18.781: Solution to Practice Questions for Final Exam

Transcription

1 18.781: Soluton to Practce Questons for Fnal Exam 1. Fnd three solutons n postve ntegers of x 6y = 1 by frst calculatng the contnued fracton expanson of 6. Soluton: We have 1 6=[, ] 6 6+ =[, ] 1 =[,, ]=[,, ]=[,, 6+] =[,, 4, ] 6 =[,, 4,, 4,, 4,...]=[,, 4]. Therefore, lookng at [, ] we get 5/, whch leads to 5 6 = 1. Therefore (5, ) s a soluton. To get two more we compute (5 + 6) = (5 + 6) 3 =(5+ 6) ( ) = Therefore (5, ), (49, 0) and (485, 198) are three solutons. Note that snce the length of the perod s (even) there are no solutons to x 6y = 1.. If θ 1 =[3, 1, 5, 9,a 1,a,...]andθ =[3, 1, 5, 7,b 1,b,...], show that θ 1 θ < 49/7095. Soluton: Let θ =[3, 1, 5] = 3/6. Then computng the next convergent to [3, 1, 5, 9] = 11/56, we see that θ 1 θ < 1/(6 55). Smlarly θ θ < 1/(6 43). So by the trangle nequalty θ 1 θ θ 1 θ + θ θ < 1/(6 55) + 1/(6 43) = 49/7095. Note: once you compute [3, 1] = 4/1 and[3, 1, 5] = 3/6, you can be a bt lazy and not compute say [3, 1, 5, 9], snce you re only nterested n q 3 = a 3 q + q 1 = = 55. Smlarly for [3, 1, 5, 7]. 3. For n = 178, fgure out the number of postve dvsors of n, and the sum of ts postve dvsors. Soluton: n = 6 3 3,sod(n) = (6 + 1)(3 + 1) = 8, and 7 1 σ(n) = = = Use multplcatvty to calculate the sum φ(d). d d 59 1

2 Soluton: Snce f(n) = φ(n)/n s a multplcatve functon of n, so s φ(d) g = U f =. d d n So we need to fgure out what t s on prme powers. We have g(1) = 1 and for e 1, g(p e p 1 p(p 1) p e 1 (p 1) 1 )=1+ ( ) =1+e p p 1. p 1 p Now 59 = We have g( 5 ) = 1 + 5/ = 7/ andg(3 4 ) = 1 + 8/3 = 11/3. So g(n) =77/6. 5. Prove that f a prme p>3 dvdes n n + 1 for an nteger n, thenp 1 (mod 6). (the orgnal problem should have sad p>3) Soluton: We have n n +1 0(modp). So 4n 4n +4 0(modp). That s, (n 1) 3(modp). So 3 s a quadratc resdue mod p (snce gcd(3,p)=1). Ths forces p 1(mod3),snce ( ) ( )( ) ( 1= = =( 1) (p 1)/ ( 1) (p 1)/ p ) ( p =, p p p 3 3) and the last expresson s +1 f p 1(mod3)and 1 fp (mod3). 6. Compute the value of the nfnte perodc fracton 1, 4. Fnd the smallest postve (.e. both x, y > 0) soluton of x 145y =1. Soluton: Let y =[4] = 4+1/y. Theny 4y 1 =0,soy = (plus sgn because y>0). Therefore x =[1, 4] = 1 + 1/y =1+1/( ) = = 145. The perod s odd. Therefore the smallest postve soluton to the Brahmagupta-Pell equaton wll come from [1, 4] = 1 + 1/4 = 89/4, and t s (89, 4). Note that [1] = 1/1 wll gve = Determne whether there s a nontrval nteger soluton of the equaton 49x +5y +38z 8xy +70xz 8yz =0. Soluton: Let s smplfy the conc. It s expedtous to frst scale x by 1/7, getttng x +5y +38z 4xy +10xz 8yz =(x y +5z) + y +13z 8yz. Therefore replacng callng (x y + 5z) our new varable x, we get So we end up wth x + y +13z 8yz = x +(y 4z) 3z. x + y 3z whch s already nce and squarefree. By Legendre s theorem, we just need to verfy whether the local condtons are satsfed. The coeffcents 1, 1, 3 don t all have the same sgn, so that one s ok. We also need to check that 1 s a square mod 3, whch s not ok. So the orgnal conc doesn t have any nontrval ratonal or nteger ponts.

3 8. Fnd a Pythagorean trangle such that the dfference of the two (shorter) sdes s 1, and every sde s at least 100. Soluton: Suppose that the sdes are r s, rs, r + s. (In general, they wll be some common multple of these, but the fact that the dfference of two of the sdes s 1, and postvty of the sdes, forces that multple to be 1 anyway). So we need r s rs =1,.e. (r s) s =1. Letx = r s and y = s, then ths s just a Brahmagupta-Pell type equaton x y = ±1. We want a soluton such that s = y and r = x+y ar e both postve, and such that mn(rs, r s ) s larger than 100. The contnued fracton of s[1, ]. So the smallest postve soluton comes from [1] = 1/1,.e. s (x, y) =(1, 1) (whch we could have guessed anyway, wthout usng contnued fractons). Now 1 1 = 1. To get all solutons, we just have to look at the ratonal and rratonal parts of (1+ ) n. The smallest one whch works s (1+ ) 3 =7+5. So we get (r, s) =(1, 5). So the Pythagorean trangle s (119, 10, 169). 9. Show that x +y =8z + 5 has no ntegral soluton. Soluton: Lookng mod 8, we see that we have x +y 5 (mod 8). Now, further lookng mod we see x must be odd. So x 1 (mod 8), whch forces y 4 (mod 8). Ths s mpossble, snce f y s even, then y 0(mod8),andfysodd, then y (mod8). 10. Defne a sequence by a 0 =, a 1 = 5 and a n = 5a n 1 4a n for n. Show that a n a n+ a n+1 s a square for every n 0. Soluton: The characterstc polynomal s T 5T +4=(T 1)(T 4). So we must have a n = A 4 n + B 1 n. Substutng n n =0and1,wegeta n =4 n +1. So whch s a perfect square. a n a n+ a n+1 =(4n + 1)(4 n+ +1) (4 n+1 +1) =4 n+ +4 n 4 n+1 n =4 ( ) = 9 4 n =( 3 n ) 11. Let p ab. Show that ax + by c (mod p) has a soluton. Soluton: Consder the (p +1)/ numberax for x =0,...,(p 1)/. These are all dstnct modulo p. Smlarly, the (p +1)/ numbersc by are also all dstnct modulo p. Snce we now have p + 1 numbers n all, and only p resdue classes mod p, by the Pgeonhole prncple, two of these must be congruent mod p. Therefore we must have ax c by (mod p) for some x, y. Therefore the congruence has a soluton. 1. How many solutons are there to x +3x (mod 8)? Fnd all of them. Soluton: We need to fgure out the number of solutons mod 4 and mod 7, and multply. Modulo 4 we have x +3x + (x +1)(x + ) (mod 4). It s easy to see ths has the solutons x 1, (mod4). Note: one must be very careful when dealng wth congruences modulo prme powers. It s not necessarly true that f you have a product of (e.g. lnear) factors, that the product wll 3

4 be zero mod p e ff one of them s. For example, x(x +) 0 (mod 8) has more than the two solutons x 0, ; n fact, any even number x wll make x(x + ) vansh mod 8, so there are 4 solutons mod 8. If p e s small enough, the best strategy s probably just to run over all the congruence classes and check. On the other hand, f you re workng mod a prme (.e. e = 1) then you can separate out factors. Modulo 7 we have x 3 +3x +18 x 4x +4=(x ) 0 (mod 7). So just one soluton x (mod 7). So the total number of solutons mod 8 s 1 =. To fnd them, we need the lnear combnaton 4+( 1) 7=1. Then to combne 1 andwehave( 1) ( 1) 7+ 4 = 3. To combne andwe have ( ) ( 1) 7+ 4=30 (mod 8). So the two solutons mod 8 are and Let a, m be postve ntegers, not necessarly coprme. Show that a m a m φ(m) (mod m). Soluton: Wrte m = p e. Enough to prove that for every. Ifp a then by Euler e a m a m φ(m) e e (mod p ) a φ(p ) 1 (mod p ). Snce φ(p ) dvdes φ(m) = φ(p ), we get that e e a φ(m) 1 (mod p e ), and then by multplyng by a m φ(m), we get the desred congruence. On the other hand, f p a, then the left and rght sdes of the congruence we wsh to prove wll be dvsble by m m φ(m) p and p respectvely. If we prove that both the exponents are at least e,thenboth sdes wll be congruent to 0 modulo p e, and we ll be done. It s obvously enough to prove m φ(m) e,sncem m φ(m). We ll assume m>1, snce f m = 1 there are no prmes dvdng t, and nothng to prove. Now note that m>φ(m) (snceφ(m) sthenumberof ntegers n {1,...,m} coprme to m, and there s at least one whch s not coprme to m, namely m). Also, snce p e m, rememberng that φ(m) =pe 1 tmes other stuff, we see that e 1 p e 1 dvdes m φ(m). So m φ(m) p e 1 e whenever e 1. (It s an easy exercse to prove by nducton that e 1 e for any e N.) 14. Parametrze all the ratonal ponts on the curve x 3y =1. Soluton: We fnd one trval pont (1, 0). So wrte y = m(x 1) and plug t n, to get So Cancellng a factor of (x 1), we get x 3m (x 1) =1 (x 1)(x +1)=3m (x 1). x +1=3m (x 1) 4

5 So x =(3m +1)/(3m 1) and then y = m(x 1) = m/(3m 1). Ths parametrzes all ratonal ponts on the conc (except for the orgnal pont (1, 0), whch s obtaned as a lmt when m ). 15. Fnd an nteger soluton of 37x +41y = 3. Soluton: We run the Eucldean algorthm on 37 and 41 to get So ( 9) = Show that f n>1thenn n 1. (Hnt: consder the smallest prme dvdng n). Soluton: By contradcton. Suppose n n 1. Let p be the smallest prme dvdng n. Then p n 1. So the order of mod p dvdes n. But t also dvdes p 1. So the order h of mod p s less than p and dvdes n, and so are the prmes dvdng h. Snce there s no prme smaller than p whch dvdes n, the order must be 1. So 1 1(modp), whch s mpossble. 17. Let p 11 be prme. Show that for some n {1,...,9}, bothn and n + 1 are quadratc resdues. If ether or 5 s a quadratc resdue mod p, we re done, consderng the pars (1, ) and (4, 5), snce 1 and 4 are squares and therefore quadratc resdues. If both and 5 are quadratc nonresdues, then 10 s a quadratc resdue. So (9, 10) does the job. 18. Show that f 3a b (mod 17) then 3a b (mod 89). Soluton: We calculate ( ) ( ) ( ) ( ) ( ) = = = = = So f 3a b (mod 17) then we clam 17 a, else we would have 3 (ba 1 ) (mod 17) whch s a contradcton to the above calculaton. So 17 3a = b and so 17 b. Then 17 3a b,so3a b (mod 89). 19. Calculate the product α ( α), where α runs over the prmtve 14 th roots of unty. Soluton: We compute the cyclotomc polynomal Φ 14 (x) notngthattmusthavedegree φ(14) = 6. x 14 1=(x 7 1)(x 7 +1). Throwng out x 7 1, we have x 7 +1=(x +1)(x 6 x 5 + x 4 x 3 + x x +1) 5

6 So Φ 14 (x) =x 6 x 5 + x 4 x 3 + x x +1. Note that Φ 14 (x) = α (x α), where α runs over the prmtve 14 th roots of unty. Substtutng x =n,weget ( α) = =43. α 0. Let f be a multplcatve functon wth f(1) = 1, and let f 1 be ts nverse for Drchlet convoluton. Show that f 1 s multplcatve as well, and that for squarefree n, wehave f 1 (n) =μ(n)f(n). Soluton: The nverse f 1 s defned by f f 1 = 1, and t exsts as long as f(1) =0,aswe showed on a problem set. To show t s multplcatve, we wll defne a functon g whch wll be multplcatve by defnton, and show that f g = 1. Then t wll follow that f 1 = g. So let g(1) = 1, and defne g on prme powers p e by nducton on e 1 by lettng 1(p e )=0=(f g)(p e )=f(1)g(p e )+f(p)g(p e 1 )+...f(p e 1 )g(e)+f(p e )g(1) = g(p e )+f(p)g(p e 1 )+...f(p e 1 )g(e)+f(p e )g(1). Snce g(1),...,g(p e 1 ) have been defned by the nducton hypothess, we can solve ths unquely for g(p e ). Then for n = p e 1 e 1...p r r, defne g(n) = e g(p ). By constructon, g s multplcatve. Therefore so s g f. By constructon (g f)(1) = 1 and (g f)(p e )=0for e 1. So (g f)(n) =0forn 1 by multplcatvty. That s, g f = 1. Therefore g s the multplcatve nverse of f. Note that there s a unque multplcatve nverse, snce f g were another nverse, then g = g 1 = g (f g )=(g f) g = 1 g = g. Fnally, we need to show g(n) = μ(n)f(n) for n squarefree. By multplcatvty of g, μ and f, t s enough to show ths when n = p, a prme (t s clearly true for n =1). Buttheng(p) s defned by 0=g(p)+f(p)g(1) = g(p)+f(p). That s, g(p) = f(p) = μ(p)f(p), whch fnshes the proof. 6

7 MIT OpenCourseWare Theory of Numbers Sprng 01 For nformaton about ctng these materals or our Terms of Use, vst: