2011/088) Find a right angled triangle with integer sides whose all three sides are fibonacci numbers.

No solution

Proof:

Let the legs be x and y

X and y cannot be same then hypotenuse shall be x qrt(2) and it is not integer.

Now let x < y and so hypotenuse >= (x+y)

as next Fibonacci number = (x+y) if x and y are consecutive Fibonacci numbers

and > (x+y) if they are not consecutive

So x^2+y^2 >= (x+y)^2 which is not possible unless x = 0

Hence no solution

2011/087) Prove Cos^4x =(1/8)(3+4cos2x+cos4x)

realizing that RHS has got cos of multiple of x

Cos^4x = 1/4(2 cos ^2 x )^2
= 1/4( cos 2x + 1)^2 knowing cos 2x = 2 cos^2 x - 1
= 1/4( cos^2 2x + 2 cos 2x + 1)
= 1/8( 2 cos ^2 2x + 4 cos 2x + 2)
= 1/8( cos 4x +1 + 4 cos 2x + 2) as 2 cos^2 2x - 1 = cos 4x)]
= 1/8( cos 4x + 4 cos 2x + 3)

2011/086) prove that the expression 2x+3y and 9x + 5y are divisible by 17 for same integral values of x and y

if 9x + 5y is divisible by 17 then
4(9x + 5y) is divisible by 17 (note below why multiply by 4)
or 36x + 20 y is divisible by 17
or 36x-34 x + 20y - 17 y or 2x + 3y is divisible by 17

multiplication by 4 was not magic

as 2 = 9m mod 17 gives m = 4

to find it

let us find inverse 0f 9 mod 17

we have 17 = 9 + 8
8 = 17-9
now 9= 8+1 = (17-9) + 1 or
2 * 9 = 17 + 1 so 2 is inverse of 9
so m = 4 as multiplying by 2 gives 1 so multiply by 4 to give 2

2011/085) The quadratic function f(x) is negative for x > 9/2 and x < -1, but for no other value of x. If f(1) = 28 find f(x)

As it is –ve for x > 9/2 and x < -1 it is of the form –A (x-9/2)(x+1) where A > 0

Or – B(2x-9)(x+1)(where B = A/2)

F(1) = -B(-7)(2) = 28A = 28 or B = 2

So we get

F(x) = – 2(2x-9)(x+1)

2011/084) If p q r are roots of equation x^3 – 3px^2 + 3q^2x – r^3 = 0 then prove that p = q = r

proof:

as sum of roots is 3p so p+ q + r = 3p

so there exists t such that

q = p- t, r = p+ t

coefficient of x = pq + pr + qr = 3q

or p(q+r) + qr = 3q^2

or p* 2p + (p – t)(p+t) = 3(p-t)^2

or 3p^2 – t^2 = 3p^2-6pt + 3t^2

or 4t^2-6pt = 0 or

t(2t-3p) = 0 ..1

constant = p(p+t)(p-t) = (p+t)^3

p+t = 0 or p(p-t) = (p+t)^2

or p^2-pt = p^2 +2 pt + t^2

or t^2 + 3pt =0

t(t+3p) = 0 ….2

fom (1) and (2) t = 0

or

2t-3p = 0 and t+ 3p = 0 which again gives t = 0 (also p = 0 this is redundant as a sub set of 1^st solution)

So t = 0

So p = q =r

hence proved

2011/083) Show that for p and q odd x^2 + 2px + 2q = 0 does not have rational solution

proof

the discriminant is

b^2-4ac = (2p)^2 – 8q = 4(p^2-2q)

as p and q are odd

p^2-2q = (2s+1)^2 – 2(2t+1) = (4s^2+ 4s+1) – (4t+2) =(4s^2 + 4s – 4t -1) which is 3 mod 4

so discriminant cannot be a perfect square so there is no rational root

2011/082) The equation x³ + px² + qx + r = 0 (where p, q, r are non zero) has roots α, β, γ such that 1/α , 1/β, 1/γ...?

are consecutive terms in an arithmetic sequence, show that β = -3r / q

f(x) = x^3+px^2+qx+r = 0 has the roots α, β, γ

so f(1/x) has the roots 1/α , 1/β, 1/γ. Which are in AP

f(1/x) = 1/x^3+p/x^2+q/x+ r = 0

or rx^3+qx^2+px+1 = 0

sum of roots = - q/r = 3/ β or β = -3r/q (sum of 3 terms of AP = 3 * middle term)

proved