Math IOQM

 Q1. S1. Method 1: a1 + a2 + a3 + a4 + a5  = ai1 + ai2 .. ai5 => (ai1 - a1) + (ai2 - a2) ... + (ai5 - a5) = 0 = even__[1] Now if the product of all the terms in [1] is odd => each of the terms is odd. But if each of the terms is odd then sum has to be odd not even. Hence proof by contradiction. Method 2: PHP WLOG at least 3 numbers of these 5 are even(odd will also work). No matter how you permute them, at least one position will have both numbers even. Hence the product will be even. Q2. S2. There are n possible counts of handshakes: 0,1,2...n-1 But 0 and n-1 can't occur together. So n pigeons and n-1 holes. H.P. Q3. S3. at least 3 of them will leave the same mod wrt 3 (which can be only 0,1) So adding those 3 will give a number divisible by 3. Q4. S4. ab(a-b)(a+b) (a-b) and (a+b) are even. Now we need a power of 5. Let's create 3 remainder groups mod 5. 0 1,4 2,3 If any of the integers is 0 mod 5. Pick that. Else at least 2 integers will belong to the same group. If th...

 Q1. S1. p is odd. why? Since q(p-1) = perfect square p is either 2 or an odd prime. p = 2 => q = perfect square but q is prime. So p has to be odd. Now what? p - q is perfect square => p >= q If p = q then p^2 - p is perfect square. => p(p-1) is perfect square which is not possible since p is a prime and (p-1) can't have p as a factor. => p > q With trial and error we find (3,2) = (p,q) as a valid solution. p - q = a^2___[2] pq - q = b^2__[1] a,b are positive integers. b^2 > a^2 why? pq - q > p - q => pq > p which is true since q >= 2 [1] - [2] => pq - p = b^2 - a^2 p(q-1) = (b-a)(b+a)___[3] p is a prime. It will divide at least one of (b-a) and (b+a). a^ = p - q < p < p^2 => 0 < a < p. b^2 = pq - q < pq < p^2 => 0< b < p. => 0 < b - a < p and => a + b < 2p So p can't divide (b-a), it will have to divide (b+a). But a + b < 2p => a + b = p since that's the only number smaller than 2p and...

Q1. Find all f: R -> R s.t. for all real x,y  f(x + y) + f(xy) = x + y + xy + 2 S1: plug in y = 0 => f(x) + f(0) = x + 2. Plugin x = 0 => f(0) = 1 => f(x) = x + 1 Check the answer in original equation. f(x+y) = x + y + 1 f(xy) = xy + 1 They satisfy the original equation. Q2. Find all f: R -> R s.t. for all real x,y f(2x + y) = f(-y) + 2f(x) + 4y S2. We want to find f(x) but other two f() are in our way. Remove them. 2x + y = -y => -y = x so use this substitution y = -x f(x) = f(x) + 2f(x) + 4y => f(x) = -2y = 2x Check answer in the original equation. It matches. Q3. Find all f: R -> R s.t. for all real x,y f(f(x) + y) = f(x-y) + 2y S3. Making inputs inside f() equal => f(x) + y = x - y  => 2y = x - f(x) => f(x) = x And this matches in the original equation. Q4. Find all f: R -> R s.t. for all real x,y 2f(x) + f(1-x) = 4x + 1 S4. Replace x with 1-z 2f(1-z) + f(z) = 4(1-z) + 1 Now we have 2 vars f(z), f(1-z) and 2 equations. Solve to get f(x)....

 Q1. S1. f(0+0) = f(0) + f(0) = 2f(0) => f(0) = 0. f(t-t) = f(0) = 0 = f(t) + f(-t) => f(-t) = -f(t) => odd function. f(1+1) = 2f(1) = f(2) Similarly, f(3) = 3.f(1) In fact, simply using (a) we can prove f(x) = cx for all rational numbers. How? Assume f(1) = a For any positive integer 'n': f(n. 1/n) = f(1) = f(1/n) + f(1/n) ... n times. using (a) f(1) = n.f(1/n) => f(1/n) = a/n For another non zero positive integer 'm': f(m/n) = f(m.1/n) = m.f(1/n) = m.a/n = a.(m/n) = f(1).(m/n) f(-m/n) = -f(m/n) as we showed earlier. So any rational number m/n can be expressed as a multiple of f(1). Hence for any rational number 'x': f(x) = a.x H.P. But how do we prove it for irrational numbers? Partially we can. 2.f(sqrt(3)/2) = f(sqrt(3)) using (a) Similarly, f(2.sqrt(3)) = 2.f(sqrt(3)) So f(q.t) = q.f(t) where q is rational and t is real. But we need to prove f(x) = cx. How? We need to use the relation (b). f(x) = (x^2).f(1/x) Earlier we assumed f(1) = a We need ...

Q1. S1. f(2) = 3 f(2^2) = f(4) = 3.3 = 9 2 < 3 < 4 => f(2) < f(3) < f(4) => 3 < f(3) < 9 => f(3) can be 4,5,6,7,8 Let f(3) = a. Now, f(3^2) = f(9) = f(3).f(3) 8 < 9 < 16 => f(8) < f(9) < f(16) => 27 < f(9) = a^2 < 81 => 27 < a^2 < 81 => a is 6,7,8 (Since 4,5 doesn't work) Now 16 < 27 < 32 f(16) < f(27) < f(32) 81 < a^3 < 243 => a = 6 (7,8 don't work) So f(3) = 6 Now, Powers of 2: 2,4,8,16,32,64,128,256 Powers of 3: 3,9,27,81,243 f(243) < f(256) 6^5 < 3^8 3^5.2^5 < 3^8 2^5 < 3^3 32 < 27 contradiction. H.P. Q2. S2. Let S be the set of all values of f(n) for n >= 1. From well ordering principle, there will be a minimum value in this set(made up of non zero integers). Let that value be d. Let that value occur at m => f(m) = d Now, there are 2 cases, m = 1 or m >= 1. If m >= 1, we can write the given inequality as: f(m) > f(f(m-1)) If f(m-1) = t then f(m) > f(t) =>...

Q1.  S1.