2010/041) If (x-k)^2 is a factor of x^3 + 3px + q

prove that 4p^3 + q^2 =0
proof:
(x-k)^2 is a factor of x^3+3px + q so x-k is factor of this as well as derivative of this

so (x-k) is a factor so k^3 + 3pk + q = 0

and (x-k) is a factor of d/dx (x^3+3px + q) or 3x^2 + 3p or x^2+p =0

so k^2 + p = 0 ... 1 or p = -k^2

sum of zeros zero so 3rd factor = -2k

product = -2k^3 = q ...2

from 1 p = - k^2 or p^3 = -k^6 or 4p = -4k^6
and q^2 = 4k^6by adding we get 4p^3+q^2 = 0

2010/040) Let f(x) be a polynomial of degree>2

if f(x) is divided by (x-1) and(x-2)...?
it leaves a remainder 0,2 respectively.what is the remainder when f(x) is divided by (x-1)(x-2).

if f(x) is divided by (x-1)(x-2) remainder shall be a polynomial of degree 1

f(x) = p(x)(x-1)(x-2) + a x +b

f(1) = a + b = 0 so a = - b is the remainder when divided by (x-1)

f(2) = 2a + b = 2 so -2b +b = 2 or b - 2 and a = 2

so remainder = 2x - 2

2010/039) A problem in fibonacci sequence

show that
Σ (n = 1 to ∞) F(n)/x^n = x/(x^2-x-1) where F(n) is n th Fibonacci number

Where "x" is any integer >= 2

proof:
this is based on Geometric series and not on relationship in Fibonacci sequence which can be found at http://en.wikipedia.org/wiki/Fibonacci_number#Power_series

We know

F(n) = (p^n – (1-p)^n)/ sqrt(5) where p is phi (value = (1+ sqrt(5))/2)

So sqrt(5) F(n)/x^n = (p/x)^n – (1-p)/x)^n

So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = Σ (n = 1 to ∞)
(p/x)^n – Σ (n = 1 to ∞) (1-p)/x)^n

As x is positive
The sum converges when (p/x) < 1 and (1-p)/x < 1

Or x > p that is x > the golden ratio and need not be integer.

The sum = 1/(1-(p/x)) – 1/(1-(1-p)/x)
= x/(x-p) – x/(x-1+p)
= x((1/(x-p) – 1/(x-1+p))
= x(x-1+p) –(x-p))/)(x^2- x –(p(p-1))

As p(p-1) = 1 this is x^2-x-1
we get
So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = x(2p-1)/ (x^2-x-1)

As p = (1+ sqrt(5))/2) so (2p-1) = sqrt(5)

So Σ (n = 1 to ∞) sqrt(5) F(n)/x^n = x sqrt(5)/ (x^2-x-1)

Or Σ (n = 1 to ∞) F(n)/x^n = x/ (x^2-x-1)
Proved

2010/038) For a natural number n, let An = n^2 + 20

If Dn denotes the greatest common divisor of An and An+1,?
then show that Dn divides 81.

proof:
GCD(An , An+1)as we need to get a constant
= GCD(n^2+20, (n+1)^2 + 20) (one of 1st or 2nd term need to be constant by proper transform)
= GCD(n^2+20, n^2+2n + 21)
= GCD(n^2+20, 2n+1) as GCD(a,b) = GCD(a,b-a)
[now we need to eliminate n^2 from 1st term so we double the 1st term as 2nd term is odd so GCD(a,b) = GCD(2a,b) when b is odd]
= GCD(2n^2+40, 2n+1)
= GCD(2n^2+40-n(2n+1),2n+21) as GCD(a,b) = GCD(a-kb,b) for any k
= GCD(40-n, 2n+1)
= GCD(80-2n, 2n+1) [ same argument above as 2n+1 is odd double the 1st term
= GCD(81, 2n+1) as GCD(a,b) = GCD(a+b.b)

now 81 is constant and GCD must devide 81

2010/037) Sum of six consecutive whole squares can never be a whole square

let the 1st number be n^2

sum of 6 consecutive squares is

n^2+(n+1)^2+(n+2)^2+(n+3)^2+(n+4)^2 + (n+5)^2

= 6n^2+ 30n + 55
= 6n(n+5) + 55

1st part 6n(n+5)is divsible by 4 as either n or n+ 5 is even

so 6n(n+5) + 55 mod 4 = 55 mod 4 = 3

it cannot be a perfect square as perfect square mod 4 = 0 or 1

2010/036) if n is an integer give an example of n consecutive composite integers

in the 1st attempt the solution appears to be

(n+1)! + 2 to (n+1)!+n+1 the kth term is divisible k+1

but it can be bettered as

LCM(2,3,... n+1) + 2 to
LCM(2,3,... n+1) + n+ 1

and LCM(2,3,... n+1) is the lower of the two

It can still be bettered to the product of all primes (<= n+1)say P(n) + 2 to that + n+ 1

as the number from 2 to n+1 if a prime the product has that prime as a factor

and if not a prime then P(n) and and k have a common factor so sum is not a prime

(one of my friends informed me)

2010/035) Show that

27 x (23^n) + 17 x (10^(2n)) is divisible by 11 for all positive integers n

proof:
23 = 1 mod 11
so 23^n = 1 mod 11

10^2 = 100 = 1 mod 11

so 10^2n =1 mod 11

so 27 x (23^n) + 17 x (10^(2n)) = 27 * 1 + 17 * 1 = 44 = 0 mod 11

so divisible by 11