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Solution Manual for A Friendly Introduction to Number Theory, 4th Edition

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    S OLUTIONS M ANUAL A F RIENDLY I NTRODUCTION TO N UMBER T HEORY F OURTH E DITION Joseph Silverman Brown University

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    Table of Contents Chapter 1 What is Number Theory? 1 Chapter 2 Pythagorean Triples 5 Chapter 3 Pythagorean Triples and the Unit Circle 11 Chapter 4 Sums of Higher Powers and Fermat’s Last Theorem 16 Chapter 5 Divisibility and the Greatest Common Divisor 19 Chapter 6 Linear Equations and the Greatest Common Divisor 25 Chapter 7 Factorization and the Fundamental Theorem of Arithmetic 30 Chapter 8 Congruences 35 Chapter 9 Congruences, Powers, and Fermat’s Little Theorem 40 Chapter 10 Congruences, Powers, and Euler’s Formula 43 Chapter 11 Euler’s Phi Function and the Chinese Remainder Theorem 46 Chapter 12 Prime Numbers 55 Chapter 13 Counting Primes 60 Chapter 14 Mersenne Primes 65 Chapter 15 Mersenne Primes and Perfect Numbers 68 Chapter 16 Powers Modulo m and Successive Squaring 75 Chapter 17 Computing k th Roots Modulo m 78 Chapter 18 Powers, Roots, and “Unbreakable” Codes 81 Chapter 19 Primality Testing and Carmichael Numbers 84 Chapter 20 Squares Modulo p 88 Chapter 21 Is –1 a Square Modulo p ? Is 2? 91 Chapter 22 Quadratic Reciprocity 95 Chapter 23 Which Primes are Sums of Two Squares? 108 Chapter 24 Which Numbers are Sums of Two Squares? 113 Chapter 25 As Easy as One, Two, Three 116 Chapter 26 Euler’s Phi Function and Sums of Divisors 122 Chapter 27 Powers Modulo p and Primitive Roots 126 Chapter 28 Primitive Roots and Indices 135 Chapter 29 The Equation X 4 + Y 4 = Z 4 138 Chapter 30 Square–Triangular Numbers Revisited 143 Chapter 31 Pell’s Equation 147 Chapter 32 Diophantine Approximation 152 Chapter 33 Diophantine Approximation and Pell’s Equation 156 Chapter 34 Number Theory and Imaginary Numbers 159 Chapter 35 The Gaussian Integers and Unique Factorization 163 Chapter 36 Irrational Numbers and Transcendental Numbers 168 Chapter 37 Binomial Coefficients and Pascal’s Triangle 176 Chapter 38 Fibonacci’s Rabbits and Linear Recurrence Sequences 180 Chapter 39 Oh, What a Beautiful Function 190 Chapter 40 Cubic Curves and Elliptic Curves 195 Chapter 41 Elliptic Curves with Few Rational Points 200 Chapter 42 Points on Elliptic Curves Modulo p 205 Chapter 43 Torsion Collections Modulo p and Bad Primes 211 Chapter 44 Defect Bounds and Modularity Patterns 215

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    Chapter 45 The Topsy-Turvy World of Continued Fractions [online] 219 Chapter 46 Continued Fractions, Square Roots, and Pell’s Equation [online] 227 Chapter 47 Generating Functions [online] 232 Chapter 48 Sums of Powers [online] 240

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    Chapter 1 What Is Number Theory? Exercises 1.1. The first two numbers that are both squares and triangles are 1 and 36 . Find the next one and, if possible, the one after that. Can you figure out an efficient way to find triangular–square numbers? Do you think that there are infinitely many? Solution to Exercise 1.1. The first three triangular–square numbers are 36, 1225, and 41616. Triangular–square numbers are given by pairs ( m, n ) satisfying m ( m + 1) / 2 = n 2 . The first few pairs are (8 , 6) , (49 , 35) , (288 , 204) , (1681 , 1189) , and (9800 , 6930) . The pattern for generating these pairs is quite subtle. We will give a complete description of all triangular–square numbers in Chapter 28, but for now it would be impressive to merely notice empirically that if ( m, n ) gives a triangular–square number, then so does (3 m + 4 n + 1 , 2 m + 3 n + 1) . Starting with (1 , 1) and applying this rule repeatedly will actually give all triangular–square numbers. 1.2. Try adding up the first few odd numbers and see if the numbers you get satisfy some sort of pattern. Once you find the pattern, express it as a formula. Give a geometric verification that your formula is correct. Solution to Exercise 1.2. The sum of the first n odd numbers is always a square. The formula is 1 + 3 + 5 + 7 + · · · + (2 n − 1) = n 2 . The following pictures illustrate the first few cases, and they make it clear how the general case works. 3 3 1 3 5 5 5 3 3 5 1 3 5 7 7 7 7 5 5 5 7 3 3 5 7 1 3 5 7 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1

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    [Chap. 1] What Is Number Theory? 2 1.3. The consecutive odd numbers 3, 5, and 7 are all primes. Are there infinitely many such “prime triplets”? That is, are there infinitely many prime numbers p such that p + 2 and p + 4 are also primes? Solution to Exercise 1.3. The only prime triplet is 3, 5, 7. The reason is that for any three odd numbers, at least one of them must be divisible by 3. So in order for them all to be prime, one of them must equal 3. It is conjectured that there are infinitely many primes p such that p + 2 and p + 6 are prime, but this has not been proved. Similarly, it is conjectured that there are infinitely many primes p such that p + 4 and p + 6 are prime, but again no one has a proof. 1.4. It is generally believed that infinitely many primes have the form N 2 + 1 , although no one knows for sure. (a) Do you think that there are infinitely many primes of the form N 2 − 1 ? (b) Do you think that there are infinitely many primes of the form N 2 − 2 ? (c) How about of the form N 2 − 3 ? How about N 2 − 4 ? (d) Which values of a do you think give infinitely many primes of the form N 2 − a ? Solution to Exercise 1.4. First we accumulate some data, which we list in a table. Looking at the table, we see that N 2 − 1 and N 2 − 4 are almost never equal to primes, while N 2 − 2 and N 2 − 3 seem to be primes reasonably often. N N 2 − 1 N 2 − 2 N 2 − 3 N 2 − 4 2 3 2 1 0 3 8 = 2 3 7 6 = 2 · 3 5 4 15 = 3 · 5 14 = 2 · 7 13 12 = 2 2 · 3 5 24 = 2 3 · 3 23 22 = 2 · 11 21 = 3 · 7 6 35 = 5 · 7 34 = 2 · 17 33 = 3 · 11 32 = 2 5 7 48 = 2 4 · 3 47 46 = 2 · 23 45 = 3 2 · 5 8 63 = 3 2 · 7 62 = 2 · 31 61 60 = 2 2 · 3 · 5 9 80 = 2 4 · 5 79 78 = 2 · 3 · 13 77 = 7 · 11 10 99 = 3 2 · 11 98 = 2 · 7 2 97 96 = 2 5 · 3 11 120 = 2 3 · 3 · 5 119 = 7 · 17 118 = 2 · 59 117 = 3 2 · 13 12 143 = 11 · 13 142 = 2 · 71 141 = 3 · 47 140 = 2 2 · 5 · 7 13 168 = 2 3 · 3 · 7 167 166 = 2 · 83 165 = 3 · 5 · 11 14 195 = 3 · 5 · 13 194 = 2 · 97 193 192 = 2 6 · 3 15 224 = 2 5 · 7 223 222 = 2 · 3 · 37 221 = 13 · 17 Looking at the even values of N in the N 2 − 1 column, we might notice that 2 2 − 1 is a multiple of 3 , that 4 2 − 1 is a multiple of 5 , that 6 2 − 1 is a multiple of 7 , and so on. 2

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    [Chap. 1] What Is Number Theory? 3 Having observed this, we see that the same pattern holds for the odd N ’s. Thus 3 2 − 1 is a multiple of 4 and 5 2 − 1 is a multiple of 6 and so on. So we might guess that N 2 − 1 is always a multiple of N + 1 . This is indeed true, and it can be proved true by the well known algebraic formula N 2 − 1 = ( N − 1)( N + 1) . So N 2 − 1 will never be prime if N ≥ 2 . The N 2 − 4 column is similarly explained by the formula N 2 − 4 = ( N − 2)( N + 2) . More generally, if a is a perfect square, say a = b 2 , then there will not be infinitely many primes of the form N 2 − a , since N 2 − a = N 2 − b 2 = ( N − b )( N + b ) . On the other hand, it is believed that there are infinitely many primes of the form N 2 − 2 and infinitely many primes of the form N 2 − 3 . Generally, if a is not a perfect square, it is believed that there are infinitely many primes of the form N 2 − a . But no one has yet proved any of these conjectures. 1.5. The following two lines indicate another way to derive the formula for the sum of the first n integers by rearranging the terms in the sum. Fill in the details. 1 + 2 + 3 + · · · + n = (1 + n ) + (2 + ( n − 1)) + (3 + ( n − 2)) + · · · = (1 + n ) + (1 + n ) + (1 + n ) + · · · . How many copies of n + 1 are in there in the second line? You may need to consider the cases of odd n and even n separately. If that’s not clear, first try writing it out explicitly for n = 6 and n = 7 . Solution to Exercise 1.5. Suppose first that n is even. Then we get n/ 2 copies of 1 + n , so the total is n 2 (1 + n ) = n 2 + n 2 . Next suppose that n is odd. Then we get n − 1 2 copies of 1 + n and also the middle term n +1 2 which hasn’t yet been counted. To illustrate with n = 9 , we group the terms as 1 + 2 + · · · + 9 = (1 + 9) + (2 + 8) + (3 + 7) + (4 + 6) + 5 , so there are 4 copies of 10 , plus the extra 5 that’s left over. For general n , we get n − 1 2 (1 + n ) + n + 1 2 = n 2 − 1 2 + n + 1 2 = n 2 + n 2 . 3

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    [Chap. 1] What Is Number Theory? 4 Another similar way to do this problem that doesn’t involve splitting into cases is to simply take two copies of each term. Thus 2(1 + 2 + · · · + n ) = (1 + 2 + · · · + n ) + (1 + 2 + · · · + n ) = (1 + 2 + · · · + n ) + ( n + · · · + 2 + 1) = (1 + n ) + (2 + n − 1) + (3 + n − 2) + · · · + ( n + 1) = (1 + n ) + (1 + n ) + · · · + (1 + n ) ︸ ︷︷ ︸ n copies of n + 1 = n (1 + n ) = n 2 + n Thus the twice the sum 1+2+ · · · + n equal n 2 + n , and now divide by 2 to get the answer. 1.6. For each of the following statements, fill in the blank with an easy-to-check crite- rion: (a) M is a triangular number if and only if is an odd square. (b) N is an odd square if and only if is a triangular number. (c) Prove that your criteria in (a) and (b) are correct. Solution to Exercise 1.6. (a) M is a triangular number if and only if 1 + 8 M is an odd square. (b) N is an odd square if and only if ( N − 1) / 8 is a triangular number. (Note that if N is an odd square, then N 2 − 1 is divisible by 8 , since (2 k +1) 2 = 4 k ( k +1)+1 , and 4 k ( k +1) is a multiple of 8 .) (c) If M is triangular, then M = m ( m +1) / 2 , so 1+8 M = 1+4 m +4 m 2 = (1+2 m ) 2 . Conversely, if 1 + 8 M is an odd square, say 1 + 8 M = (1 + 2 k ) 2 , then solving for M gives M = ( k + k 2 ) / 2 , so M is triangular. Next suppose N is an odd square, say N = (2 k + 1) 2 . Then as noted above, ( N − 1) / 8 = k ( k +1) / 2 , so ( N − 1) / 8 is triangular. Conversely, if ( N − 1) / 8 is trianglular, then ( N − 1) / 8 = ( m 2 + m ) / 2 for some m , so solving for N we find that N = 1+4 m +4 m 2 = (1 + 2 m ) 2 , so N is a square. 4

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    Chapter 2 Pythagorean Triples Exercises 2.1. (a) We showed that in any primitive Pythagorean triple ( a, b, c ) , either a or b is even. Use the same sort of argument to show that either a or b must be a multiple of 3 . (b) By examining the above list of primitive Pythagorean triples, make a guess about when a , b , or c is a multiple of 5 . Try to show that your guess is correct. Solution to Exercise 2.1. (a) If a is not a multiple of 3 , it must equal either 3 x + 1 or 3 x + 2 . Similarly, if b is not a multiple of 3 , it must equal 3 y + 1 or 3 y + 2 . There are four possibilities for a 2 + b 2 , namely a 2 + b 2 = (3 x + 1) 2 + (3 y + 1) 2 = 9 x 2 + 6 x + 1 + 9 y 2 + 6 y + 1 = 3(3 x 2 + 2 x + 3 y 2 + 2 y ) + 2 , a 2 + b 2 = (3 x + 1) 2 + (3 y + 2) 2 = 9 x 2 + 6 x + 1 + 9 y 2 + 12 y + 4 = 3(3 x 2 + 2 x + 3 y 2 + 4 y + 1) + 2 , a 2 + b 2 = (3 x + 2) 2 + (3 y + 1) 2 = 9 x 2 + 12 x + 4 + 9 y 2 + 6 y + 1 = 3(3 x 2 + 4 x + 3 y 2 + 2 y + 1) + 2 , a 2 + b 2 = (3 x + 2) 2 + (3 y + 2) 2 = 9 x 2 + 12 x + 4 + 9 y 2 + 12 y + 4 = 3(3 x 2 + 4 x + 3 y 2 + 4 y + 2) + 2 . So if a and b are not multiples of 3 , then c 2 = a 2 + b 2 looks like 2 more than a multiple of 3 . But regardless of whether c is 3 z or 3 z +1 or 3 z +2 , the numbers c 2 cannot be 2 more than a multiple of 3 . This is true because (3 z ) 2 = 3 · 3 z, (3 z + 1) 2 = 3(3 z 2 + 2 z ) + 1 , (3 z + 2) 2 = 3(3 z 2 + 4 z + 1) + 1 . 5

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    [Chap. 2] Pythagorean Triples 6 (b) The table suggests that in every primitive Pythagorean triple, exactly one of a , b , or c is a multiple of 5 . To verify this, we use the Pythagorean Triples Theorem to write a and b as a = st and b = 1 2 ( s 2 − t 2 ) . If either s or t is a multiple of 5 , then a is a multiple of 5 and we’re done. Otherwise s looks like s = 5 S + i and t looks like 5 T + j with i and j being integers in the set { 1 , 2 , 3 , 4 } . Next we observe that 2 b = s 2 − t 2 = (5 S + i ) 2 − (5 T + j ) 2 = 25( S 2 − T 2 ) + 10( Si − T j ) + i 2 − j 2 . If i 2 − j 2 is a multiple of 5 , then b is a multiple of 5 , and again we’re done. Looking at the 16 possibilities for the pair ( i, j ) , we see that this accounts for 8 of them, leaving the possibilities ( i, j ) = (1 , 2) , (1 , 3) , (2 , 1) , (2 , 4) , (3 , 1) , (3 , 4) , (4 , 2) , or (4 , 3) . Now for each of these remaining possibilities, we need to check that 2 c = s 2 + t 2 = (5 S + i ) 2 + (5 T + j ) 2 = 25( S 2 + T 2 ) + 10( Si + T j ) + i 2 + j 2 is a multiple of 5 , which means checking that i 2 + j 2 is a multiple of 5 . This is easily accomplished: 1 2 + 2 2 = 5 1 2 + 3 2 = 102 1 + 1 2 = 5 2 2 + 4 2 = 20 (2.1) 3 1 + 1 2 = 103 2 + 4 2 = 254 2 + 2 2 = 204 2 + 3 2 = 25 . (2.2) . 2.2. A nonzero integer d is said to divide an integer m if m = dk for some number k . Show that if d divides both m and n , then d also divides m − n and m + n . Solution to Exercise 2.2. Both m and n are divisible by d , so m = dk and n = dk ′ . Thus m ± n = dk ± dk ′ = d ( k ± k ′ ) , so m + n and m − n are divisible by d . 2.3. For each of the following questions, begin by compiling some data; next examine the data and formulate a conjecture; and finally try to prove that your conjecture is correct. (But don’t worry if you can’t solve every part of this problem; some parts are quite difficult.) (a) Which odd numbers a can appear in a primitive Pythagorean triple ( a, b, c ) ? (b) Which even numbers b can appear in a primitive Pythagorean triple ( a, b, c ) ? (c) Which numbers c can appear in a primitive Pythagorean triple ( a, b, c ) ? Solution to Exercise 2.3. (a) Any odd number can appear as the a in a primitive Pythagorean triple. To find such a triple, we can just take t = a and s = 1 in the Pythagorean Triples Theorem. This gives the primitive Pythagorean triple ( a, ( a 2 − 1) / 2 , ( a 2 + 1) / 2) . (b) Looking at the table, it seems first that b must be a multiple of 4 , and second that every multiple of 4 seems to be possible. We know that b looks like b = ( s 2 − t 2 ) / 2 with 6

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    [Chap. 2] Pythagorean Triples 7 s and t odd. This means we can write s = 2 m + 1 and t = 2 n + 1 . Multiplying things out gives b = (2 m + 1) 2 − (2 n + 1) 2 2 = 2 m 2 + 2 m − 2 n 2 − 2 n = 2 m ( m + 1) − 2 n ( n + 1) . Can you see that m ( m + 1) and n ( n + 1) must both be even, regardless of the value of m and n ? So b must be divisible by 4 . On the other hand, if b is divisible by 4 , then we can write it as b = 2 r B for some odd number B and some r ≥ 2 . Then we can try to find values of s and t such that ( s 2 − t 2 ) / 2 = b . We factor this as ( s − t )( s + t ) = 2 b = 2 r +1 B. Now both s − t and s + t must be even (since s and t are odd), so we might try s − t = 2 r and s + t = 2 B. Solving for s and t gives s = 2 r − 1 + B and t = − 2 r − 1 + B . Notice that s and t are odd, since B is odd and r ≥ 2 . Then a = st = B 2 − 2 2 r − 2 , b = s 2 − t 2 2 = 2 r B, c = s 2 + t 2 2 = B 2 + 2 2 r − 2 . This gives a primitive Pythagorean triple with the right value of b provided that B > 2 r − 1 . On the other hand, if B < 2 r − 1 , then we can just take a = 2 2 r − 2 − B 2 instead. (c) This part is quite difficult to prove, and it’s not even that easy to make the correct conjecture. It turns out that an odd number c appears as the hypotenuse of a primitive Pythagorean triple if and only if every prime dividing c leaves a remainder of 1 when divided by 4 . Thus c appears if it is divisible by the primes 5 , 13 , 17 , 29 , 37 , . . . , but it does not appear if it is divisible by any of the primes 3 , 7 , 11 , 19 , 23 , . . . . We will prove this in Chapter 25. Note that it is not enough that c itself leave a remainder of 1 when divided by 4 . For example, neither 9 nor 21 can appear as the hypotenuse of a primitive Pythagorean triple. 2.4. In our list of examples are the two primitive Pythagorean triples 33 2 + 56 2 = 65 2 and 16 2 + 63 2 = 65 2 . Find at least one more example of two primitive Pythagorean triples with the same value of c . Can you find three primitive Pythagorean triples with the same c ? Can you find more than three? 7

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    [Chap. 2] Pythagorean Triples 8 Solution to Exercise 2.4. The next example is c = 5 · 17 = 85 . Thus 85 2 = 13 2 + 84 2 = 36 2 + 77 2 . A general rule is that if c = p 1 p 2 · · · p r is a product of r distinct odd primes which all leave a remainder of 1 when divided by 4 , then c appears as the hypotenuse in 2 r − 1 primitive Pythagorean triples. (This is counting ( a, b, c ) and ( b, a, c ) as the same triple.) So for example, c = 5 · 13 · 17 = 1105 appears in 4 triples, 1105 2 = 576 2 + 943 2 = 744 2 + 817 2 = 264 2 + 1073 2 = 47 2 + 1104 2 . But it would be difficult to prove the general rule using only the material we have developed so far. 2.5. In Chapter 1 we saw that the n th triangular number T n is given by the formula T n = 1 + 2 + 3 + · · · + n = n ( n + 1) 2 . The first few triangular numbers are 1 , 3 , 6 , and 10 . In the list of the first few Pythagorean triples ( a, b, c ) , we find (3 , 4 , 5) , (5 , 12 , 13) , (7 , 24 , 25) , and (9 , 40 , 41) . Notice that in each case, the value of b is four times a triangular number. (a) Find a primitive Pythagorean triple ( a, b, c ) with b = 4 T 5 . Do the same for b = 4 T 6 and for b = 4 T 7 . (b) Do you think that for every triangular number T n , there is a primitive Pythagorean triple ( a, b, c ) with b = 4 T n ? If you believe that this is true, then prove it. Otherwise, find some triangular number for which it is not true. Solution to Exercise 2.5. (a) T 5 = 15 and (11 , 60 , 61) . T 6 = 21 and (13 , 84 , 85) . T 7 = 28 and (15 , 112 , 113) . (b) The primitive Pythagorean triples with b even are given by b = ( s 2 − t 2 ) / 2 , s > t ≥ 1 , s and t odd integers, and gcd( s, t ) = 1 . Since s is odd, we can write it as s = 2 n + 1 , and we can take t = 1 . (The examples suggest that we want c = b + 1 , which means we need to take t = 1 .) Then b = s 2 − t 2 2 = (2 n + 1) 2 − 1 2 = 2 n 2 + 2 n = 4 n 2 + n 2 = 4 T n . So for every triangular number T n , there is a Pythagorean triple (2 n + 1 , 4 T n , 4 T n + 1) . (Thanks to Mike McConnell and his class for suggesting this problem.) 2.6. If you look at the table of primitive Pythagorean triples in this chapter, you will see many triples in which c is 2 greater than a . For example, the triples (3 , 4 , 5) , (15 , 8 , 17) , (35 , 12 , 37) , and (63 , 16 , 65) all have this property. (a) Find two more primitive Pythagorean triples ( a, b, c ) having c = a + 2 . 8

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    [Chap. 2] Pythagorean Triples 9 (b) Find a primitive Pythagorean triple ( a, b, c ) having c = a + 2 and c > 1000 . (c) Try to find a formula that describes all primitive Pythagorean triples ( a, b, c ) having c = a + 2 . Solution to Exercise 2.6. The next few primitive Pythagorean triples with c = a + 2 are (99 , 20 , 101) , (143 , 24 , 145) , (195 , 28 , 197) , (255 , 32 , 257) , (323 , 36 , 325) , (399 , 40 , 401) . One way to find them is to notice that the b values are going up by 4 each time. An even better way is to use the Pythagorean Triples Theorem. This says that a = st and c = ( s 2 + t 2 ) / 2 . We want c − a = 2 , so we set s 2 + t 2 2 − st = 2 and try to solve for s and t . Multiplying by 2 gives s 2 + t 2 − 2 st = 4 , ( s − t ) 2 = 4 , s − t = ± 2 . The Pythagorean Triples Theorem also says to take s > t , so we need to have s − t = 2 . Further, s and t are supposed to be odd. If we substitute s = t + 2 into the formulas for a, b, c , we get a general formula for all primitive Pythagorean triples with c = a + 2 . Thus a = st = ( t + 2) t = t 2 + 2 t, b = s 2 − t 2 2 = ( t + 2) 2 − t 2 2 = 2 t + 2 , c = s 2 + t 2 2 = ( t + 2) 2 + t 2 2 = t 2 + 2 t + 2 . We will get all PPT’s with c = a + 2 by taking t = 1 , 3 , 5 , 7 , . . . in these formulas. For example, to get one with c > 1000 , we just need to choose t large enough to make t 2 +2 t + 2 > 1000 . The least t which will work is t = 31 , which gives the PPT (1023 , 64 , 1025) . The next few with c > 1000 are (1155 , 68 , 1157) , (1295 , 72 , 1297) , (1443 , 76 , 1445) , obtained by setting t = 33 , 35 , and 37 respectively. 2.7. For each primitive Pythagorean triple ( a, b, c ) in the table in this chapter, compute the quantity 2 c − 2 a . Do these values seem to have some special form? Try to prove that your observation is true for all primitive Pythagorean triples. Solution to Exercise 2.7. First we compute 2 c − 2 a for the PPT’s in the Chapter 2 table. a 3 5 7 9 15 21 35 45 63 b 4 12 24 40 8 20 12 28 16 c 5 13 25 41 17 29 37 53 65 2 c − 2 a 4 16 36 64 4 16 4 16 4 9

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    [Chap. 2] Pythagorean Triples 10 all the differences 2 c − 2 a seem to be perfect squares. We can show that this is always the case by using the Pythagorean Triples Theorem, which says that a = st and c = ( s 2 + t 2 ) / 2 . Then 2 c − 2 a = ( s 2 + t 2 ) − 2 st = ( s − t ) 2 , so 2 c − 2 a is always a perfect square. 2.8. Let m and n be numbers that differ by 2 , and write the sum 1 m + 1 n as a fraction in lowest terms. For example, 1 2 + 1 4 = 3 4 and 1 3 + 1 5 = 8 15 . (a) Compute the next three examples. (b) Examine the numerators and denominators of the fractions in (a) and compare them with the table of Pythagorean triples on page 18. Formulate a conjecture about such fractions. (c) Prove that your conjecture is correct. Solution to Exercise 2.8. (a) 1 4 + 1 6 = 5 12 , 1 5 + 1 7 = 12 35 , 1 6 + 1 8 = 7 24 . (b) It appears that the numerator and denominator are always the sides of a (primitive) Pythagorean triple. (c) This is easy to prove. Thus 1 N + 1 N + 2 = 2 N + 2 N 2 + 2 N . The fraction is in lowest terms if N is odd, otherwise we need to divide numerator and de- nominator by 2 . But in any case, the numerator and denominator are part of a Pythagorean triple, since (2 N + 2) 2 + ( N 2 + 2 N ) 2 = N 4 + 4 N 3 + 8 N 2 + 8 N + 4 = ( N 2 + 2 N + 2) 2 . Once one suspects that N 4 + 4 N 3 + 8 N 2 + 8 N + 4 should be a square, it’s not hard to factor it. Thus if it’s a square, it must look like ( N 2 + AN ± 2) for some value of A . Now just multiply out and solve for A , then check that your answer works. 2.9. (a) Read about the Babylonian number system and write a short description, includ- ing the symbols for the numbers 1 to 10 and the multiples of 10 from 20 to 50 . (b) Read about the Babylonian tablet called Plimpton 322 and write a brief report, in- cluding its approximate date of origin. (c) The second and third columns of Plimpton 322 give pairs of integers ( a, c ) having the property that c 2 − a 2 is a perfect square. Convert some of these pairs from Baby- lonian numbers to decimal numbers and compute the value of b so that ( a, b, c ) is a Pythagorean triple. Solution to Exercise 2.9 . There is a good article in wikipedia on Plimpton 322. Another nice source for this material is www.math.ubc.ca/˜cass/courses/m446-03/pl322/pl322.html 10

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    Chapter 3 Pythagorean Triples and the Unit Circle Exercises 3.1. As we have just seen, we get every Pythagorean triple ( a, b, c ) with b even from the formula ( a, b, c ) = ( u 2 − v 2 , 2 uv, u 2 + v 2 ) by substituting in different integers for u and v . For example, ( u, v ) = (2 , 1) gives the smallest triple (3 , 4 , 5) . (a) If u and v have a common factor, explain why ( a, b, c ) will not be a primitive Pytha- gorean triple. (b) Find an example of integers u > v > 0 that do not have a common factor, yet the Pythagorean triple ( u 2 − v 2 , 2 uv, u 2 + v 2 ) is not primitive. (c) Make a table of the Pythagorean triples that arise when you substitute in all values of u and v with 1 ≤ v < u ≤ 10 . (d) Using your table from (c), find some simple conditions on u and v that ensure that the Pythagorean triple ( u 2 − v 2 , 2 uv, u 2 + v 2 ) is primitive. (e) Prove that your conditions in (d) really work. Solution to Exercise 3.1. (a) If u = dU and v = dV , then a , b , and c will all be divisible by d 2 , so the triple will not be primitive. (b) Take ( u, v ) = (3 , 1) . Then ( a, b, c ) = (8 , 6 , 10) is not primitive. 11

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    [Chap. 3] Pythagorean Triples and the Unit Circle 12 (c) u \ v 1 2 3 4 5 6 7 8 9 2 (3 , 4 , 5) 3 (8 , 6 , 10) (5 , 12 , 13) 4 (15 , 8 , 17) (12 , 16 , 20) (7 , 24 , 25) 5 (24 , 10 , 26) (21 , 20 , 29) (16 , 30 , 34) (9 , 40 , 41) 6 (35 , 12 , 37) (32 , 24 , 40) (27 , 36 , 45) (20 , 48 , 52) (11 , 60 , 61) 7 (48 , 14 , 50) (45 , 28 , 53) (40 , 42 , 58) (33 , 56 , 65) (24 , 70 , 74) (13 , 84 , 85) 8 (63 , 16 , 65) (60 , 32 , 68) (55 , 48 , 73) (48 , 64 , 80) (39 , 80 , 89) (28 , 96 , 100) (15 , 112 , 113) 9 (80 , 18 , 82) (77 , 36 , 85) (72 , 54 , 90) (65 , 72 , 97) (56 , 90 , 106) (45 , 108 , 117) (32 , 126 , 130) (17 , 144 , 145) 10 (99 , 20 , 101) (96 , 40 , 104) (91 , 60 , 109) (84 , 80 , 116) (75 , 100 , 125) (64 , 120 , 136) (51 , 140 , 149) (36 , 160 , 164) (19 , 180 , 181) (d) ( u 2 − v 2 , 2 uv, u 2 + v 2 ) will be primitive if and only if u > v and u and v have no common factor and one of u or v is even. (e) If both u and v are odd, then all three numbers are even, so the triple is not primitive. We already saw that if u and v have a common factor, then the triple is not primitive. And we do not allow nonpositive numbers in primitive triples, so we can’t have u ≤ v . This proves one direction. To prove the other direction, suppose that the triple is not primitive, so there is a number d ≥ 2 that divides all three terms. Then d divides the sums ( u 2 − v 2 ) + ( u 2 + v 2 ) = 2 u 2 and ( u 2 − v 2 ) − ( u 2 + v 2 ) = 2 v 2 , so either d = 2 or else d divides both u and v . In the latter case we are done, since u and v have a common factor. On the other hand, if d = 2 and u and v have no common factor, then at least one of them is odd, so the fact that 2 divides u 2 − v 2 tells us that they are both odd. 3.2. (a) Use the lines through the point (1 , 1) to describe all the points on the circle x 2 + y 2 = 2 whose coordinates are rational numbers. (b) What goes wrong if you try to apply the same procedure to find all the points on the circle x 2 + y 2 = 3 with rational coordinates? Solution to Exercise 3.2. (a) Let C be the circle x 2 + y 2 = 2 . Take the line L with slope m through (1 , 1) , where m is a rational number. The equation of L is y − 1 = m ( x − 1) , so y = mx − m + 1 . To find the intersection L ∩ C , we substitute and solve: x 2 + ( mx − m + 1) 2 = 2 ( m 2 + 1) x 2 − 2( m 2 − m ) x + ( m − 1) 2 = 2 ( m 2 + 1) x 2 − 2( m 2 − m ) x + ( m 2 − 2 m − 1) = 0 12

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