2014/045) Solve for x: (ab)^2 =(bc)^ 4 =(ca)^ x =abc

we have ab = (abc)^(1/2)
bc = (abc)^(1/4)
ca = (abc)^(1/x)
multiply and get
(abc)^2 = (abc)^(1/2 + 1/4 + 1/x)

hence 1/2 + 1/4 + 1/x = 2 or x = 4/5

2014/044) Find all positive integers a and b such that a(a+2)(a+8)=3^b

all of a , a + 2 and a + 8 have to be power of 3( power 0 included)

a cannot be  power of 3 >= 1 or >= 3 then a+ 2 and a+ 8 are not divisible by 3

so check a =3^ 0 = 1

 that give a + 2 = 3 and a + 8 = 9 both power of 3

so we get a(a+2)(a+8)=3 3  

so a = 1 and b= 3 is the only solution

2014/043) For any polynomial P(x) show that P(a) - P(b) is divisible by a-b

 Proof:
Let p(x) =  t(n)x^n + t(n-1)x^{n-1} + ...  + t(0)
Then
 p(a) =  t(n)a^n + t(n-1)a^{n-1} + ...  + t(0)
 p(b) =  t(n)b^n + t(n-1)b^{n-1} + ...  + t(0)

So  p (a) – p(b) =  t(n)(a^n- b^n)  +t(n-1)(a^(n-1)-b^(n-1)  + ....  + t(1)(a-b)

As each of the a^k-b^k  is divisible by a- b so  p(a) – p(b) is divisible by a-b.

As a corollary


If p(x) has integer coefficients and P(0) and P(1) are odd it does not have any integer root.
This is so because P(even) – p(0) is even and P(odd) – p(1) is even so neither can be zero

2014/0042) show that x = 2 + 2^(1/3) + 2^(2/3) is a root of x^3 - 6x^2 + 6x -2 = 0

Let x = 2 + 2^(1/3) + 2^(2/3)
so x -  2 =  2^(1/3) + 2^(2/3)  ... (1)

cube both sides to get (x-2)^3 = 2 + 4 + 3 *2 * (2^(1/3) + 2^(2/3)) = 6 + 6 (x-2) ( from 1)
or x^3 - 6x^2 + 12 x - 8 = 6x - 6
or x^3 - 6x^2 + 6x - 2 = 0

so x is a root of above equation
proved

2014/0041) If a,b≥3,prove that 2^b−1 does not divide 2^a+1

There are 2 cases
case 1
b is a factor of a say a = mb

then 2b−1 is a factor of 2mb−1 so 2b−1 is not a factor of 2a+1 as remainder = 2
case 2
b is not a factor of a so a = mb + c where c < b

2a+1=2mb+c+1
= 2c(2mb−1)+2c+1
now 2b devides 2c(2mb−1) but as c < b
2c+1<2b−1as2c+2<2c+1

so it does not devide

2014/040) Let x,y,z be not necessarily distinct integers between 1 and 2011, inclusive. Find the smallest possible value of (xy+z)/(x+y+z).

xy+zx+y+z
= 1+xy−x−yx+y+z
= 1+(x−1)(y−1)−1x+y+z

(x-1)(y-1) - 1 is positive for all x and y except for x=1 or y=1 ( in the condition x, y <= 2011)

so x =1 , and y = 1

so we get given expression
= 1−12+z
z = 1 shall make it lowest

so x = 1 = y = z shall give the value 23

2014/039) Find x,y

(x+1)^(1/2) + (y+1)^(1/2) = 3

                xy – x – y + 15 = 0

I copy the 1st equation

(x+1)^(1/2) + (y+1)^(1/2) = 3

The second equation given

xy–x–y+1+14=0
Or
(x−1)(y−1)=−14...(2)

One one of (x-1) and (y-1) is –ve
As it is symmetric in xand y we can assume x to be so
So x – 1 = - 14 or -2 or – 1
From the 1st equation x + 1 >= 0 so x >= -1
So x = -1 , 0
X = 0 => y- 1 = 15 so does not satisfy (1)
X= -1 => y = 8 and it satisfiles (1)
So x = -1 , y = 8 or x = 8, y = -1