RI 2022 H3 Test 1 (Questions and Solutions)

RI 2022 RAFFLES INSTITUTION 2022 Year 6 H3 9820 Lecture Test 1 Time allocated: 1 hour Total Marks: 25 Instructions: Write your name and CT group on all the work you hand in. Answer all questions. 1 The positive integers a, b, c and d satisfy the equation 2( ) ( )( ).ad bc a b c d− = + + (a) Show that there are positive integers x, y and z, with x and y coprime, such that 22 and a b x z c d y z+ = + = . [4] (b) Find a quadratic equation satisfied by y x and hence, or otherwise, prove that 41ac+ is a perfect square. [6] (a) [4] Let gcd( , )z a b c d= + + . Then clearly z is a positive integer, and we have also , a b pz c d qz+ = + = , where p and q are coprime. Substituting these into the equation we have 2 22() ad bcad bc pqz pq z −− =  =  . Since p and q are coprime and their product is a perfect square, they must be perfect squares themselves. Hence there are positive coprime integers x and y such that 22,p x q y== and thus positive integers x, y and z such that 22 and a b x z c d y z+ = + = . (b) [6] We note that 22 .ad bc a b c dz xy x y − + += = = So we have 2 2 ,y c d x a b += + .y ad bc x a b −= + Note also that 2 2 ()y y a c d ad bc ac bcacx x a b a b a b + − +− = − = =+ + + which is the quadratic equation we are looking for. Solving the quadratic equation we have 1 1 4 .2 y ac xa += Since x and y are coprime positive integers, 14 ac+ has to be an integer (otherwise it is irrational since a and c are positive integers), and thus 41ac+ is a perfect square.

RI 2022 2 Let u and v be quadratic functions of x and let .uy v= (a) Use mathematical induction to prove that 2 1 2 2 1 2 2d d d d d( 2) 0, 2d d d d d n n n n n n ny v y v yvn x x x x x ++ ++ ++ + + =  for 1n . [8] (b) Now assume that 2()vx =− for some real number  and, for all positive integers n, define 2( ) d .!d nn n n xyz nx  +−= Use the result of (a) to prove that 1 2 3, , ,...z z z is an arithmetic progression. By writing y as partial fractions, or otherwise, show that the common difference is ()u  . [7] (a) [8] Let nP be the statement 2 1 2 2 1 2 2d d d d d( 2) 0 2d d d d d n n n n n n ny v y v yvn x x x x x ++ ++ ++ + + =  for positive integers n. From uy yv uv=  = and thus differentiating with respect to x we have d d d .d d d y v uvyx x x+= Differentiating once more with respect to x we have 2 2 2 2 2 2 d d d d d2.d d d d d y y v v uvyx x x x x+ + = Differentiating once more with respect to x and using the fact that u is a quadratic expression implies that 3 3 d 0,d u x = (since d d u x linear, 2 2 d d u x constant) 3 2 2 2 2 3 3 3 2 2 2 2 3 3 d d d d d d d d d d d2 2 0.d d d d d d d d d d d y y v y v y v y v v uvyx x x x x x x x x x x+ + + + + = = This simplifies to 3 2 2 3 3 2 2 3 d d d d d d3 3 0.d d d d d d