2012/027) What is (p - 1)! mod p when p is not prime? Wilson's Theorem (p - 1)! ≡ -1 (mod p) when p is prime

There are 2 cases.
1) P is not a perfect square of a prime. Then there are 2 unequal factors and product of them mod p = 0 hence 0

2) P is a square of a prime say q. Then product of q and next multiple of q ( that is 2q) mod p = 0

unless 2q = p where 2q goes outside the range For that 2q= p = q^2 => q = 2

P = 4 and (p-1)!mod p = 3! Mod 4 = 2

else if p is composite it is zero

2012/026) Solve: (sqrt(1+3x)-sqrt(2x-1)) = sqrt(x+2)

given
(sqrt(1+3x)-sqrt(2x-1)) = sqrt(x+2) ..1

we know
(1+3x) - (2x-1) = (x+ 2)

so given
(sqrt(1+3x)-sqrt(2x-1))(sqrt(1+3x) +sqrt(2x-1)) = (x+2) ..2

divide (2) by (1) but before dividing check that x + 2 is zero is not a solution

x = 0 => x = - 2

putting in (1) we see the solution is satisfied.

Assume x is not 2 and we devide to get

(sqrt(1+3x) +sqrt(2x-1)) = sqrt(x+2) ..3

from (1) and (3)

(sqrt(1+3x) +sqrt(2x-1)) = (sqrt(1+3x) -sqrt(2x-1))

so sqrt(2x-1) = 0 or x = 1/2

check:
LHS = (sqrt(1+3x)-sqrt(2x-1)) = (sqrt(5/2)-sqrt(0))= sqrt(5/2))
RHS = sqrt(5/2)

x = {1/2 ,-2} is the solution set

2012/025) factor a^2(b-c)^3+b^2(c-a)^3+c^2(a-b)^3.

This is cyclic and we can keep one unchanged and collect the a terms and constants together
= a^2(b-c)^3 + b^2(c^3-3ac^2 + 3a^2c – a^3) + c^2(a^3-3a^2b + 3a b^2 – b^3)
= a^2(b-c)^3 + a^3(c^2-b^2 ) + 3a^2(b^2c – bc^2) + b^2c^3- c^2b^3
= a^3(c^2-b^2) + a^2(b^3 – c^3) + b^2c^2(c-b)
= (c-b)(a^3(c+b) - a^2(b^2 + c^2 + bc) + b^2c^2)

Now
(a^3(c+b) - a^2(b^2 + c^2 + bc) + b^2c^2) is cubic in a but we can make it quadratic in c by decreasing on order of c (easier to factor)
= c^2(b^2-a^2) + c( a^3 – a^2b) + a^3b - a^2b^2
= c^2(b-a)(b+a) + ca^2(a-b) + a^2b(a-b)
= (b-a)(c^2(b+a) + ca^2 – a^2b)

Now we need to factor
(c^2(b+a) - ca^2 – a^2b)
= c^2b + c^2 a – ca^2 – a^2 b
= b(c^2-a^2) + ca(c-a)
= b(c-a) (c+ a) + ca(c-a)
= (c-a) (ab+ bc + ca)

So we get the complete factorization as
(c-a)(c-b)(b-a)(ab+bc + ca)
= (c-a)(b-c)(a-b)(ab + bc +ca) multiplying 2nd and 3rd term by 1 to get in cyclic form

2012/024) Prove there are no positive integer solutions to x^2 + x + 1 = y^2?

we know x^2 < x^2 + x + 1 < x^2 + 2x + 1 as x is positive

so x^2 < x^2 + x + 1 < (x+1)^2

as x and x +1 are successive integers there cannot be an integer whose square is between x^2 and (x+1)^2,
hence proved

2012/023) if GCD(a,b) and LCM(a,b) are given, how can we find a,b?

Let m = GCD(a,b)
n = LCM(a,b)

find k = n/m

now prime factorize k = p1^q1p2^q2... pr^q2

from that factorize k into product of 2 co-primes x and y.( or 1, k) with x < y

then numbers are mx and my

to illustrate

let GCD = 12 LCM = 144

k = 144/12 = 12 = 2^2*3

so the factors x, y are (1, 12),( 2^2,3)

so numbers are (12,144), (48,36)

2012/022) Prove that tanA + tan(π/3 +A) - tan(π/3 -A) =3tan3A

LHS
= tan(A) + tan(A+60°) - tan(60-A)

= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) - (tan 60 - tan(A))/(1 + tan60tan(A))

= tan(x) + (tan(x) + tan(60))/(1 - tan(x)tan(60)) + (tan A - tan 60)/(1 + tan60tan(A))

= tan(x) + (tan(x) + √3)/(1 - √3tan(x)) + (tan(x) - √3)/(1 + √3tan(x))

= tan(x) + (tan(x) - √3)/(1 + √3tan(x)) + (tan(x) + √3)/(1 - √3tan(x))

= tan(x) + [(tan(x) - √3)(1 - √3tan(x)) + (tan(x) + √3)(1 + √3tan(x))]/(1-3tan²(x))

= tan(x) + [tan(x) - √3tan²(x) - √3 + 3tan(x) + tan(x) + √3tan²(x) + √3 + 3tan(x)]/(1-3tan²(x))

= tan(x) + 8tan(x)/(1-3tan²(x))

= (tan(x) - 3tan³(x) + 8tan(x))/(1-3tan²(x))

= (9tan(x) - 3tan³(x))/(1-3tan²(x))

= 3(3tan(x) - tan³(x))/(1-3tan²(x))

= 3tan(3x)

= RHS

2012/021) If the expression E= a(sin^6t + cos^6t) - b(sin^4t + cos^4t) + 1 vanishes for all values of θ, then (a + b) .

sin ^ 6 t + cos^ 6 t= ( sin ^2 t+ cos^2 t) ^3 - 3 sin ^2 t cos^ 2 t( sin ^2 t + cos^2 t) = 1 - 3 sin ^2 t cos ^2 t

sin ^4 t + cos^ 4t = (sin ^2 t + cos ^2 t)^2 - 2 sin^2 t cos^2 t

a(sin^6t + cos^6t) - b(sin^4t + cos^4t) + 1
= a(1- 3 sin^2 t cos^2 t) - b(1- 2 sin^2 t cos ^2 t) + 1
= ( a + 1 - b) - sin^2 t cos ^2)(-3a + 2b)

as it is zero for any t so

a + 1 = b
- 3a + 2b = 0
or -3a + 2(a+1) = 0
a= 2 and b = 3 hence a+ b = 5

2012/020) find(1+1/2)(1+1/2^2)(1+1/2^4)-----… product of infinite rems

(1+x)(x+x^2)(1+ x^4) ,,, (1+x^(2n) n goes to infinite and x= 1/2

multiply by (1-x) to get

(1-x)((1+x)(1+x^2)(1+ x^4) ,,, (1+x^(2^n))
=(1-x^2)(1+x^2)(1+ x^4) ... (1+x^(2^n))
= (1-x^(2^(n+1))

so (1-x)((1+x)(x+x^2)(1+ x^4) ,,, (1+x^(2^n)) = ( 1- (x^(2^(n +1))/(1-x) = 1/(1-x) as n-> infinte and |x| < 1
as x = 1/2
product = 2

2012/019) If a and b are unequal and x^2+ax+b and x2+bx+a have a common factor ,then show that a+b+1=0

they have a common factor x- m

so m^2 + ma + b = 0 ...1
m^2 + mb + a = 0 ..2

subtract 2 from 1

m(a-b) + (b-a) = 0

m = 1

this is the common factor and putting in any of above equation we get the result

2012/018) If 2^x=3^y=6^-z, then find the value of (1/x+1/y+1/z)

we have 2 = 6^(-z/x)
3= 6^(-z/x)

multiply to get 6 = 6^(-z/x-z/y)

or 6^1 = 6^(-z/x-z/y)
or 1= - z/x -z/y

or 1 + z/x + z/y = 0

deviding by z we get 1/z + 1/x + 1/y = 0