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 y y v y v vvyx x x x x x+ + + = Since 2( 2) 3 2 nn ++ = = when n = 1, the statement 1P is true. Now suppose nP is true for some positive integer n, i.e. for some positive integer n, 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 ++ ++ ++ + + =  .

RI 2022 Differentiating once more with respect to x we have 3 2 2 2 1 3 2 2 2 1 3 2 1 3 2 1 3 2 2 1 3 2 2 1 d d d d d d d ( 2) ( 2)d d d d d d d 22 d d d d 0 22 d d d d 2d d d d d( 3) 2 2d d d d d n n n n n n n n nn nn n n n n n n y v y v y v yv n nx x x x x x x nn v y v y x x x x ny v y v yv n nx x x x x + + + + + + + + + + + + + + + + + + + + + ++   + + =        +   + + + + +   0= since 3 3 d 0d v x = as v is a quadratic. Since 2 ( 2)( 1)22 2 2 2( 2) ( 2)( 1) 2 ( 2)(2 1) ( 3)( 2) 22 3 2 n nnnn n n n n n n n n + +++ + = + + + + + += + + + + +== += We thus have 3 2 2 1 3 2 2 1 3d d d d d( 3) 0 2d d d d d n n n n n n ny v y v yvn x x x x x + + + + + + ++ + + =  . Since 1P is true and 1nnPP + , by Mathematical Induction, nP is true for all positive integers n. (b) [7] To show 1 2 3, , ,...z z z is an arithmetic progression, we will show that 2 1 1n n n nz z z z+ + +−=− or equivalently 21 20n n nz z z++− + = . From the definition of 2( ) d ,!d nn n n xyz nx  +−= we have 4 2 3 1 2 21 21 2 2 1 2 21 ( ) d ( ) d ( ) d22 ( 2)! d ( 1)! d ! d ( ) d d d ( ) 2( )( 1) ( 2)( 1)( 2)! d d d n n n n n n n n n n n n n n n n n n n x y x y x yz z z n x n x n x x y y y x x n n nn x x x      + + + + + ++ ++ + + + ++ − − −− + = − + ++ −= − − − + + + ++  Here, we note that with the choice of 2()vx =− , we have d 2( )d v xx =− − and 2 2 d 2d v x = . Therefore ( ) 21 2 2 1 21 2 2 1 2 2 1 2 2 ( ) d d d ( 2)( 1) d ( 1) 2( 2)! d d d 2 d 2( ) d d d d d ( 1) 0. from (a) 2( 2)! d d d d d n n n n n n n n n n n n n n n n n z z z x y v y n n yvnn x x x x nx y v y v yvnn x x x x x   ++ + + + ++ + + + ++ −+ − + += + + + +   +  −= + + + = +  Hence 1 2 3, , ,...z z z is an arithmetic progression.

RI 2022 2 .() uuy vx == − Since u is a quadratic, we can write 2( ) ( ) BCyA xx= + + −− . Thus 2 22 ( ) ( ) ( ) ( ) A x B x C uy xx   − + − +== −− . Substituting x = we have ()Cu = . We also have 2 2 3 2 3 4 d 2 d 2 6 d ( ) ( ) d ( ) ( ) y B C y B C x x x x x x   = +  = +− − − − . The common difference of the arithmetic progression is 4 2 3 21 2 4 3 3 4 2 3 ( ) d ( ) d 2! d 1! d ( ) 2 6 2 ()2 ( ) ( ) ( ) ( ) ( ) 3 ( ) 2 () x y x yzz xx x B C B C xx x x x x B C x B C Cu         −−− = −    −= + − − +   − − − −    = − + − − − == as required. [END OF PAPER]