the setting of Pythagorean triples, but other values of d are important for other problems. Turning now to a leg and hypotenuse which differ by 1, the story is much simpler. The hypotenuse is odd, so it can only differ by 1 from the even leg. This difference is k + l kl = (k l), which is an odd square. This is 1 only if k = l + 1 (recall k > l by convention), leading to the triple (l + 1, l + l, l + l + 1). The first four examples are in Table 3. l l + 1 l + l l + l Table 3. Consecutive Leg and Hypotenuse Now we look at a connection between Pythagorean triples and reducible quadratic polynomials, taken from []. While we have x + 4x + 3 = (x + 1)(x + 3) and x + 4x 3 is irreducible, x + 5x + 6 = (x + )(x + 3) and x + 5x 6 = (x 1)(x + 6).

7 PYTHAGOREAN TRIPLES 7 Are there more examples like this second one, where integral polynomials x + mx + n and x + mx n both factor? Here m and n are nonzero. If we ask what happens when the sign of m changes, the answer is not interesting. Indeed, if x + mx + n = (x r 1 )(x r ) then x mx + n = (x + r 1 )(x + r ), so either both of these factor (with integral coefficients) or both don t. Therefore in our question on x + mx ± n, we may assume m > 0. Since the issue is about sign changes on n, we may take n > 0 also. By the quadratic formula, the roots of x + mx ± n are m ± m ± 4n, which are integers exactly when m ± 4n =. (Note m ± 4n m mod, so the numerator is even when the discriminant is a perfect square.) So we can factor x + mx + n and x + mx n if and only if m 4n = d, m + 4n = e, d and e Z. Then d + e = m, so d e mod. Solving, m = d + e ( ) e + d ( ) e d = +. Thus we have a Pythagorean triple (without a specified even term) ( e d, e + d ), m e d, < e + d < m. As an exercise, show this Pythagorean triple is primitive if and only if (m, n) = 1. There is a one-to-one correspondence given by Pythagorean triples (a, b, c) with a < b < c reducible x + mx ± n, (a, b, c) x + cx ± ab ( e d, x + mx ± n, e + d ), m, with m 4n = d and m + 4n = e. The table below shows some corresponding Pythagorean triples and reducible x + mx ± n. In particular, the example x + 5x ± 6 corresponds to the (3, 4, 5) triple and thus is the simplest example. a b c m n x + mx + n x + mx n (x + )(x + 3) (x 1)(x + 6) (x + 3)(x + 10) (x )(x + 15) (x + 5)(x + 1) (x 3)(x + 0) (x + 14)(x + 15) (x 6)(x + 35) Using the formula for primitive Pythagorean triples, we can now write a formula for all the reducible x + mx ± n when (m, n) = 1: x + (k + l )x ± kl(k l ), where k > l > 0, (k, l) = 1, and k l mod. As an exercise, work out the factorization of this polynomial explicitly in terms of k and l. 5. Generalizations The two derivations of the formula for primitive Pythagorean triples, one algebraic and the other geometric, are each worthwhile: they let us extend Theorem 1. in different directions.

8 8 KEITH CONRAD The algebraic proof of Theorem 1. carries over to Pythagorean triples of polynomials in R[T ]. These are polynomials f(t ), g(t ), and h(t ) in R[T ] which satisfy f(t ) + g(t ) = h(t ). One example of such a triple is (T 1, T, T + 1). Can we describe all polynomial Pythagorean triples? There are two features of the Pythagorean triples of integers which play no role when we look at Pythagorean triples of polynomials: (1) The special attention to positive solutions of a + b = c is ignored. We allow any polynomial solutions to f + g = h with the only proviso being that f, g, and h are all nonzero. () The number is an invertible constant in R[T ], so the even/odd aspect that cropped up with integral Pythagorean triples simply drops out of consideration. Being divisible by is not an important feature in R[T ] since we can always divide by in R[T ]. We will call a Pythagorean triple of polynomials (f, g, h) primitive if the terms in it are pairwise relatively prime. This is the same as any two of the polynomials being relatively prime, and is also the same as (f, g, h) not being a non-constant multiple of another triple (polynomial analogue of Lemma 1.1). Theorem 5.1. The primitive Pythagorean triples (f(t ), g(t ), h(t )) in R[T ] are given by the formulas f(t ) = c(k(t ) l(t ) ), g(t ) = ±ck(t )l(t ), h = ±c(k(t ) + l(t ) ), where c R and k(t ) and l(t ) are relatively prime in R[T ]. Proof. This is left as an exercise in adapting the techniques in the algebraic proof of Theorem 1.. Note that if two polynomials are relatively prime and multiply to a squared polynomial, the two polynomials have to be squares up to constant multiple. This follows from unique factorization in R[T ]. The geometric proof of Theorem 1. has a benefit in calculus. First of all, this proof leads to a non-trigonometric parametrization of the points on the unit circle: any point (x 0, y 0 ) satisfying x + y = 1 other than ( 1, 0) has the form (5.1) x 0 = 1 m 1 + m, y 0 = m 1 + m where m = y 0 /(x 0 + 1). The formulas in (5.1) for x 0 and y 0 arise by looking at the intersection points of x + y = 1 and y = m(x + 1), where this linear equation describes any line through ( 1, 0) other than the vertical line. The correspondences ( ) 1 m (5.) m 1 + m, m 1 + m, (x, y) m = y 1 + x, between real numbers m and points on the unit circle (x, y) other than ( 1, 0) are inverses of each other, as the reader can check. Moreover, the formulas have arithmetic content: m is rational if and only if x 0 and y 0 are both rational: the formulas in (5.) each one have rational output if there is rational input. So (5.1) is well-suited to describe the rational points on x + y = 1. See Table 4. Compare this to the trigonometric parametrization (cos θ, sin θ), where nearly all rational points have messy corresponding angles θ.

9 PYTHAGOREAN TRIPLES 9 m x y 1/ 3/5 4/5 1/3 4/5 3/5 /5 1/9 0/9 7/9 16/65 63/65 Table 4. Rational points from slopes on x + y = 1 Comparing the rational parametrization of the unit circle in (5.1) and the trigonometric parametrization, we can write (with some assistance from (5.)) cos θ = 1 m m, sin θ = 1 + m 1 + m, where m = sin θ/(1 + cos θ). A point on the unit circle at angle θ relative to the origin and the positive x-axis makes an angle of θ/ relative to the point ( 1, 0) and the half-line to its right. Therefore m = tan(θ/). Now take a look in any calculus book for the method of integrating rational functions in sin θ and cos θ. The trick of the method is the u- substitution u = tan(θ/), which turns cos θ and sin θ into rational functions of u and any integral of a rational function of cos θ and sin θ into an integral of a rational function of u. Now we see the origin of this trick: use two different parametrizations of the unit circle to express an integral involving one parameterization (trig functions) as an integral involving the other parametrization (rational functions). If we take any conic in the plane (a curve of degree : an ellipse, hyperbola, or parabola) given by an equation having rational coefficients and we can find one rational point on it, then we can use slopes of lines through that point to parametrize all the other rational points on the conic. Here is an example using another circle. Theorem 5.. The rational points on the circle x + y = other than ( 1, 1) can be described by the formula ( ) 1 + m m m, m + m m + 1 where m Q. A rational point (x, y) other than ( 1, ±1) arises this way using m = (y + 1)/(x + 1), and ( 1, 1) arises this way using m = 1. Proof. The point ( 1, 1) lies on the circle. A non-vertical line through ( 1, 1) has the form y = m(x + 1) 1. Any line through ( 1, 1) intersects the circle in a second point, except for the tangent line y = x (where m = 1). y ( 1, 1) ( 1, 1) x

10 10 KEITH CONRAD To find the second point of intersection, substitute the equation of the line into the equation x + y = and solve for the two roots of the resulting quadratic in x: one root is 1, and we can find the other root by the same method as in the geometric proof of Theorem 1.. This will give the x-coordinate formula above, and substituting this into y = m(x + 1) 1 gives the y-coordinate. The only rational point on x + y = which we don t find by this method is ( 1, 1), which is the second intersection point of the circle with the vertical line through ( 1, 1). This vertical line is not described by an equation of the form y = m(x + 1) 1. The parameter value that gives the point ( 1, 1) itself is m = 1, which is the slope of the tangent line to x + y = at ( 1, 1) (touching the point ( 1, 1) twice, in a sense). Remark 5.3. Since x + y = if and only if ((x y)/) + ((x + y)/) = 1, the geometric proof of Theorem 1. yields (x y)/ = (1 m )/(1 + m ) and (x + y)/ = m/(1 + m ). Adding and subtracting these equations recovers the formulas in Theorem 5.. Armed with the description of the rational points on x + y = in Theorem 5., we can get a description of the primitive integral solutions of a + b = c, where primitive means a, b, and c have no common factor (equivalently, this means any two of a, b, and c are relatively prime, i.e., an analogue of Lemma 1.1 holds). In a primitive triple (a, b, c), a and b both must be odd, so c mod 8, hence c is odd as well. For Pythagorean triples, the middle term was chosen to be the even one. The analogue of that here is that we take c > 0 and pick the signs on a and b so that a b mod 4. With these conventions, we have a = k + kl l, b = l + kl k, c = k + l where (k, l) = 1, k l mod, and k > 0. Conversely, any triple defined by these formulas is a primitive solution to a + b = c with b a mod 4. k l a b c Table 5. Primitive Solutions to a + b = c Integers satisfying a + b = c are not the sides of a right triangle, but they have an arithmetic interpretation: since c = 1 (a + b ), c is the average of a and b. In other words, a, c, b is an arithmetic progression of squares (taking a < b ). From Table 5, squaring the entries in the a, b, and c columns yield three such progressions: 1, 5, 49 (common difference 4), 49, 89, 59 (common difference 40), and 1, 841, 1681 (common difference 840). There are infinitely many 3-term arithmetic progressions of squares, and we can find them all using the rational parametrization of the rational points on x +y =. What about 4-term arithmetic progressions of squares? References [1] W. Casselman, cass/courses/m446-03/pl3/pl3.html. [] J. L. Poet and D. L. Vestal, Jr., Curious Consequences of a Miscopied Quadratic, College Math. J. 36 (005),

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