Math IOQM

S1. Let a = sqrt(2014) => 4029 = 2a^2 + 1 Equation becomes: ax^3 - (2a^2 + 1).x^2 + 2 = 0 ax^3 - x^2 - 2a^2.x^2  + 2 = 0 x^2(ax -1) - 2(a^2x^2 - 1) = 0 x^2(ax -1) - 2(ax + 1)(ax - 1) = 0 (ax - 1)(x^2 - 2ax - 2) = 0 x = 1/a is one root. x^2 - 2ax - 2 = 0 gives x = [2a +-sqrt(4a^2 + 8)]/2 = a +- sqrt(a^2 + 2) So the 3 roots are: a > 1 1/a (middle) a + sqrt(2 + a^2) (biggest) a - sqrt(2 + a^2) (smallest since negative) So x2 = 1/a Answer = 2 Q2. S2. Take 4 to LHS and get: x/x-3 + x/x-5 ... = x^2 - 11x => x = 0 is one solution => x/x-3 + x/x-5 ... = x - 11 Now combine 3,19 and 5,17 terms 1/x-3 + 1/x-19 = 2x - 22/(x-3)(x-19) 1/x-5 + 1/x-17 = 2x - 22/(x-5)(x-17) => 2(x-11)[1/(x-3)(x-19) + 1/(x-5)(x-17)] = x-11 => x = 11 is another root. Now 1/(x-3)(x-19) + 1/(x-5)(x-17) = 1/2 1/(x^2 - 22x + 57) + 1/(x^2 - 22x + 85) = 1/2 Let x^2 - 22x + 57 = u => 1/u + 1/(u + 28) = 1/2 => 2(2u + 28) = u^2 + 28u = 4u + 56 => u^2 + 24u - 56 = 0 u = [-24 +-sqrt(24*24 + 4*56)]/2...

Q1. For any integer a, show the following: (a) gcd(2a+1,9a+4)=1. (b) gcd(5a+2,7a+3)=1. (c) If a is odd, then gcd(3a,3a+2)=1.  S1. (a) Using Euclid algorithm: (2a+1,9a+4) = (2a+1,a) = (1,a) = 1 (b) (5a+2,7a+3) = (5a+2,2a+1) = (a,2a+1) = (a,1) = 1 (c) (3a,3a+2) = (3a,2) Since 3a is odd => gcd = 1 Q2. If  a∣bc, then show that a∣ gcd(a,b) gcd(a,c) S2. Using Bezout's identity: gcd(a,b) = d1 = m1a + m2b gcd(a,c) = d2 = n1a + n2c d1.d2 = m1.n1.a^2 + m1.n2.ac + n1m2ab + m2n2bc We need to show that d1d2 = ak. Each term in d1.d2 is div by a except m2n2bc but we know that bc is div by a. H.P. Q3. Use the Euclidean Algorithm to obtain integers x and y s.t. (a)  gcd(56,72)=56x+72y. (b) gcd(24,138)=24x+138y. (c) gcd(119,272)=119x+272y. (d) gcd(1769,2378)=1769x+2378y. S3. (a) (56,72) = (56,72-56.1=16) = (56-16*3 = 8,16) = 8 8 = 56-(72 - 56.1)*3  = 56*4 - 3*72 = 224-216 (b) gcd(24,138) = 24,138-24*5 = 24,18 = 24-18,18 = 6,18 = 6 (138-24*5)*(-1) + 24 = 24*6 - 138*1 = 144-138 ...

Q2. For any arbitrary integer a, 3 | a (2a^2 + 7) True or false S2. a(2a^2 + 7) mod 3 = a(2a^2 + 1) = 2a^3 + a Case 1: a = 0 mod 3 => True Case 2: a = 1 mod 3 => 2 + 1 True Case 3: a = 2 mod 3 => 4 + 2 = 6 True True = answer Q3. If a,b are both odd integers then a^4 + b^4 - 2 | 16 True or false. S3. For an even integer a^4 mod 16 = 0. Clearly. For an odd integer a^4 mod 16 = 1. Why? a = 2m + 1 a^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1 m(m+1) is even = 2k => a^2 = 8k + 1 a^4 = 64k^2 + 16k + 1 H.P. So a^4 + b^4 - 2 = 0 mod 16 H.P. True Q4. If a is an odd integer then 24 | a ( a^2 - 1) S4. True Method 1: a(a^2 - 1) = (a-1).a.(a+1) In 3 consecutive integers, one is always div by 3. (a-1) and (a+1) are 2 consecutive even integers. In 2 consecutive even integers, one is div by 2, other by 4. So in total, div by 8. So div by both 8,3 H.P. Method 2: a(a^2 - 1) = (a-1).a.(a+1) = (2m)(2m+1)(2m+2) = 4m(2m+1)(m+1) = 4m(m+1)(2m+1) We need to show that m(m+1)(2m+1) is divisible by 6. Clearly m(m...

 

Q1.  Find the number of integer solutions for | |x| - 2020| < 5 S1. |x| - 2020 = y => -5 < y < 5 => -5 < |x| - 2020 < 5 Part 1: -5 < |x| - 2020 => 2015 < |x| => x < -2015 or x > 2015 AND Part 2: |x| - 2020 < 5 => |x| < 2025 => -2025 < x < 2025 Combine both to get x = -2024... -2016 and x = 2016 to 2024 Answer: 18 Q2.  Find the sum of positive integers (n) for which |2^n + 5^n - 65|  is a perfect square. S2. for n = 1, y = 58 for n = 2, y = 36 = one solution found, n = 2 for n >= 3 y > 0 so we can remove absolute sign. 2^n + 5^n - 65 = k^2 Take mod 3 (-1)^n + (-1)^n = -1 + k^2 mod 3 If n is odd: -1 = 2 = k^2 mod 3 But k^2 mod 3 can be only 0,1. => n has to be even Let n = 2m and m>=2 y = 4^m + 25^m - 65 m = 2 => y = 16 + 625 - 65 = 576 = 24^2 => n = 4 is a solution. m = 3 => 4^3 + 25^3 - 65 = k^2 = 25^3 - 1 = 125^2 - 1 so not a square since you won't get 124^2 from here. For m>=4 => 4^m - 6...

Q1. Can product of eight consecutive integers be the perfect 4-th power of an integer? S1. n(n+1)(n+2)...(n+7) rearrange: n(n+7)(n+1)(n+6)(n+2)(n+5)(n+3)(n+4) Let a = n^2 + 7n + 6 Then product becomes: P = (a-6).a.(a+4).(a+6) = a^4 + 4a^3 -36a^2 - 144a (a+1)^4 = a^4 + 4a^3 + 6a^2 + 4a + 1 This is clearly more than the above product. Why? Since a is positive. (a+1)^4 > P Can we show a^4 < P? P = a^4 + 4a(a^2 - 9a - 36), a^2 - 9a - 36 = (a-12)(a+3) > 0  since a >= 14 for n = 1 => a^4 < P < (a+1)^4 H.P. that it can't be 4th power. Q2. S2. Evaluate the expression: Denominator would become: k(k+1) Numerator = k^4 + 2k^3 + 3k^2 + 2k + 1 = (k^2+k+1)^2 So each term is: (k^2 + k + 1)/(k^2 + k) = 1 + 1/k(k+1) 1/k(k+1) = 1/k - 1/k+1 Now upon addition: 1s add upto 40. And telescopic sum for the rest: 1/1 - 1/2 1/2 - 1/3 ..... 1/40 - 1/41 Add all to get: 1/1 - 1/41 = 40/41 So the final sum is 40 + 40/41 a + b = 80 = answer Another method: if x + y + z = 0 then: (1/x + 1/y ...