MATH 304: MIDTERM EXAM SOLUTIONS

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1 MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest commo divisor of 746 ad The last ozero remaider is the greatest commo divisor of the iitial pair, so GCD746, 46) 14.. How may of the itegers betwee 0 ad 600 are relatively prime to 600? By defiitio, the Euler φ-fuctio of is the umber of umbers betwee 0 ad that are relatively prime to. Usig the fact that φm ) φm)φ) wheever GCDm, ) 1, ad that for prime powers φp e ) p e p e 1, we compute that: φ600) φ ) φ 3 )φ3)φ5 ) 3 )3 3 0 )5 5) Therefore there are 160 umbers betwee 0 ad 600 that are relatively prime to Fid the remaider whe 3 39 is divided by 1. By Fermat s little theorem with p 7, we have the cogruece: ) mod 7). O the other had, 3 clearly divides the differece as well. product also divides the differece, so we have: Therefore the mod 1). It follows that whe you divide 3 39 by 1 the remaider is Suppose that a 4 1 mod 19). What are the possible orders of a modulo 19? By the order divisibility property, if a k 1 mod 19), the the order of a modulo 19 must divide k. I our situatio where k 4, this meas that the order of a must be oe of the followig umbers: e p a) 1,, 3, 4, 6, 1, 4.

2 O the other had, by Fermat s little theorem a 18 1 mod 19). Thus by the order divisibility property agai, the order of a modulo 19 must divide 18, ad thus must be oe of the followig umbers: e p a) 1,, 3, 6, 9, 18. The umbers that occur i both lists are the possible values for the order of a modulo 19: e p a) 1,, 3, Do there exist iteger solutios x, y to the equatio 36x + 56y 1? Explai your reasoig. Recall that the smallest positive iteger that ca be writte as a liear combiatio ax + by of a ad b is the greatest commo divisor GCDa, b). I our situatio, we have GCD36, 56) 4, so there exist itegers x 0 ad y 0 such that 36x y 0 4. Multiplyig by 3, we see that x, y) 3x 0, 3y 0 ) is a iteger solutio to the give equatio: 363x 0 ) + 563y 0 ) I fact, x 0, y 0 ) 11, 7) are solutios to the first equatio, so x, y) 33, 1) are solutios to the desired equatio. But you DON T eed to fid a solutio to aswer the give questio!) 6. Let p be a prime umber ad let x ad y be itegers. Prove that x + y) p x p + y p Let z be a iteger. If p divides z, the z p z mod p) is true because p divides both sides. O the other had, if p does ot divide z, the z p z mod p) holds by Fermat s little theorem. This shows that z p z mod p) is is always true. Applyig this fact i the cases z x + y, z x ad z y, we see that: x + y) p x + y x p + y p Remark: we proved this fact affectioately called the freshma s dream because the equatio x + y) x + y is a commo bit of wishful thikig amog the youg) usig Fermat s little theorem. But i fact, we could have doe thigs i the opposite order! That is, there is a proof that does ot rely o Fermat s little theorem, ad the we ca prove Fermat s little theorem usig this fact. Here s how this goes.

3 The biomial coefficiet proouced choose k ) is the umber of ways of choosig k uordered) objects from a collectio of objects. Oe ca show that: )! k k!!. So i particular this fractio is a iteger!) Cosider the expressio x + y). I the fial product, the coefficiet of y k is exactly the umber of ways to choose k of the differet terms x + y) to take a factor of y from. It follows that: ) ) ) ) x + y) x + x 1 y + x y + x y + xy 1 + y 1 1 i0 i ) x i y i. Notice that by the symmetry of this expressio i x ad y, this argumet shows that, a fact which is also apparet from the formula ) k!.) k!! Now, we wish to use this formula i the case where p is a prime umber. The key fact is that p divides p whe 1 k p 1. To prove this, otice that p does ot divide either k! or p!, sice each term i these products is less tha p ad hece relatively prime to p. It follows that i the fractio ) p p! k k!p!, there is o factor of p i the deomiator. Sice there is a factor of p i the umerator, it follows that p divides the whole expressio. This proves that p divides p whe 1 k p 1. It follows that i the biomial expasio x + y) p p i0 ) p x p i y i, i all terms are divisible by p except for the summads correspodig to i 0 ad i p. Therefore, x + y) p x p + y p mod p) as claimed. Here is the proof of Fermat s little theorem startig from the freshma s dream. We will prove that a p a mod p) for all positive itegers a by iductio o a. It is clearly true for a 1. Assume iductively that the statemet holds for the iteger a 1. By the freshma s dream, a p a 1) + 1) p a 1) p + 1 p a 1) p + 1 O the other had, a 1) p a 1 by the iductive hypothesis. It follows that a p a 1) + 1 a

4 Havig proved the geeral statemet, we ca derive Fermat s little theorem by takig a relatively prime to p. I this case, we may divide p from both sides of the cogruece, yieldig a p 1 1 Thus cocludes our alterative proof of Fermat s little theorem. 7. Let p be a odd prime umber ad suppose that g ad g are both primitive roots modulo p. Prove that their product gg is ot a primitive root modulo p. Sice p is odd, the fractio p 1 is a iteger. I claim that for a primitive root g, g p 1)/ 1 To see this, write x g p 1)/, ad otice that x g p 1)/ ) g p 1) 1 Therefore, x ±1 Sice g is a primitive root, it caot be the case that x g p 1)/ 1 It follows that g p 1)/ 1 Applyig this argumet for both of the primitive roots g ad g, we see that: gg ) p 1)/ g p 1)/ g p 1)/ 1) 1) 1 This is a istace of a cogruece g k 1 mod p) with k < p 1, which shows that gg is ot a primitive root. 8. a) Let p be a prime umber ad let g be a primitive root mod p. Show that p 1)! g pp 1)/ Sice g is a primitive root, the residue classes g, g, g 3,..., g p 1 mod p) are all distict. Hece the above list of residue classes is the same as the list 1,,..., p 1 mod p), but i a differet order. Takig the product of all residue classes i the list, we see that: p 1)! g g g p 1 g p 1) Usig the idetity k k i i1 k + 1)k

5 with k p 1, we coclude that: as desired. b) Use part a) to deduce that p 1)! g p 1) g pp 1) mod p), p 1)! 1 [Note: to receive credit o this problem, you must prove the result usig part a). It is ot fair to just quote the HW where you proved this i a differet way!] If p, the so the cogruece holds. p 1)! 1! 1 1 mod ), Now assume that p is odd. By the proof give i problem 7, we kow that g p 1)/ 1 Applyig part a), we have: p 1)! g pp 1) g p 1)/) p 1) p We kow that 1) p 1 because p is odd. Therefore, p 1)! 1

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