Math IOQM

 

Q1. Let a, b, c ∈ ℝ. Prove the inequalities: 3(ab + bc + ca) ≤ (a + b + c)² ≤ 3(a² + b² + c²) S1. First see the RHS one. Cauchy shwartz. (1^2 + 1^2 + 1^2)(a^2 + b^2 + c^2) >= (1.a + 1.b + 1.c)^2 Since (a1.b1 + a2.b2 + a3.b3)^2 <= (a1^2 + a2^2 + a3^2).(b1^2 + b2^2 + b3^2) And equality happens when: a1/b1 = a2/b2 = a3/b3 => a/1 = b/1 = c/1 => a = b = c Now the LHS one, we need to do effectively this: ab + bc + ca <= a^2 + b^2 + c^2 WLOG a <= b <= c Which is simple using rearrangement inequality. And in RE also a = b = c gives the equality. Q2. S2. Now expanding it will give us the required inequality. Q3. S3. Q4. S4. Q5. S5. Method 2:

 V_p(n) = k => highest power of prime p in number 'n' is k. This is known as p-adic valuation. For e.g. V2(12) = 2 => p-adic valuation of 2 wrt 12 is 2. Also it's additive, ie V_p(m.n) = V_p(m) + V_p(n) V2(36) = V2(18.2) = V2(18) + V2(2) = 1 + 1 = 2 Legendre's formula:

 for any integer k > 1, there will always be a prime p s.t. k < p < 2k.

Q1. Find all positive integers n for which 2^n + 7^n is a perfect square. S1. for n = 1, it's 9. Ok. for n even  2^n = (-1)^n mod 3 = 1 mod 3 7^n = 1^n mod 3 = 1 mod 3 2^n + 7^n = 2 mod 3 for all even 'n'. But any positive number square = 0,1 mod 3. So no even n is possible. for n odd and n>=3 2^n = 0 mod 4 3^n = (-1)^n = -1 mod 4 = 3 mod 4 2^n + 7^n = 3 mod 4 Any positive square  = 0,1 mod 4 So only answer is n = 1 Q2. Given five distinct integers, consider the ten differences formed by pairs of these numbers. (Note that some of these differences may be equal.) Determine the largest integer that is certain to divide the product of these ten differences, regardless of which five integers were originally given. S2. 1,2,3,4,5 => 288 = 2^5.3^2 Now show 288 will always divide any such product. 3k,3k+1,3k+2 3 buckets, 5 objects. No matter how you distribute, you will get 2 factors of 3. Because at least 1 bucket will have more than 1 object. Case 1: 1 bucket with 3 objects...