# Pythagorean Triples and Rational Points on the Unit Circle

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1 Pythagorean Triles and Rational Points on the Unit Circle Solutions Below are samle solutions to the roblems osed. You may find that your solutions are different in form and you may have found atterns not listed here. Exloring Triangles Algebraically Take a moment and list some Pythagorean triles that you know. Are there any atterns in your list? For instance, in any Pythagorean trile, how many even entries are ossible? Are 0, 1,, and 3 all ossible? Can the hyotenuse be the only even side length? Justify your answer. The key to this roblem is that the square of an even number is even and the square of an odd number is odd. So, we cannot have all of a, b, and c being odd since if the sum or difference of two odd numbers must be even. So, 0 even numbers is imossible. Clearly, having all of a, b, and c as even numbers is ossible as 6, 8, 10) is a Pythagorean trile. Also, a single even is ossible since 3, 4, 5) is a trile. Since the sum or difference of even numbers is also even, it is imossible to have exactly two numbers in the trile be even. Thus, there can only be triles with a single even number or three even numbers. In the case of a single even number in a Pythagorean trile, it is imossible for the hyotenuse to be even. If a = j + 1 and b = k + 1 are the odd triangle lengths and c = m is the even hyotenuse length, then we have j + 1) + k + 1) = m) = 4j + 4j k + 4k + 1 = 4m = 4j + j + k + k) + = 4m The left-hand side is two more than a multile of 4 while the right-hand side is a multile of 4; this is imossible. 1

2 Notice that 3, 4, 5) and 6, 8, 10) are two triles that are, in some sense, very closely related. How? Can you find another trile that is related to 3, 4, 5) and to 6, 8, 10)? Can you use this to show that there are infinitely many right triangle triles? Exlain why your result is true. Multilying a given Pythagorean trile by a ositive integer, k yields another Pythagorean trile. That is, if a, b, c) is a Pythagorean trile, then so is ka, kb, kc) since ka) + kb) = k a + b ) = k c = kc). Since there are infinitely many ositive integers, each Pythagorean trile generates an infinite family of triles. Consider the following table of Pythagorean triles. a b c Send some time determining how these triles are related. Can you determine the next three triles in this attern? Be sure to check that your triles are Pythagorean triles!) The next three triles are 15, 11, 113), 17, 144, 145) and 19, 180, 181). It is obvious that c = b + 1 for each of these triles and that a is running through the odd numbers in order. Then, or b = 1 a 1). a + b = b + 1) = a = b + 1, Use the attern you found in the revious roblem to create a formula for these triles and will hence generate an infinite number of Pythagorean triles. Prove that your formula always gives a Pythagorean trile. From the revious roblem, we see that given an odd number a, we can create a Pythagorean trile: a, 1 a 1), 1 ) a 1) + 1 = a, 1 a 1), 1 ) a + 1).

3 We can even do a bit better and actually enumerate these triles by writing the odd number a as a = j + 1, with j = 1,, 3,... Then, we have j + 1, 1 j + 1) 1), 1 ) j + 1) + 1) = j + 1, j } {{ } + j, j } {{ } + j + 1 } {{ } a b c We check that this always is a Pythagorean trile: a + b = j + 1) + j + j) = 4j + 4j j 4 + 8j 3 + 4j = 4j 4 + 8j 3 + 8j + 4j + 1 = j + j + 1) = c A natural question at this oint is: Have now found ALL the Pythagorean triles? Unfortunately or fortunately, deending on your oint of view), the following table gives another attern of Pythagorean triles. a b c Send some time determining how these triles are related. Can you determine the next three triles in this attern? Be sure to check that your triles are Pythagorean triles!) The next three triles are 8, 195, 197), 3, 55, 57), and 36, 33, 35). The first thing we notice is that the a values are running through the ositive multiles of 4. We also notice that c = b + and that b and c are, resectively, 1 more than and 1 less than a erfect square. What erfect square? Well, c = b + imlies that a = b + ) b = 4b + 4 = b = 1 4 a 1. Then, for k = 1,, 3,..., we have a = 4k and so b = 1 4 4k) 1 = 4k 1 = k) 1 and c = k)

4 Use the attern you found in the revious roblem to create a formula for these triles and will hence generate another infinite list of Pythagorean triles. Prove that your formula always gives a Pythagorean trile. As we saw in the revious roblem, these triles are given by We check: }{{} 4k, k) 1, k) + 1). } {{ } } {{ } a b c a + b = 4k) + k) 1) = 16k + 16k 4 8k + 1 = 16k 4 + 8k + 1 = k) + 1) = c A General Formula for Pythagorean Triles We will see how the following formula is derived in the Rational Points on the Unit Circle section, but for now we will work with this. q, q, + q ) is a Pythagorean trile whenever and q are ositive integers with q >. Furthermore, every Pythagorean trile is similar to a trile of this form. Use this formula to find a few Pythagorean triles that you have not yet seen in this investigation. Many ossible answers, such as Prove that, if, q and k are ositive integers with q >, then kq ), kq, k + q )) is a Pythagorean trile. kq )) + kq) = k q ) + 4k q = k q 4 q q ) = k q 4 + q + 4 ) = k q + ) = k + q )) 4

5 Rational Points on the Unit Circle In this section, we investigate rational oints on the unit circle and make the link to Pythagorean triles. A oint x, y), on the unit circle is rational if x and y are rational numbers satisfying x + y = 1. Similarly, x, y) is an integer oint on the unit circle if x and y are integers with x + y = 1. Find all integer oints on the unit circle. How do you know that these are all of them? The only integer oints on the unit circle are 1, 0), 0, 1), 1, 0) and 0, 1). These are the only integer oints there are only four integer oints within distance 1 of the origin. Find some rational oints on the unit circle. For instance, can you have x = 1/ for a rational oint on the unit circle? Exlain why or why not. If x = 1/, then y = ± 1 1/) = ± 3/ is not rational. Some rational oints include 3/5, 4/5), 5/13, 1/13), and 8/17, 15/17). Do you notice anything familiar about these oints? Now, consider the integer oint 1, 0) on the unit circle. Draw a line through 1, 0) and any other oint x, y) on the unit circle. Go to htt://faculty.ithaca.edu/dabrown/docs/ythagorean/rational/ and download the file shown. Double-click to decomress and then load the file in a browser window. Drag the sloe slider to change the sloe of the this line. If the sloe of this line is m, then write down the equation of this line. The line between 1, 0) and x, y), with sloe m has equation y = mx + 1). Show that if x, y) is a rational oint on the unit circle, the sloe of the line above must also be rational if x 1. The sloe, m, of this line is m = y 0 x 1) = y x + 1. So, as long as x 1, this is the quotient of two rational numbers with non-zero denominator) and hence is rational. Set the sloe equal to 0.5 and observe the x and y coordinates of the intersection oint. If you lug these coordinates into the equation of the unit circle and clear the denominators, what do you get? Click the the box show triangle for a clue. Try this with a few other oints. 5

6 When m =.5, the oint of intersection with the unit circle is 3/5, 4/5). Plugging into the equation of the circle and clearing the denominator yields ) which is a Pythagorean trile! ) 4 = 1 = = 5 5 Show that if a, b, c) is a Pythagorean trile, then the oint a/c, b/c) is a rational oint on the unit circle. a + b = c = a c + b a ) ) b c = 1 = + = 1. c c Since a, b, and c are integers, a/c and b/c are rational, and a/c, b/c) lies on the unit circle. Now, show that if x, y) is the oint of intersection of the above line with the unit circle and the sloe m is rational, then x and y are also rational. You can do this concretely by showing that x = 1 m 1 + m and y = m 1 + m. We intersect y = mx + 1) with x + y = 1 in the first quadrant. x + y = 1 = x + mx + 1)) = 1 = x + m x + m x + m 1 = 0 = 1 + m )x + m )x + m 1) = 0 = x = m ± 4m m )m 1) 1 + m ) = x = m ± 4m 4 4m 4 1) 1 + m ) = x = m ± m ) = x = m ± 1 + m ) = x = m m, 1 m 1 + m = x = 1, 1 m 1 + m 6

7 Working in the first quadrant, x = 1 m 1 + m and ) 1 m y = mx + 1) = m 1 + m + 1 = m 1 m 1 + m ) m = m 1 + m 1 + m. Recall that a rational number m can be exressed as /q for integers and q 0. Use this fact and the revious roblem to exress rational oints x, y) on the unit circle in terms of and q. Reduce the fractions as much as ossible and recover the Pythagorean Triles Formula from earlier. Setting m = q, x = 1 m 1 + m = ) q q ) = q + q and Thus, ) y = m 1 + m = q 1 + q ) = q + q. ) q q + ) = 1 = q ) + q) = + q ). + q + q So, q, q, + q ) gives the general formula for a Pythagorean trile. And, the geometric construction shows that there is a one-to-one corresondence between the Pythagorean triles and the rational oints on the unit circle in the first quadrant). 7

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