Some Own Problems In Number Theory
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1 Some Own Problems In Number Theory Mathmdmb April 16, 2011 Here are some problems proposed by me, and the problems or their solutions have not been approved by someone else. So if any fault occurs, I shall take the whole responsibility. In this case, please inform me. Among the problems,many were posted by me on AoPS fora, so I thank the users who posted replies and solutions there. A notable fact is that I put the problems not in order to difficulty, just randomly - which is, in my opinion, more interesting. 1 Notations N = {1, 2, 3,..., n,...} : the set of positive integers. Z = {1, 2, 3,..., n,...} : the set of integers. N 0 = {0, 1, 2,...} : the set of nonnegative integers. a A : a is an element of the set A. a b : b is divisible by a. a b : b is not divisible by a. a b c : b and c are both divisible by a. gcd(a, b) : the Greatest Common Divisor of a and b. a and b are coprime : gcd(a, b) = 1. ϕ(m) : the number of positive integers x, not exceeding m with the property gcd(x, m) = 1. n i=1 a i = a 1 a 2...a n. x : the largest integer not exceeding x. x : for all x. a is squarefree : there does not exist x N such that x 2 a. a N is a perfect number : sum of positive divisors of a is 2a. 1
2 ord m (a) = x : x is the order of a modulo m. v p (x) = α : α is the greatest positive integer such that p α x W 5 : Which is What We Wanted. W LOG : Without Loss Of Generality. 2
3 2 Problems 1. Prove that there does not exist a pair(n, m) N so that n+3m and n 2 +3m 2 are both perfect cubes. Find all such pairs if (m, n) Z. 2. Find all primes p such that the number 11 p + 10 p is a perfect power of a positive integer. 3. Let F n = 2 2n + 1 be the n th Fermat number. Prove that 2 2m +2 n F Fm 1 n 1, m, n N. 4. Let a > 2 be an integer. Show that a a a is never square-free. 5. Show that there are infinitely many pairs of positive integers (a, b) such that if the number a5 +b 5 a 3 b 3 +1 is an integer, then it is a perfect 5th power. 6. Let p 2 (mod 3) be a prime number. Show that there exists a complete set of residue class of p such that the sum of its elements is divisible by p Prove that n n for all n N Find all positive integers n such that n 2 n! Let p be a prime. Determine all perfect numbers having p factors. 10. Prove that a number which has only one prime factor, can t be a perfect number. 11. Solve in positive integers the equation 12. Find all n N such that a 7 + b 7 = (ac) n 2 27n is a perfect square. 2. n 2 27n is a perfect square 13. Find all (a, b) N 0 such that the number 7 a + 11 b is a perfect square. 14. Consider a complete set of residues modulo p. Show that we can partition this set into two subsets with equal number of elements such that the sum of elements in each set is divisible by p. 15. Let a i, m, and n be positive integers such that a i + m is a prime for all 1 i n. Let N = n i=1 pai i and let S be the number of ways of expressing N as a product of m positive integers. Prove that m n S. 16. Find all integers (a, b, c, d) such that abc d = 1, bcd a = 2 3
4 17. Find all positive integer values that x2 +y 2 +1 xy+1 can take where x and y positive integers. 18. Solve in positive integers : 19. Find all n such that ϕ(n) N. n n n 10 8 =
5 3 Solutions 1. If n + 3m, n 2 + 3m 2 both are cubes,then their product too. Then (n + 3m)(n 2 + 3m 2 ) = a 3 = n 3 + 3m 2 n + 3mn 2 + 9m 3 = a 3 = (m + n) 3 + (2m) 3 = a 3 By Fermat s Last Theorem, the last equation has no solution. If (m, n) Z, then we have two cases: 1. m+n = 0 Then we get n+3m = 2m is a perfect cube which gives m = 2x 3. Also n 2 + 3m 2 = 4m 2 From this we conclude that m = 4t 3, n = 4t m = 0 or m = 0 In this case, n + m = n = u 3 and n 2 + 3m 2 = n 2 = (u 2 ) 3 Thus, (m, n) = (4t 3, 4t 3 ), (0, u 3 ) are only solutions. 2. First Solution : First look at p = 2,which is not a solution. Now p odd, so 11 p + 10 p is divisible by = 21.So 3, 7 11 p + 10 p According to the Lifting The Exponent Lemma(LTE), we get that p + 10 p or p + 10 p if p = 3 or p = 7.Checking, we easily get p 3, 7. So, it can t be a perfect power. Second Solution : Again p 2.So 11 p + 10 p 0 (mod 9) The latter implies 2 p 1 (mod 9), since 2 6n+3 1 (mod 9) it follows that 3 p, p = 3. But, = = 2331 is not a perfect power. Third Solution : First check with p = 2, 3 manually which don t work. So let p 5. We know that every prime greater than 3 is of the form 6k ± 1. So, two cases: 1. p = 6k + 1,then 11 p + 10 p 2 p + 1 (2 6 ) k (mod 9).But any integer of the form 9m + 3 = 3(3m + 1) is not a perfect power because 9 does not divide it. 2. p = 6k +5, then 2 p (mod 9) Since it is of the form 9m 3, a similar logic works here too. Therefore, no solution exists. 5
6 3. First Solution : From LTE, if v 2 ( a2 b 2 2 ) = α, v 2 (n) = β,then v 2 (a n b n ) = α + β Applying this,we get that,v 2 (Fm r 1) = v 2 (F m 1) + v 2 (r) Since v 2 (F m 1) = 2 m, v 2 (F n 1) = 2 n, now it follows that Then,2 2m +2 n F Fn 1 m 1 v 2 (F Fn 1 m 1) = 2 m + 2 n Second Solution : In this solution, we use the repeated use of the identity a 2 1 = (a + 1)(a 1). Then, a 2k 1 = (a 2k 1 + 1)(a 2k 1 ) = (a 2k 1 + 1)(a 2k 2 + 1)((a 2k 2 1) = (a 2k 1 + 1)(a 2k 2 + 1)...(a + 1)(a 1) This equation says that v 2 (a 2k 1) = v 2 (a 1) + v 2 (2 k ) = v 2 (a 1) + k. Applying the result,it easily follows that v 2 (Fm Fn 1 1) = v 2 (F m 1)+v 2 (2 2n ) = 2 m + 2 n. Note. Not only 2 2m +2 n Fm Fn 1 1, also 2 2m +2 n mostly divides Fm Fn 1 1. Also 2 2m +2 n Fn Fm a a a = a(a a 1 1) and we claim that a a 1 1 must have a square factor. Let s eliminate the trivial case a = 3 first, since then = 2 3 So a > 3.If a 1 has no odd prime factor, then a 1 must be a power of 2.Since a > 3, a and thus it has a square factor. Assume to the contrary that, a 1 has an odd prime factor, say p. Then from the LTE, we find that v p (a a 1 1) is at least 2, implying that p 2 a a 1 1, as desired. 5. This seems to be like the famous IMO-1988,6 which stated that if a2 +b 2 ab+1 is an integer, then it is a perfect square.but this is not that hard at all. See, we are asked to prove the existence of such an infinite couple,not to prove that,if a 5 +b 5 a 3 b 3 +1 is an integer,then it is a perfect 5th power. This suggests us to find a suitable parameter for a, b such that the condition satisfies. Let s recall the parameters of IMO-1988,6 where, we could find an infinite couples of a, b with a = c, b = c 3. Then, our intution says to choose a = b k The problem is reduced to find k such that b 5k = b 3k which gives k = 4. Indeed, a 5 + b 5 a 3 b = b5 Thus, (a, b) = (c 4, c), (c, c 4 ) gives such an infinite couple. W 5 6
7 6. First we prove the following fact. Lemma : If a 2 + ab + b 2 is divisible by p 2 mod 3, then p a, p b Proof : Let p = 3k + 2. So, a 2 + ab + b 2 0 mod p = a 3 b 3 (mod p) Now from Fermat s Little Theorem a p 1 b p 1 (mod p) It is well-known that if a m b m (mod p) and a n b n (mod p) then, a gcd(m,n) b gcd(m,n) (mod p) = a gcd(3,p 1) b gcd(3,p 1) (mod p) = a gcd(3,3k+1) b gcd(3,3k+1) (mod p) = a b (mod p) = p a b From this, we see that p a 2 2ab + b 2, a 2 + ab + b 2 and we can say p a 2 + ab + b 2 (a 2 2ab + b 2 ) = 3ab Since gcd(p, 3) = 1, p ab implying that p a, p b. Now back to the original problem. Let s consider the complete residue class 1, 2,..., p 1.Then according to the lemma, 1 3, 2 3,..., (p 1) 3 is also a complete residue class of p.if two of its elements were congruent modulo p. Let i 3 j 3 mod p = (i j)(i 2 +ij+j 2 ) 0 mod p. Because i j < p, gcd(i j, p) = 1 and so i 2 + ij + j 2 0 mod p which would imply p i, j which is clearly absurd. Hence, any two terms are incongruent modulo p and it forms a complete residue class. And the sum of them is p2 (p 1) 2 4, it is obviously divisible by p First Solution : We use induction on n. The base case n 0 = 0 is true since = 0 Let it be true for some n.then we require to prove that n+2 9n n+2 9n 19 = 10(10 n n) + 81(n + 1) This is now obviously divisible by 81. Second Solution : This is solution is more elegant than the previous one. Consider the sum, S = upto n th terms containing n 1 s. Let s try to find a closed formula for S. 9S = = = 10( ) n = 10( 10n 1 9 ) n So, S = 10n n 81. Since, S is a positive integer, 81 must divide 10 n n. 7
8 8. Firstly, n odd. Now, 2 ϕ(n) 1 (mod n) since gcd(n, 2) = 1. But, ϕ(n) < n, so n! must contain ϕ(n) as a factor since n! is the product of integers from 1 to n. Thus, ϕ(n) n!. 2 n! 1 mod n,so true for all n odd. 9. We claim that, if N has p factors, then N must be a perfect power of only one prime. This is because if N = p r 1p s 2 with p 1, p 2 distinct primes, then the number of divisors of N is (r + 1)(s + 1) can t be a prime unless r = 0 or s = 0. Because, otherwise the number of divisors of N = (r + 1)(s + 1) which is not a prime. So, our claim is true. If the sum of proper divisors is equal to N,say N = q p 1.Thus, 1 + q + q q p 2 = N = qp 1 1 q 1 But, it is not possible as qp 1 1 q 1 < q p 1. This proves the desired fact. = q p This problem is actually a special case of the previous one. Note. In fact, we can say more specifically that the sum of some divisors of a prime power or a number having a prime number of divisors. 11. This is an equation involving very high exponents. So, it can t have many solutions. In fact, we shall prove that it has no positive integer solution. First Solution : At first note that and the consequence is obvious,we need to search for Fermat s Last Theorem. So, we need to express as a perfect 7 th power. Let s try a bit. The given number is not divisible by 3 because the sum of its digits 29 is not divisible by 3.So see if it is divisible by the next prime 7. A number is divisible by 7 iff the number obtained by the subtraction of the three digit number formed by its last three digit and the number that is left. Example is divisible by 7.See that, = 399.Since,399 is divisible by 7,the original number is divisible by 7. Now, let s check whether is divisible by 7 or not = 280 which is divisible by 7. This leads us to factorize Note that = Again, = 532 = 7 76.Thus, = = = = 7 7 The initial equation becomes a 7 + b 7 = (7a 285 c 285 ) 7 having no solution in positive integers. 8
9 Second Solution : This is an easier and more elementary,elegant one. Again,since a 7 a 1995, a 7 b 7. Let b = ak. Th equation stands k = (7a 284 c 285)7 But, no two perfect 7 th powers don t differ by First Solution : We shall use two very useful fact in solving this problem. It is a widely known tactic. There is no positive integer between two consecutive positive integers. So there is no perfect square which lies between two consecutive perfect squares. The product of two consecutive integers is not a perfect square. 1. Just note that n 2 27n = (n 13)(n 14). Since n 13 and n 14 does not share any common factor,n 13 and n 14 both must be squares. But then the difference of two squares is 1,contradiction! 2. Again, n 2 27n = (n 13)(n 14) + 1. Let n 14 = m giving us m(m + 1)) + 1 = x 2 If m is a positive integer,. If m is a negative integer, m 2 < m 2 + m + 1 < m 2 + 2m + 1 = (m + 1) 2 m 2 > m 2 + m + 1 > m 2 + 2m + 1 Thus, n 2 27n+183 can t be a perfect square except 1 yielding n = 14, 13. Second Solution : This time we shall use discriminant of a quadratic equation. 1. Let, n 2 27n = z 2. Then, its discriminant is (182 z 2 ) = 4z is a perfect square. But (2z) is a square only for z = 0. Then n = 13, 14. 9
10 2. The discriminant of n 2 27n = z 2 is 729 4(183 z 2 ) = 4z 2 3. Let, 4z 2 3 = x 2 = (2z + x)(2z x) = 3 = 2z + x = 3, 2z x = 1 = z = x = 1 which gives us n = 13, Let 7 a + 11 b = x 2. Let s make several cases : Case 1 : a = 0, b > 0. Then, the equation is reduced to 11 b = (x + 1)(x 1) which means that 11 x + 1 x 1 = 11 2 Contradiction! Case 2 : b = 0, a > 0. This is similar to the preceeding one. Case 3 : a, b > 0. Then 7 a + 11 b must be even. So 7 a + 11 b 0 (mod 4). Take modulo 4 in the equation. So, x 2 0 (mod 4) = 7 a + 11 b 0 (mod 4) = ( 1) a + ( 1) b 0 (mod 4) From the latter,we may conclude that a and b are of different pairity. Otherwise x 2 2 (mod 4) which is not true. Again, take modulo 3. If y even, 7 a + 11 b 2 (mod 3) but no square gives remainder 2 upon division by 3. This makes b odd and a even. Let a = 2k. Thus, 11 b = (x + 7 k )(x 7 k ) = 11 x + 7 k x 7 k = k Contradiction! The only case left is b = 1 which does not produce any solution. Therefore, no such a, b exists. 14. Lemma : If p > 3 a prime, then 24 p 2 1. Proof : Since p > 3, obviously p odd and not divisible by 3. Thus p ±1 (mod 3) and p ±1 (mod 2). Combining theses two p ±1 (mod 6). Let, p = 6k ± 1 = p 2 = 36k 2 ± 12k + 1 = 12k(3k + 1) = p
11 k and 3k + 1 are of opposite pairity since their difference is not divisible by 2. So, one of them must be even and thus, 12 2 = 24 p 2 1. The complete residue class is {1, 2,..., p 1} Now, recall that there are exactly p 1 2 quadratic residues mod p. The sum of the residues is p(p 1) 2 which is divisible by p. So, we seek for only a set with p 1 2 elements, since then the sum of the elements of the other set will be divisible by p automatically. And consider the quadratic residues 1 2, 2 2,..., ( p 1 2 )2. Note that, their sum is Now, we are done if 24 p 2 1 W ( p 1 2 )2 = p(p2 1) Lemma 1 : For p prime and a positive integer r < p, r ( p 1 r 1). Proof : Note the identity : ( ) p = p ( ) p 1 r r r 1 Here, gcd(r, p) = 1 but ( ) ( p r is a positive integer. So r must divide p 1 r 1). Lemma 2 : If N = n i=1 pai i and the number of ways to write N as a product of m positive integers, then n ( ) ai + m 1 S = m 1 Proof : Take any a i. Thus, there are n i=1 i=1 From the balls in urns principle, there are ( a i+m 1) m 1 ways. ( ai+m 1) n ( m 1 ways and S = ai+m 1) i=1 m 1. Let, a i + m = q i where q i are primes. Then, from lemma 1 above, m ( q i 1 m 1) for all q i. Multiplying them together we get, m n n ( ) ai + m 1 = S m 1 i=1 16. From the equations, abcd = d 2 + d and abcd = a 2 + 2a implying that d 2 + d + 1 = (a + 1) 2 If d N, d 2 + d + 1 is not a perfect square from problem 12. So, we have 1 d 0. If d = 0, a = 2 and abc = 1 = 2bc = 1, no solution. 11
12 If d = 1, abc = 0 and bc = (a + 2). Now, a(a + 2) = d(d + 1) = 0 giving a = 0 or a = 2. If a = 0, bc = 2 = (b, c) = (1, 2), ( 1, 2), ( 2, 1), (2, 1) 17. We can use the famous Vietta Jumping to solve this problem. We claim that (x, y) = (c, c+1) is the only pair for which the fraction x2 +y 2 +1 xy+1 is a positive integer and thus the only integer value it can assume is 2. Let x 2 + y xy + 1 = k and the smallest solution of be (x, y). Since is symmetric over x, y we can assume W LOG x y. Note that x = y yields x x x = x < 2 x 2 + 1, contradiction!! Therefore, x > y. x 2 kxy + y k = 0 Fix in y, k and consider all values of (x, y). will be quadratic in x and let the other root of be x 1. From Vietta s formula, Thus x + x 1 = ky xx 1 = y k x 1 = y2 + 1 k x < y2 x < y2 y = y But then (x, x 1 ) < (x, y) would be a smaller solution than the smallest solution. Contradiction! 18. From Fermat s little theorem, a 11 1 = a 10 0 or 1 (mod 11). n n n , 1, 2, 3, 4, 5, 6, 7, 8 (mod 11) But (mod 11) which isn t possible by any combination of the above. 19. Let, the canonical prime factorization of N is, N = p a1 1...pan n well known that ϕ(n) = p a p an 1 n.(p 1 1)...(p n 1) Then it is. Therefore, ϕ(n) N implies (p 1 1)...(p n 1) p 1...p n. So, one of the primes, say p 1 = 2. Moreover, N can t have more than two prime factors, otherwise 2p 2...p n would be divisible by 4, but p 2,..., p n are distinct odd prime factors. This yields a contradiction! Now we have N = 2 a p b. Then p 1 2p. Since Gcd(p 1, p) = 1, they can not share any common factor. So, p 1 2 and because p odd, we conclude that 12
13 p = 3. Thus, we found an infinite numbers satisfying this property with N = 2 a 3 b. Note that a is necessarily positive integer. Note. We can re-state the problem as : Find all N such that there are N k positive numbers less than N and co-prime to N. Then the value of k would be 3. 13
14 References [1] AoPS topic #393335, Lifting The Exponent Lemma (Containing PDF file), posted by Amir Hossein Parvardi (amparvardi on AoPS). [2] Santiago Cuellar, Jose Alejandro Samper, A nice and tricky lemma (lifting the exponent), Mathematical Reflections [3] AoPS topic #356847, n+3 perfect cube (Problem 1 in this article), posted by mathmdmb. [4] AoPS topic #36224, perfect power (Problem 2 in this article), posted by mathmdmb. [5] AoPS topic #393335,Lifting The Exponent Lemma (Problem 5 in this article), posted by amparvardi. [6] AoPS blog #85314,The Law Of Nature-Number Theory (Problem 8 in this article), posted by mathmdmb. [7] AoPS topic #356079, Find all n (Problem 16 in this article), posted by mathmdmb. [8] AoPS topic #304361, Problem 202 in Number Theory Marathon (problem 17 in this article), posted by mathmdmb. [9] AoPS topic #365172, Solve equation (problem in this article), posted by mathmdmb. [9] AoPS topic #362222, equation (problem in this article),posted by mathmdmb. 14
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