(2mn, m 2 n 2, m 2 + n 2 )

Transcription

1 MATH 16T Homewk Solutons 1. Recall that a natural number n N s a perfect square f n = m f some m N. a) Let n = p α even f = 1,,..., k. be the prme factzaton of some n. Prove that n s a perfect square f and only f α s b) Let a, b N 0 such that gcda, b) = 1 and ab s a perfect square. Prove that both a and b are perfect squares. Proof : a) Suppose frst that n s a perfect square. n = m f some m N wth m. Let m = the prme factzaton of m. Hence n = m = k p β ) = So ths must be the prme factzaton of n. Thus α = β s even f = 1,,... k. Suppose next that α s even f = 1,..., k. α = β f some β N 0 f = 1,,..., k. Put m = n = p α = p β = k p β p β ) = m So n s a perfect square. b) If a = 1 resp. b = 1) then b = ab resp. a = ab) s a perfect square. So we may assume that a, b > 1. Let a = p α1 1 pα k k resp. b = q β1 1 qβ l l ) be the prme factzaton of a resp. b). Snce gcda, b) = 1, we have that p q j f = 1,,..., k and j = 1,,..., l. Hence ab = p α1 1 pα k k qβ1 1 qα l l s the prme factzaton of ab. Snce ab s a perfect square, t follows from a) that α and β j are even f = 1,,..., k and j = 1,,..., l. Agan usng a), we get that a and b are perfect squares. p β be p β.. Let m, n N 0 such that gcdm, n) = 1, m > n and not both m and n are odd. Prove that s a prmtve Pythagean trple. Proof : Note that mn, m n, m + n ) mn) + m n ) = m n + m m n + n = m + m n + n = m + n ) So mn, m n, m + n ) s a Pythagean trple. Suppose that gcdmn, m n ) 1. there exsts a prme p such that p mn) and p m n ). Snce p s a prme, we get that p p m p n. Suppose frst that p. p =, a contradcton snce p m n ) but m n s odd snce gcdm, n) = 1 and not both m and n are odd). Suppose next that p m. p m. Snce p m n ), we get that p m m n )). So p n. Snce p s a prme, we get that p n, a contradcton snce gcdm, n) = 1. Suppose fnally that p n. Smlarly as n the case p m, we get a contradcton. Hence gcdmn, m n ) = 1. So gcdmn, m n, m + n ) = 1 and mn, m n, m + n ) s a prmtve Pythagean trple. 1

2 3. Fnd all prmtve Pythagean trples such that one of the components s 15. Soluton : We may assume that the prmtve Pythagean trple x, y, z) s of the fm mn, m n, m + n ) where m, n N 0 such that gcdm, n) = 1, m > n and not both m and n are odd. ether m n = 15 m + n = 15. Suppose frst that m + n = 15. We easly get by checkng whether any of the numbers 15, 15, 15 6, s a perfect square) that m = 11 and n =. So x, y, z) =, 117, 15). Suppose next that m n = 15. m n)m + n) = 15. Snce m > n, we get { m n = 1 m + n = 15 { m n = 5 m + n = 5 So m, n) = 63, 6) m, n) = 15, 10). Snce gcdm, n) = 1, we get m, n) = 63, 6). So x, y, z) = 781, 15, 7813). x, y, z) {, 117, 15), 117,, 15), 781, 15, 7813), 15, 781, 7813)}. Use unque prme factzaton to fnd all couples a, b) N 0 such that a > b and a b = b a. Soluton : If b = 1, we easly get that a = a 1 = a b = b a = 1 a = 1, a contradcton snce a > b. So b > 1. we can wrte a = p α and b = p β where p 1 < p < < p n are prmes, α, β N and not both α and β are zero f = 1,,..., n. Substtutng ths nto a b = b a, we fnd p bα = p aβ Usng unque prme factzaton, we get bα = aβ f = 1,,..., n ) Snce α, β ) 0, 0) f = 1,,..., n, we get that α 0 β f = 1,,..., n. Hence t follows from *) that So β < α f = 1,,..., n. Hence α = a > 1 f = 1,,..., n ) β b b = p β dvdes So a = kb f some k N wth k snce a > b). Hence by **), α = kβ f = 1,,..., n. So a = p α = Puttng everythng together, we fnd kb = a = b k and so p kβ = b k 1 = k n p α p β = a ) k = b k If k = then b = b 1 = and so a = bk = =. So, ) s a soluton. We prove by nducton on k) that b k 1 > k f k 3. F k = 3, we get that b 3 1 = b = > 3. Assume that b k 1 > k f some k 3. b k+1) 1 = b k = b b k 1 > k > k + 1. a, b) =, )

3 5. Fnd all the ratonal ponts on the parabola y y = x + 3. How many ntegral ponts are on ths parabola? Soluton : We easly get that P = 3 ), 0 s a ratonal pont on the curve. We know that we wll get all the ratonal ponts on the curve by ntersectng the curve wth a lne l through P that s ether vertcal has a ratonal slope. If l s vertcal then an equaton f l s x = 3 The ponts of ntersecton between the lne and the parabola are solutons of { x = 3 y y = x + 3 We easly get that these ponts of ntersecton are P = 3 ), 0 and Q = 3, 1 ). If l has a ratonal slope then l has an equaton of the fm y = m x + 3 ) where m Q The ponts of ntersecton between the lne and the parabola are solutons of y = m x + 3 ) 1) y y = x + 3 ) Substtutng 1) nto ), we get m x + ) 3 m x + 3 ) = x + 3 m x + 3m m )x + 9 m 3 m 3 = 0 ) If m = 0, there s only one pont of ntersecton : P = 3 ), 0. So assume that m 0 so the lne l s not hzontal). We know that x = 3 s a root of the quadratc equaton *). Snce the sum of the roots of the quadratc equaton ax + bx + c = 0 s b, we get that the second root of the quadratc equaton *) s a So we get a second pont of ntersecton : x = 3m m m 3 ) + m 3m = m y = m x + 3 ) + m 3m = m m + 3 ) = + m m + m 3m Q = m, + m ) m All ratonal ponts on the parabola y y = x + 3 are gven by 3 ) { + m 3m, 1 m, + m ) } : m Q 0 m 3

4 There s another way to get ths. If l s not hzontal then l has an equaton of the fm x + 3 = ky where k Q. all the ratonal ponts on the parabola are gven by {k + k 3 ) }, k + 1 : k Q Ths cresponds to the substtuton m = 1 k. Suppose that x, y) s an ntegral pont on the parabola. So y y = x + 3. However, y y = yy 1). So y y s always even, a contradcton snce x + 3 s odd. There are no ntegral ponts on the parabola y y = x Consder the cubc curve y = x Fnd an ntegral pont x 0, y 0 ) on ths curve. Use the tangent lne to the curve at ths pont to fnd a non-trval ratonal pont on ths curve. Soluton : We easly get that 1, ) s an ntegral pont on the cubc curve. To fnd the equaton f the tangent lne to the cubc curve at the pont 1, ), we use mplct dfferentaton close to the pont 1, ) the equaton y = x determnes y as a functon of x) : ydy = 3x dy dx and so dx = 3x y Hence Thus an equaton f that tangent lne s y 1) = 3 1 = 3 y = 3 x 1) To fnd the ponts of ntersecton between ths tangent lne and the cubc curve, we have to solve { y = x ) Solvng ) f y, we get Substtutng ths nto 1), we fnd y = 3 x 1) ) y = 3 x x + 5 ) = x x 3 9x 30x + 3 = 0 Note that x = 1 s a double root of ths equaton. Snce the sum of the roots s 9, we get that the thrd root s 16 x = = 3 16 y = 3 x + 5 = 3 3 ) = , 11 ) s a ratonal pont on the cubc curve y = x

5 7. Let x, y, z) N 3 0 wth x + y = z and gcdx, y, z) = 1. a) Prove that gcdx, z) = 1. b) Prove that x and z are odd whle y s even. b) Mmc the proof of Theem 1.7 about prmtve Pythagean trples) to come up wth fmulas f x, y and z smlar to those n Theem 1.7). Note that there are two cases here! You don t need to prove all the detals, I m manly nterested n the fmulas and an explanaton of how you got them). Proof : a) Suppose that gcdx, z) > 1. there exsts a prme p wth p x and p z. So p x and p z. Hence p z x ). So p y ). If p = then p y and so p y; f p then p y and agan p y. But ths s a contradcton snce gcdx, y, z) = 1. Hence gcdx, z) = 1. b) Suppose that x s even. z = x + y s even and so z s even, a contradcton snce gcdx, z) = 1 by a). So x s odd. z = x + y s odd. Hence z s odd. Suppose that y s odd. Consderng modulo, we get a contradcton. Hence y s even x + y z 1 mod c) Snce x + y = z, we get that z x)z + x) = z x = y. From b), we see that z x, z + x and y are even. So ) ) z x z + x y ) = z x By a), gcdx, z) = 1. So gcd, z + x ) = 1. Smlarly as n Exercse 1, we can prove : If a, b, c N such that gcda, b) = 1 and ab = c, then there exst m, n N such that a, b) {m, n ), m, n )}. Hence there exst m, n N such that Solvng f z, x and y, we fnd z x = n z + x = m x = m n y = mn z = m + n z x = n z + x = m x = m n y = mn z = m + n Note that we have condtons on m and n. In the frst case, m, n N 0 wth gcdm, n) = 1, m > n and n s odd; n the second case, m, n N 0 wth gcdm, n) = 1, m > n and m s odd. 5