2011/037) If α is a root of equation x² + x + 1 = 0 then show that α is a root of equation x^(3p) + x^(3q+1) + x^(3m+ 2) = 0

Proof:
α is a root of equation x² + x + 1 = 0

so α^2 + α + 1 = 0 and hence α^3 =1

now there are 2 approaches

1)

α ^(3p) + α ^(3q+1) + α ^(3m+ 2)
= (α ^3)^p + (α ^3)^3q* α +( α ^3)3m* α ^2
= 1 + α+ α^2
= 0
Hence α is a root

2)

α ^(3p) + α ^(3q+1) + α ^(3m+ 2)
= α ^(3p) + α ^(3p+1) + α ^(3p+ 2) + α ^(3q+1) - α ^(3p+ 1) + α ^(3m+ 2) - α ^(3p+ 2)
= α ^(3p)( 1+ α + α ^ 2) + α (α ^(3q) - α ^(3p)) + + α^2 (α ^(3m) - α ^(3p))
= 0 + 0+ 0 = 0 as each part is zero
Hence α is a root


as a corollary we see that x² + x + 1 devides x^(3p) + x^(3q+1) + x^(3m+ 2)

2011/036) Find z^2010 where The complex number z is such that z^9 = zz', where z and z' are complex conjugates.

let z = r e^(it)

then z^9 = r^9 e^(9it)

zz' = r^2 = r^9 e^(9it)

so e^(9it) = r^7 or r = 1 and hence zz' = 1

so z^9 = 1 so z^2007 = 1 and so z^2010 = z^3 = cube root of 1

z = r = 0 is also another solution

so solutions are 3 cube roots of 1 and zero

2011/035) Prove that if the sides of a triangle are prime numbers its surface can not be whole number.

proof:
Let the sides are a,b c, and a <=b <= c

now there are 2 cases

a = 2 or all are odd

now A^2 = ( a+b-c)(a+b+c)(a-b+c)(b+c-a)/ 4
if all are odd then all 4 terms on the RHS are odd then A^2 cannot be integer so A cannot be whole number

case 2:
for a= 2 and b = 2 or a= 2 and b != 2

a = 2 then b =2 => c = 2 or 3

a =2 b = 2 c = 2 => A^2 = 6*2^3/4 = 12 so A is not integer

a =2 , b= 2 c = 3 => A^2 = 7 * 1 * 3 * 3/4 so A is not integer


if b != 2 then b= c because if c >b then c>= b+2 or a+b= c

so we get A^2 = (2+2b)* (2b-2)* b^2 /4 = b^2(b^2-1)/4 cannot be a perfect square


so no solution

2011/034) show that 3^2n+ 3^n + 1 is divisible by 13 if n is not multiple of 3

proof:

we know 3^3 = 27 = 1 mod 13

so we proceed

if n is of the form 3k + 2 then 2n of form 3m + 1 and if n is of the form 3k + 1 the 2n of the form 3m + 2

so we have
3^2n+ 3^n + 1 = 3^ (3k + 2) + 3^(3m + 1) + 1

= 27^k * 9 + 27^m * 3 + 1
= 9 + 3 +1 mod 13 = 13 mod 13 = 0 mod 13 do divisible by 13

2011/033) Prove: If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq

q is prime

so p^(q-1) mod q = 1 as per FLT


so p^(q-1) = mq + 1

multiply by p on both sides p^q = mpq + p = p mod pq

similarly as q is prime q^p = p mod pq

adding we get (p^q + q^p) mod pq = (p+q)

proved

2011/032) factorize x^4+2x^3y-3x^2y^2-4xy^3-y^4?

seeing coefficient of x^4 as 1 and y^4 is -1 and this is homogeneous in x y

that is power of x and y sum is 4 that is same

we see that if shall be (x^2+axy + y^2)(x^2+bxy-y^2)

= x^2 + x^3y(a+b) + x^2y^2(ab) + (b-a) xy^2 - y^4

so a + b = 2, ab = -3 and a-b = 4

a + b = 2, and a-b = 4 give a = 3 and b = -1 and this satisfies
ab = -3

so we get (x^2+axy + y^2)(x^2+bxy-y^2)
= (x^2+3xy + y^2)(x^2-xy-y^2)

2011/031) If f(x) + 2f(2002/x) = 3x then find f(2)?

ans: this type of problem we need to eliminate f(2002/x)and put x = 2

(x) + 2f(2002/x) = 3x --- 1
we realize that x and 2002/x are reciprocal so putting 2002/x for x we get

f(2002/x) + 2 f(x) = 3(2002/x) ... 2

multiply 2nd by 2 and subtract (1) fom it

3 f(x) = 6(2002/x) - 3 x

or f(x) = 4004/ x - x

putting x = 2 we get

f(2) = 2000

2011/030) prove that

(1....1) = (22....2)+ (3333....3)^2
2n times ntimes + n times

Proof:

LHS = (99999...9)/9 = (10^2n -1 )/ 9
RHS = 2/9( 10^n-1)+ (10^n-1)/9

so we need to prove that (RHS = LHS as RHS is more complex)

(10^n-1 )^2 + 2(10^n-1) = (10^2n- 1)

LHS = 10^2n - 2* 10^n +1 + 2*10^n - 2 = (10^2n- 1)