Number Theory. Everything else. Misha Lavrov. ARML Practice 10/06/2013

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Transcription:

Number Theory Everything else Misha Lavrov ARML Practice 10/06/2013

Solving integer equations using divisors PUMaC, 2009. How many positive integer pairs (a, b) satisfy a 2 + b 2 = ab(a + b)?

Solving integer equations using divisors PUMaC, 2009. How many positive integer pairs (a, b) satisfy a 2 + b 2 = ab(a + b)? 1 Let p be a prime. Let p x be the highest power of p dividing a, and p y be the highest power of p dividing b.

Solving integer equations using divisors PUMaC, 2009. How many positive integer pairs (a, b) satisfy a 2 + b 2 = ab(a + b)? 1 Let p be a prime. Let p x be the highest power of p dividing a, and p y be the highest power of p dividing b. 2 Suppose x < y. Then p 2x is the highest power dividing a 2 + b 2, p x+y is the highest power dividing ab, and p x is the highest power dividing a + b.

Solving integer equations using divisors PUMaC, 2009. How many positive integer pairs (a, b) satisfy a 2 + b 2 = ab(a + b)? 1 Let p be a prime. Let p x be the highest power of p dividing a, and p y be the highest power of p dividing b. 2 Suppose x < y. Then p 2x is the highest power dividing a 2 + b 2, p x+y is the highest power dividing ab, and p x is the highest power dividing a + b. 3 So p 2x = p 2x+y, which means y = 0. But x < y, so this is impossible. So we can t have x < y; we can t have x > y for the same reason, so x = y.

Solving integer equations using divisors PUMaC, 2009. How many positive integer pairs (a, b) satisfy a 2 + b 2 = ab(a + b)? 1 Let p be a prime. Let p x be the highest power of p dividing a, and p y be the highest power of p dividing b. 2 Suppose x < y. Then p 2x is the highest power dividing a 2 + b 2, p x+y is the highest power dividing ab, and p x is the highest power dividing a + b. 3 So p 2x = p 2x+y, which means y = 0. But x < y, so this is impossible. So we can t have x < y; we can t have x > y for the same reason, so x = y. 4 This is true for all p, so a = b. Then 2a 2 = a 2 (a + a) = 2a 3, so a = b = 1.

Competition-level problems AIME, 1991. How many fractions a b are there, for which ab = 20! (when written in simplest terms)? How many of these satisfy 0 < a b < 1? Ukrainian MO, 2002. Solve for integers m, n. n 2002 = m(m + n)(m + 2n) (m + 2001n) British MO, 2002. Find all solutions in positive integers a, b, c to the equation a! b! = a! + b! + c!. Putnam, 2000. Prove that the expression gcd(n,k) ( n ) n k is an integer for all pairs of integers n k 1.

Competition-level problems Solutions AIME, 1991. We can factor 20! = 2 18 3 8 5 4 7 2 11 13 17 19. (What s important here is that there are 8 primes that appear in the factorization of 20!, which are the 8 primes 20.) If ab = 20! and a b is in simplest terms (that is, gcd(a, b) = 1) then each prime number must go entirely in a or entirely in b. There are 2 possibilities for each prime, and eight primes, so that s 2 8 = 256 choices. How many are between 0 and 1?

Competition-level problems Solutions AIME, 1991. We can factor 20! = 2 18 3 8 5 4 7 2 11 13 17 19. (What s important here is that there are 8 primes that appear in the factorization of 20!, which are the 8 primes 20.) If ab = 20! and a b is in simplest terms (that is, gcd(a, b) = 1) then each prime number must go entirely in a or entirely in b. There are 2 possibilities for each prime, and eight primes, so that s 2 8 = 256 choices. How many are between 0 and 1? We always have a b > 0, and either a b < 1 or b a < 1. Therefore the answer is 128: half of the total number of fractions.

Competition-level problems Solutions Ukrainian MO, 2002. Let p be a prime. Then: If p divides m, then p divides the RHS, so p divides the LHS, which is n 2002. Therefore p divides n. If p divides n, then p divides the LHS, so p divides the RHS, which means p divides m + kn for some k. Since p also divides kn, p must divide m.

Competition-level problems Solutions Ukrainian MO, 2002. Let p be a prime. Then: If p divides m, then p divides the RHS, so p divides the LHS, which is n 2002. Therefore p divides n. If p divides n, then p divides the LHS, so p divides the RHS, which means p divides m + kn for some k. Since p also divides kn, p must divide m. Normally, we d refine this approach to show that the same power of p divides m and n. Here, there is a shortcut: If m and n are solutions, so are m p and n p. Unless m = n = 0, we can keep dividing by p until one is no longer divisible by p; but then the other can t be divisible by p either. In any case, we prove m = n; but the only solution of this kind is m = n = 0.

Competition-level problems Solutions British MO, 2002. Ruling out 0 a 2 and 0 b 2, a! b! is much larger than a! or b!, so c is the largest of the three integers. Next, we show that a! = b!. Suppose a! < b!: then b! is divisible by (a + 1)!, and if we write a! b! b! c! = a! then everything on the left is divisible by (a + 1)!, while a! is not. This is impossible. Now we have a! 2 = a! + a! + c!, or a!(a! 2) = c!. Since a! 2 is not divisible by 3, a! and c! must have the same number of factors of 3, so c = a + 1 or c = a + 2. Checking both, we get a single solution: 3! 3! = 3! + 3! + 4!

Competition-level problems Solutions Putnam, 2000. Our goal is to show that gcd(n, k) ( n k) is divisible by n. For all primes p, suppose p x divides n and p y divides k. If x y then all is good, because at gcd(n, k) ( n k) is divisible by p x. If x > y, we can use the following trick: ( ) ( n = n n 1 k k 1), and so we can rewrite gcd(n, k) n ( ) n = k gcd(n, k) k k ( ) n 1. k 1 Now we have only a power p y in the denominator, and at least p y in the numerator, so no power of p is left in the denominator, and we are done.

The totient function The totient, or Euler s φ, is defined to be: φ(n) = The number of k, 1 k n, so that gcd(n, k) = 1. Exercise. Find φ(10000). PUMaC, 2010. Find the largest positive integer n such that nφ(n) is a perfect square.

The totient function The totient, or Euler s φ, is defined to be: φ(n) = The number of k, 1 k n, so that gcd(n, k) = 1. Exercise. Find φ(10000). Easy answer: gcd(10000, k) = 1 if k ends in 1, 3, 7, or 9. There are 4000 such numbers between 1 and 10000. General answer: Out of 10000 integers, 1 2 are divisible by 2, and 1 5 are divisible by 5, so there are 10000 ( 1 1 ( ) 2) 1 1 5 = 4000 left. PUMaC, 2010. Find the largest positive integer n such that nφ(n) is a perfect square. Using the general answer above, it s easy to see nφ(n) can t be a perfect square for n > 1.

Rule for raising something to a power mod m Theorem (Euler s theorem) For all positive integers a, n with gcd(a, n) = 1, a φ(n) 1 (mod n) and therefore a m m mod φ(n) a (mod n). Intuition: If gcd(a, 10) = 1, then there are φ(10) = 4 digits a can end in: 1, 3, 7, and 9. The powers of a will cycle through these digits: for example, when a = 3, we have 3 0 = 1, 3 1 = 3, 3 2 = 9, 3 3 = 27 7, 3 4 = 81 1,...

Rule for raising something to a power mod m Theorem (Euler s theorem) For all positive integers a, n with gcd(a, n) = 1, a φ(n) 1 (mod n) and therefore a m m mod φ(n) a (mod n). Intuition: If gcd(a, 10) = 1, then there are φ(10) = 4 digits a can end in: 1, 3, 7, and 9. The powers of a will cycle through these digits: for example, when a = 3, we have 3 0 = 1, 3 1 = 3, 3 2 = 9, 3 3 = 27 7, 3 4 = 81 1,... If gcd(a, n) 1, then powers of a eventually repeat every φ(n) steps, but this is trickier to use.

Competition problems (Note: this theorem is also useful for small things, like knowing that 1 4 2 4 3 4 4 4 1 (mod 5) last week. These are problems where Euler s theorem is the main focus.) Exercise. Compute 100 100 mod 13. Texas A&M, 2008. Find the last three digits of 2007 2008. VTRMC, 2012. Find the last two digits of 3 3... 3 }{{}. 2012 HMMT, 2011. Determine the remainder when is divided by 7. 2 1 2 2 + 2 2 3 2 + + 2 2011 2012 2

Competition problems Solutions Exercise. 100 100 ( 4) 100 ( 4) 4 9 (mod 13). Texas A&M, 2008. 2007 2008 7 2008 7 8 (mod 1000). A shortcut for this: 7 2 = 49 = 50 1, so 7 8 = (50 1) 4 = 50 4 4 50 3 + 6 50 2 4 50 + 1. But here, the first three terms are all divisible by 1000, so all we need to worry about is 4 50 + 1 801 (mod 1000).

Competition problems Solutions VTRMC, 2012. Write 3 n for 3 3... 3 with n 3 s. We use Euler s theorem recursively: for 100 we need φ(100) = 40, for which we need φ(40) = 16, for which we need φ(16) = 8, for which we need φ(8) = 4, for which we need φ(4) = 2. Since 3 is odd, 3 2007 1 (mod 2). So 3 2008 3 1 3 (mod 4). So 3 2009 3 3 27 3 (mod 8). So 3 2010 3 3 27 11 (mod 16). So 3 2011 3 11 27 (mod 40). So 3 2012 3 27 87 (mod 100).

Competition problems Solutions HMMT, 2011. We know 2 n mod 7 is determined by n mod 6. But actually, more is true: 2 3 1 (mod 7), so n mod 3 is enough. When looking at n(n+1) 2 mod 3, we know either n 1, n, or n + 1 is divisible by 3. Unless it s the first, n(n+1) 2 is also divisible by 3, in which case 2 n(n+1) 2 1 (mod 7). However, when n 1 is divisible by 3, n(n+1) 2 1 (mod 3), and 2 n(n+1) 2 2 (mod 7). Therefore 2 1 2 2 + 2 2 3 2 + + 2 2011 2012 2 mod 7 simplifies to 2 } + 1 + 1 + 2 + {{ 1 + 1 + + 2 } mod7 2011 which is 2010 3 (2 + 1 + 1) + 2 1 (mod 7).

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