2009/032) factor f(x) = x^3-3x^2+4

by taking factors of 4 that is 1,-1.2,-2,4,-4 we see that

f(-1) = 0

so (x+1) is a factor

so x^3-3x^2+4
= x^3+x^2-4(x^2-1)
= x^2(x+1)-4(x+1)(x-1)
= (x+1)(x^2-4(x-1)
= (x+1)(x^2-4x+4)
= (x+1)(x-2)^2

2009/031) simplify sin10*sin50*sin60*sin70*sin90

we know sin 90 = 1 and sin 60 = sqrt(3)/2
hence
sin10*sin50*sin60*sin70*sin90 = sqrt(3/2) sin 10 sin 50 sin 70
= sqrt(3)/2 sin 10 sin (90-40) sin (90-20)
= sqrt(3)/2 sin 10 cos 40 cos 20 ( as sin (90-x) = cos x)
= sqrt(3)/2 sin 10 cos 20 cos 40 ..1

Knowing sin 2a =2 sin a cos a and 20 is double of 10 and 40 is double od 20 we proceed

now sin10 cos 20 cos 40 = (cos 10 sin 10 cos 20 cos 40)/cos 10
= (2 cos 10 sin 10 cos 20 cos 40)/(2 cos 10)
= ( sin 20 cos 20 cos 40)/( 2 cos 10
= (2 sin 20 cos 20 cos 40)/( 4 cos 10)
= ( sin 40 cos 40)/( 4 cos 10)
= ( 2 sin 40 cos 40)/( 8 cos 10)
= sin 80 /( 8 cos 10)
= cos 10/(8 cos 10) = 1/8 ...2

from 1 and 2 we get
sin10*sin50*sin60*sin70*sin90 = sqrt(3)/16

2009/030) Factorize the polynomial 2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4}

Factorize the polynomial
2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4}?

We need to factorize
{a^4+b^4+c^4} – 2(bc)^2 - 2(ca)^2 - 2(ab)^2 and multiply by -1
Now
{a^4+b^4+c^4} – 2(bc)^2 - 2(ca)^2 - 2(ab)^2
= a^4 – 2a^2(b^2+c^2) + b^4+c^4-2ab^2c^2
= a^4 – 2a^2(b^2+c^2) + b^4+c^4+2b^2c^2- 4b^2c^2
= a^4 – 2a^2(b^2+c^2) + (b^2+c^2)^2- 4b^2c^2
= (a^2-b^2-c^2)^2 – (2bc)^2
= (b^2+c^2 – a^2)^2- (2bc)^2
= (b^2+c^2-a^2+2bc)(b^2+c^2-a^2-2bc)
= (b^2+c^2+2bc-a^2)(b^2+c^2-2bc-a^2)
= ((b+c)^2-a^2)((b-c)^2 – a^2)
= (b+c+a)(b+c-a)(b-c+a)(b-c-a)

So given expression
2{(bc)^2+(ca)^2+(ab)^2} - {a^4+b^4+c^4} = - (b+c+a)(b+c-a)(b-c+a)(b-c-a)
= (a+b+c))b+c-a)(a+b-c)(a+c-b) multiplying last term by -1 to get symmetrical form

2009/029) prove a+b+c=0 ==> 2a⁴ +2b⁴ +2c⁴ = n²

a+b+c = 0

the a+b = - c
square both sides

(a^2+b^2+2ab) = c^2
so (a^2+b^2) = c^2 -2ab

now again square both sides

a^4+b^4 + 2a^2b^2 = (c^2-2ab)^2
or a^4 + b^4 + 2a^2b^2 = c^4 + 4a^2b^2 - 4abc^2
or a^4+b^4 = c^4 + 2a^2b^2 - 4abc^2
add c^4 on both sides to get

a^4+b^4+c^4 = 2c^4 + 2a^2b^2 - 4abc^2

or 2(a^4+b^4+c^4) = 4c^4+4a^2b^2 - 8abc^2
= 4(c^4-2abc^2+a^2b^2)
= 4(c^2-ab)^2

which is a perfect quare putting n = 2 | c^2-ab|
proved

2009/028) Rationalize the denominator of expression? [1 / {1 + x^1/5}]

Rationalize the denominator of expression?
[1 / {1 + x^1/5}]


ANS:
we know that (x^(1/5))^5 = x

say x^(1/5) = a

1+ x^ 1/5 = (1+ a)

as (1+a^5)/(1+a) = (1-a+a^2-a^3+a^4) and 1+a^5 in terms of x is rationalised

so
[1 / {1 + x^1/5}] = (1-x^(1/5) + x^(2/5) - x(^3/5) + x^(4/5))/(1+x)

2009/027) If α = 2 arctan [(1 + x)/(1 - x)] and β = arcsin [(1 - x^2)/(1 + x^2)], then what is α + β

we know tan(pi/4+y) = tan (pi/4 + tan y)/(1 tan pi/4-tan y)

so tan (pi/4+y) = (1+ tan y)/(1-tan y))

so pi/4 + y = tan ^-1(1+tan y)/(1-tan y)

put x = tan y

so pi/4 + y = arctan ((1+x)/(1-x))

α = 2 arctan ((1+x)/(1-x)) = pi/2 + 2 tan ^- 1 x .1

again cos 2y = cos ^2 y - sin ^2 y

= cos^2 y(1- tan ^2 y)
= (1- tan ^2 y)/sec^2 y
= (1- tan ^2 y)/(1+ tan ^2y)

so put tan y = x

cos 2y = (1-x^2)/(1+x^2)
2y = arccos ((1-x^2)/(1+x^2))

so arcsin ((1-x^2)/(1+x^2)) = pi/2 - 2y = pi/2 - 2 tan ^- x .2

addimg 1 and 2 we get
α + β = π

2009/026) If a , b , c are in Arithmetic Progression prove: 1/(b1/2+c1/2),1/(c1/2+a1/2),1/… also in AP.

If a , b , c are in Arithmetic Progression prove : 1 / (b1/2+c1/2) , 1 / (c1/2+a1/2) , 1 / (a1/2+b1/2) in AP?

to get rid of ^1/2

Let a = x^2 , b= y^2 and c = z^2

We need to prove

1/(y+z) , 1/(x+z) and 1/(x+y) are in AP


As x^2 y^2 and z^2 are in AP

So x^2+xy + yz + xz , y^2 + +xy + yz + xz , z^2+xy + yz + xz (adding same to all terms) are in AP

So (x+y)(x+z), (y+x)(y+z), (z+x)(z+y) are in AP by factoring

So 1/(y+z) , 1/(x+z) and 1/(x+y) are in AP by deviding each term by (x+y)(x+z)(y+z)

2009/025) Let t_n denote the nth triangular number. For what values of n does t_n divide the sum t_1+t_2+...+t_n?

Well we know t_m = (m²+m)/2

We want S_n = Σ_n=1_m [ t_m ] = Σ_n=1_m [ (m²+m)/2 ]
Applying Faulhaber's Formulas,
S_n = [ (2n³+3n²+n)/6 + (n²+n)/2 ] /2
= (2n³+3n²+n + 3n²+3n) /12
= (2n³+6n²+4n) /12
= n(n+1)(n+2) /6

So the question is for what values of n does
n(n+1)/2 | n(n+1)(n+2)/6

RHS/LHS = (n+2)/3 and LHS is a factor if and only if this is an integer

so n + 2 = 0 mod 3 or n = 1 mod 3

so for n = 1 mod 3 the value t_n devides the sum

2009/024) Resolve into factors X^4-3x+20

if this can be factored with real coefficients then it can be as product of 2 quadratic
we shall get
(x^2+ax+b)(x^2+cx+d)

now coeffcient of x^3 = 0 so a +c = 0 or c = -a

so we get(x^2+ax+b)(x^2-ax+d)
= x^4 + (d + b - a^2) x^2 + x(ad-ab) + bd

comparing coefficients

a^2 = b + d
bd = 20
so ad -ab = - 3
by trial and error we see that a = 3 , b= 5 and d= 4 (this is using 20 = 1 * 20 = 2(*10 - 4 * 5 outof which only 4 +5 = 9 we get perfect square) satisfies
this can be factored with rational coefficient as
(x^2+3x+5)(x^2-3x+4)
and not further this can be checked by quadaric factorisation

hence X^4-3x+20 = (x^2+3x+5)(x^2-3x+4)

2009/023) Find values of x: √x + √(x+2) + √[x(x+2)] = 3-x

as product of 1st term and 2nd term is 3rd I would take the 3rd term to the right and get

√x + √(x+2) = (3-x)- √[x(x+2)]

square both sides to get
x + (x+2) + 2√[x(x+2)] = (3-x)^2 + x(x+2)- 2(3-x) √[x(x+2)]

or
2x + 2 + 2√[x(x+2)] = 9-6x+ x^2 + x^2 + 2x - 2(3-x) √[x(x+2)]
= 9-4x+ 2x^2 +(2x-6) √[x(x+2)]

now keeping redicals on one side we get
2x + 2 - (2x^2-4x+9) = (2x-8) √[x(x+2)]
(2x-8) √[x(x+2)] = -2x^2 +6x -7
now by squaring again you can get rid of redicals and proceed

No one has solved so far so I continue

(2x-8)^2x(x+2) = (-2x^2+6x-7)^2

or expanding we get after simplfication

64x^2-212x + 49 = 0

or(4x-1)(16x-49) = 0

x = 1/4 or 49/16
x has to be less than 3 as LHS > 0

so x = 1/4 need to be checked and this is indeed a solution

so x = 1/4

2009/022) Prove that c(r,r)+c(r+1,r)+.....+c(n,r)=c(n+1,r+1)

The right hand side is choosing r+1 objects from n+1 objects,

let us count another way

as we need to chose r+1 objects if we order the numbers from 1 to n+1 and chose r+1 objects the last object shall be in position r+1 to n+1

let last object be in position m ( r+1 <= m <=n + 1)

then we chose r objects from m-1 objects

this can be done in c(m-1,r)

m varies from 1+1 to n+1

so total number of ways = sum c(m-1.,r) m from r+1 to n+ 1

or sum c(m,r) (m from r to n)
= c(r,r) +c(r+1,r)+ c(r+2,r)+ … c(n,r)

As by 2 methods there are 2 results(it has been counted correctly) so both are same and hence

c(r,r) +c(r+1,r)+ c(r+2,r)+ … c(n,r) = c(n+1,r+1)

hence proved

2009/021) Factorize the polynomial x^4 + 4 with real coefficient?

x^4 + 4
= x^4 + 4x^2 + 4 - 4x^2
= (x^2 + 2)^2 - (2x)^2
= (x^2 + 2x + 2) (x^2 - 2x + 2)

2009/020) Pythagorean triplets.- Algebraic way

There is an algebraic way getting Pythagorean triplets

we know

x^2+y^2 = z^2 become a Pythagorean triplet if we have x y and z all to be integers.

If we can find x,y and z all to be rational then multiplying them by LCM of the denominator we are through.

To sort it we start with

n^2+(2n+1) = (n+1)^2 ..1

we know that n^2 and (n+1)^2 are squares and if we chose 2n+1 to be a square then we are through by proper transformation.

We cannot choose n to be an integer as we shall loose some values in the process so we shall put n = p/q but that is la

However we must have to chose 2n+1 to be a whole square say m^2 to get a Pythagorean triplet

So 2n = m^2 – 1

To avoid denominator in 1 we multiply (1) by 4 to get

4n^2 + 4(2n+1) = (2n+2)^2

Or (m^2-1)^2 + 4m^2 = (m^2+1)^2

so
2m, (m^2-1) and (m^2+1) satisfy the condition.(a^2+b^2 = c^2)

But m is not necessarily an integer and by choosing m = p/q and multiplying by q^2 we get

2pq, (q^2-p^2) and (q^2+p^2) satisfy the equation and if p and q are relatively prime then we get triplet with no common factor and hence it is a primitive triple..

Multiply this by any other integer and we shall get another triple(not primitive)

2009/019) Pythagorean triplets.- trigonometric way

There is a trigonometric way of getting Pythagorean triplets.

we know x^2+y^2 = z^2

is homogeneous expression and dividing it by z^2 we get

(x/z)^2 + (y/z)^2 = 1 or a^2+b^2 =1

This is an identity when we put a = sin t and b = cos t

In case we chose a and b both rational we are through. But how to we guaranty that, that is how to choose both sin t and cos t to be rational. There is a way out.

That is by using double angle formula( writing sin t and cos t in terms of tan t/2)

We know sin t = 2 sin t/2 cos t/2
= 2 tan t/2 cos ^2 t/2
= 2 (tan t/2)/(1+ tan ^2 (t/2))

And cos t = (2 cos^2(t/2) – 1) = 2/(1+ tan ^2 t/2) – 1 = (1- tan ^2 t/2)/(1+ tan ^2 t/2)

So instead of chosing 2 expressions that is sin t and cos t we can chose tan t/2 as rational and so sin t and cos t both come out to be rational

Let tan t/2 = m and we get

(2m/(1+m^2)), (1-m^2)/(1+m^2) and 1 satisfy the condition and so

2m, (1-m^2) and (1+m^2) satisfy the condition.(a^2+b^2 = c^2)

But m is not necessarily an integer and by choosing m = p/q and multiplying by q^2 we get

2pq, (q^2-p^2) and (q^2+p^2) satisfy the equation and if p and q are relatively prime then we get triplet with no common factor and hence it is a primitive triple..

Multiply this by any other integer and we shall get another triple(not primitive)

2009/018) Prove that (a+b+c)/3>=3/(1/a+1/b+1/c)? if a,b,c are positive real number.

we know

(a+b+c)/3 >= (abc)^(1/3) AM GM enaquality

(1/a+1/b+1/c)/3 >= (1/(abc))^(`1/3) AM GM enaquality

as both are positive

multiplying

(a+b+c)/3 * (1/1a+1/b+ 1/c)/3 >= 1

or (a+b+c)/3 >= 3/(1/a + 1/b+ 1/c) multiplying both sides by 3/(1/a + 1/b+ 1/c) as this is > 0

proved

2009/017) A cute little integration to find the lower and upper bound for pi

Integrate

x^4(1-x)^4/(1+x^2) from 0 to 1

expanding we get

x^6-4x^5+5x^4-4x^2+4-[4/(1+x^2)]

integrating we get

x^7/7 – 2/3x^6+ x^5- 4/3x^2 + 4x – 4tan ^-x

x=1 gives 1/7-2/3+1-4/3 + 4 = 22/7 – 4 arctan(1) = 22/7- pi

x=0 gives 0

so definite integral = 22/7 – pi

now as the LHS is positive at each point integral > 0 so 22/7 –pi or pi < 22/7

now for the lower limit let us find the higher limit of LHS

x(1-x) is highest at x= ½ and x(1-x) = ¼

so x^4(1-x)^4 highest is 1/256

and lowest of (1+x^2) is 1

so integral of LHS < 1/256

so 22/7-1/256 < pi < 22/7

gives pi(which is 3.14159..) between 3.1389 and 3.142857

pretty good is it not ?

2009/016) If a cos² x + b sin² x = c, express tan² x in terms of a, b and c

a cos² x + b sin² x = c
= c(sin ^2 x + cos ^2x) (knowing that 1 = sin ^2 x + cos ^ 2 x)
= c sin ^2 x + c cos ^2 x
or a cos^ 2 x - c cos^2 x = c sin ^2 x - b sin^ 2x
or (a-c) cos ^2x = (c-b) sin ^2 x

or (a-c)/(c-b) = sin ^2 x/ cos ^2 x = tan ^2 x

so tan ^2x = (a-c)/(c-b)

2009/015) Find a Polynomial with integral coefficients of lowest degree with 2^1/3 + 3^1/2 as a zero

Let x = 2^1/3 + 3^1/2. You want to do enough algebra to make the square and cubes roots go away, until you only have integer (or rational) coefficients.

x - 3^1/2 = 2^1/3

Now, cube both sides:

2 = (x-3^(1/2))^3

we can use the formula

(A - B)^3 = A^3 - 3*A^2*B + 3*A*B^2 - B^3

to get

2 = x^3-3x^2(3^1/2)+9x-3^(3/2) = (x^3+9x -3^3/2(x^2+1)

or 3^(3/2)(x^2+1) = (x^3+9x-2)
now squaring both sides you get

27(x^2+1)^2 = (x^3+9x-2)^2
expanding it we should get the result

27(x^4+2x^2+1) = (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x)
or 27x^4+54x^2+27 = (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x)
or (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x) - (27x^4+54x^2+27) = 0
or x^6 -9 x^4 - 4x^3 + 27x^2 - 36x - 23 = 0 after simplifying

2009/014) sum of the reciprocals 1+1/2+1/3+...+1/n is never an integer for n >1

prove that the sum of the reciprocals 1+1/2+1/3+...+1/n is never an integer for n >1


let S = 1+1/2+1/3+...+1/n

because n > 1 so from 1 to n there is only one number 2^k such that

2^k <= n < 2^(k+1) that is the higest power of 2 less than <=n

2^k devides that number (as 2^k devides 2^k) and no other number

now take LCM(2,3,...n ) = p and mutlipy S by p

now 2^k devides p
each term of RHS except 1/2^k as denominator term is not divisble by 2^k gives an even number but 1/2^k gives an odd number p is not divisible by 2^(k+1)

so RHS = odd

pS = odd

so S = odd/p

p is even

so S = odd/even and hence not an integer

2009/013) Solve x^4+3x^3-8x^2+3x+1=0 for x

Generally 4th order equation is not easy to solve but this is a special case as this is symetrical that is coefficent of x^4 is same as x^0( and of x^3 same as of x)

so we devide by x^2 to get

x^2+3x-8+3/x+1/x^2 =0
or (x^2+1/x^2) + 3(x+1/x) - 8 = 0

put x+1/x = t to get

t^2-2 + 3t - 8 =0
or t^3+3y-10 = 0
(t-2)(t+5) = 0

t- 2 = 0 =>
x+1/x-2 = 0
or x^2+1-2x = 0
(x-1)^2 = 0

or x = 1 ( a double root)
2nd part
t+-5 = 0
=>x^2+5x + 1 = 0
x= (-5 +/-sqrt(21)/2 using (-b+/-sqrt(b^2-4ac))/2

= (-5 + sqrt(21))/2, (-5 - sqrt(21))/2,

so 4 roots are (-5 + sqrt(21))/2, (-5 - sqrt(21))/2, 1(Double root)

2009/012) If the pth, qth, rth and sth terms of an AP are in GP, show that (p-q),(q-r),(r-s), are also in GP

Let 1st term be a+ t and difference be t

We need to prove
(p-q)/(q-r) = (q-r)/(r-s)
So pth term = a + pt
Qth terem= a + qt
Rth term = a + rt
And sth term = a + st

Thery are in GP then

(a+st)/(a+rt) = (a+rt)/(a+qt) = (a+qt)/a+pt) ( common ratio)

Now if (a/b) = (c/d) then both = (a-c)/(b-d)

Using this we get

From (a+st)/(a+rt) = (a+rt)/(a+qt) both = (st-rt)/(rt-qt) = (s-r)/(r-q)

From (a+rt)/(a+qt) = (a+qt)/a+pt) both = (r-q)/(q-p) same as aboth

Hence
(s-r)/(r-q) = (r-q)/(q-p)
Or
(r-s)/(q-r) = (q-r)/(p-q) multiplying numerator and denominator of both sides by -1

Proved

2009/011) Prove that (sin x)/x = cos(x/2)*cos(x/4)*cos(x/8)*cos(x/16)..

Proof:
Sin x = 2 cos (x/2) sin (x/2)
= 2 cos (x/2) (2 cos(x/4) sin (x/4))
= 2^n cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)

So (sin x)/x = 2^n cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)/ x
= cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)/ (x/2^n)
as n goes to infinite sin (x/2^n)/ (x/2^n) goes to 1 and we get

(sin x)/x = cos (x/2) cos(x/4) cos(x/8)cos(x/16) ….


Proved

2009/010) Prove Cos pi/7 + cos 3pi/7 + cos 5pi/7=1/2

To prove
Cos pi/7 + cos 3pi/7 + cos 5pi/7=1/2

let z = cos pi/7 + i sin pi/7

z^7 = cos pi + i sin pi = - 1

so z^7+1 = 0
(z+1) (z^6-z^5+z^4 - z^3+z^2 -z + 1) = 0

as z is not - 1 so

z^6-z^5+z^4 - z^3+z^2 -z + 1 = 0

so z^6-z^5+z^4 - z^3+z^2 -z= -1

so z+z^3+ z^5 = 1 + (z^2+z^4+z6)
equating the real part

cos pi/7 + cos 3pi/7 + cos 5pi/ 7 = -1 - (cos 2pi/7+ cos 4pi/7 + cos 6pi/7)

but cos 6pi/7 = - cos(pi-6pi/7) = - cos pi/7
cos 4pi/7 = - cos 3pi/7
cos 2pi/7 = cos 5pi7

so
cos pi/7 + cos 3pi/7 + cos 5pi/ 7= 1 - (\cos pi/7 + cos 3pi/7 + cos 5pi/ 7)

or 2 (cos pi/7 + cos 3pi/7 + cos 5pi/ 7) = 1
or cos pi/7 + cos 3pi/7 + cos 5pi/ 7 = 1/2

2009/009) Prove that tan (A+B) = (tan A+ tan B)/(1-tan A tan B)

Proof:

We know that

the argument of a complex number z = x + yi is the angle to the real axis

then if we take the 1st number 1 + ai and argument is A and 2nd number 1+bi and argument is B

then a = tan A and b = tan B

1st number = 1 + ai = 1 + i tan A
2nd number = 1+ bi = 1 + i tan B


When we multiply the arguments add so angle is A + B

Now (1+ ai)(1+bi) = (1-ab + i(a+b))

As argument is A+B hence

tan (A+B) = (a+b)/(1-ab)= (tan A + tan B)/(1- tan A tan B) (note tan(arg) = y/x

2009/008) If a,b,c are in AP then prove that (b+c)^2 - a^2 , (c+a)^2 - b^2 , (a+b)^2 - c^2 are in AP

let the terms be a1, a2,a3
a1 = (b+c)^2 - a^2 = (a+b+c)(b+c-a) = (a+b+c)(a+b+c - 2a)
a2 = (c+a)^2 - b^2 = (a+b+c)(a+c-b) = (a+b+c)(a+b+c - 2b)
a3 = (a+b)^2 - c^2 = (a+b+c)(a+b-c) = (a+b+c) (a+b+c - 2c)

as a b c are in ap so are -a , -b , -c so are -2a, -2b, - 2c

addiing constant to them we have

(a+b+c-2a), (a+b+c-2b), (a + b - 2c) are in AP

multiplying by (a+b+c) we have
(a+b+c)(a+b+c-2a), (a+b+c)(a+b+c-2b), (a+b+c) (a + b - 2c) are in AP

hence proved

2009/007) prove sin5A = 5sinA - 20 sin^3A + 16 sin ^5 A

We Know
Sin (A + B) = sin A cos B + cos A sin B
Putting B = 4A we get
sin 5A = sin ( A + 4 A)
= sin A cos 4A + cos A sin 4A
= sin A (1- 2 Sin^2 2A) + cos A 2. Sin 2A cos 2A
( putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = 2A)
= sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)
(putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = A)
= sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)
= sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 Sin ^2 A)
= sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^4 A)
= 5 SIn A - 20 sin ^3 A + 16 sin ^5 A

(note: edited the above based on 3rd comment below to keep the flow)

2009/006) integers x = a3+ b3+ c3-3abc for some integers a,b,c. prove that if x,y € S then xy €S.

Let S be set of integers x such that x = a3+ b3+ c3-3abc for some integers a,b,c. prove that if x,y € S then xy €S.


Proof:

we know
a^3+ b^3+ c^3-3abc = (a+bw+cw^2)(a+bw^2+cw) where w = cube root of -1

let

f(a,b, c) = a+b+c
g(a,b,c) = (a+bw+cw^2)
and h(a,b,c) = (a+bw^2+cw)


then a^3+ b^3+ c^3-3abc = f(a,b,c) g(a,b,c) h(a,b,c)

and x^3+ y^3+ z^3-3xyz = f(x,y,z) g(x,y,z) h(x,y,z)

so (a^3+ b^3+ c^3-3abc)( x^3+ y^3+ z^3-3xyz) = f(a,b,c) g(a,b,c) h(a,b,c) f(x,y,z) g(x,y,z) h(x,y,z)

now let us evaluate f(a,b,c) f(x,y,z) , g(a,b,c) g(x,y,z) and h(a,b,c) h(x,y,z)

f(a,b,c) f(x,y,z) = (a+b+c)(x+y+z) = (ax+ay+ az + bx +by + bz + cx + cy+ cz)

g(a,b,c) g(x,y,z) = (a+bw+cw^2) (x+yw+zw^2)

= (ax +ayw+azw^2 + bxw + byw^2 + bzw^3 + cxw^2 + cyw^3 + czw^4)
= (ax +ayw+azw^2 + bxw + byw^2 + bz + cxw^2 + cyw^3 + czw) knowing w^3 = 1 and hence w^4 = w
= (ax +bz+cy + (ay + bx + cz)w+ (az + by+cx) w^2)
similarly
h(a,b,c) h(x,y,z) = (a+cw+ bw^2) (x+zw+ yw^2)

= (ax+bz+cy) + (az+by+cx)w+ (ay+bx+cz) w^2

If we put ax + bz+ cy = p, ay+bx+cz = q, az+by+cx = r

We get

f(a,b,c) f(x,y,z) = p + q + r = f(p,q,r)

g(a,b,c) g(x,y,z) = p + qw + r w^2 = g(p,q,r)

h(a,b,c) h(x,y,z) = p + rw + q w^2 = h(p,q,r)

so (a^3+ b^3+ c^3-3abc)( x^3+ y^3+ z^3-3xyz) = f(a,b,c) g(a,b,c) h(a,b,c) f(x,y,z) g(x,y,z) h(x,y,z) = f(p,q,r) g(p,q,r) h(p,q,r) = ( p^3+ q^3+ r^3-3pqr)

hence proved

2009/005) parametric form for Pythagorean triplet (Gemoetric way)

To generate canonical form of Pythagorean triplet(geometric way)

Introduction

First it makes sense to define what is Pythagorean Triplet.

A Pythagorean triple is a triple of positive integers a, b, and c such that a right angled triangle exists with legs a , b and hypotenuse c. By the definition of Pythagorean Theorem , this is equivalent to finding positive integers a,b and c satisfying

a2+b2 = c2 ….1

The smallest and best-known Pythagorean triple is a =3, b =4, c = 5

It is not unusual to look for primitive Pythagorean triples.

The (1) is equivalent to finding real solution of

(a/c)2+(b/c)2 = 1
or x2+y2= 1 … 2

In case we get rational solutions in x and y the set (x,y,1) satisfy the condition. We can convert them to integer by proper multiplication that is LCM of denominator.
So we need to find rational x and y to satisfy the equation 2
We can put y as a function of x as following
Y = 2(1-x2)
We can choose x to be a rational number <1 all="" be="" br="" but="" cases.="" cases="" does="" for="" general="" generate="" get="" in="" may="" not="" of="" one="" rational.="" rational="" shall="" some="" the="" this="" to="" x="" y="">Now the problem is can we generate x and y to be rational or integer roots for a b c.

There are a couple of methods to find the rational x y.

the equation x^2+y^2 = 1 is a circle at centre (0,0) and radius 1

If we can find one point in the circle which is a rational point and draw a straight line y= mx +c through that point with m and c to be rational then it shall intersect the circle at another point which has got rational point. By choosing different m we can get more and more rational points.

We know one of the point in the circles is
x = 0 and y = -1 (people have used x = -1 and y =0 but this gives me simpler approach. The choice is arbitrary and no specific choice)
putting in y = mx + c we get y = mx-1 ….3
now put it in the equation 2 to get
x2 + (mx-1)2 = 1
or x2+m2x2-2mx=0
or x(1+m2) -2m = 0
x = 2m/(1+m2) ..4
put x = 2m/(1+m2) in 3 to get
y = 2m2/(1+m2) – 1 or (m2-1)/(m2+1)
so (2m/(1+m2), (m2-1)/(m2+1),1) satisfies x^2+y^2 = 1.
multiplying by m2+1 we get

(2m,.(m2-1),(m2+1)) satisfies x^2+y^2 = z^2 but they are not integers

By choosing different m we get different values basically in parametric form.
Now putting m =u/v which is ratio of integers and multiplying by v2
We get

x = 2uv
y = u2- v2
and z = u2 + v2

so (2uv, (u^2-v2) and (u^2+v^2)) form a triplet

2009/004) Prove the identity sin^2Acos^2B-cos^2Asin^2B = sin^2A-sin^2B

sin^(2)Acos^(2)B-cos^(2)Asin^(2)B
= sin ^2 A(1- sin ^2 B) - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 A sin ^2 B - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 B(sin ^2 A + cos ^2 A)
= sin ^2 A - sin ^2 B

2009/003) factor (x-y)^5+(y-z)^5+(z-x)^5

if a + b+ c = 0

then (a+b)^5 = -c ^5

so a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b + b^5 = - c^5

so a^5 + b^5 +c^5 = - (5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b)
= -5 ab(a^3 + 2 a^2b + 2 a b^2 + b^3)
= - 5 ab((a+b)^3 - (a^2 b + ab^2)
= - 5ab((a+b)3 - ab(a+b))
= - 5ab(a+b)((a+b)^2 - ab)
= 5abc(c^2-ab) as a+b = - c

as (x-y)+ (y-z) + (z-x) = 0

we get 5(x-y)(y-z)(z-x)((z-x)^2 - (x-y)(y-z))
= 5(x-y)(y-z)(z-x)(z^2 + x^2- 2xz -xy + xz +-y^2 -xy)
= 5(x-y)(y-z)(z-x)(x^2+y^2+z^2 - xy - yz - zx)

2009/002) show: arcsin[(4/√41)]+arcsin[(1/√82)=pi/4?

the things become easy in case we convert arc sin to arc tan

let x = arcsin 4/√41

so sin x = 4/√41

cos x = sqrt(1- sin ^2 x) = sqrt(1-16/41) = 5/√41

so tan x = 4/5

now let sin y = 1/√82

so cos y =√(1-1/82) = 9/√(82)

so tan y = 1/9

we heed to find tan (x+y) when tan x = 4/5 and tan y = 1/9

tan (x+y) = (tan x+ tan y)/(1-tan x tan y) = (4/5+1/9)(1-4/5*1/9) = (41/45)/(41/45) = 1

so x + y = tan ^-1 1 = pi/4 ( this is so because x < pi/4 and y < pi/4 so sum <= pi/2

2009/001) Find the smallest positive integer x for which 7x^25 - 10 is completely divisible by 83.

We are given
7x^25 = 10 mod 83We should make the coefficient of x^25 as 7
to get rid of 7 multiply by inverse of 7we have GCD(7,83) = 1using extended eulers algorithm 6 = 83- 7 *11 1 = 7- 6 = 7*12 – 83
so inverse of 7 is 12multiply by 12 on both sides knowing 7*12 = 1 mod 83 we get x^25 = 10 * 12 mod 83 = 37
now we need to raise a power so that x^82 = 1 mod 1now we need to find reciprocal of 25 mod 82again using extended eulers algorithm82 = 3*25 + 7
7 = 82 - 3* 2525 = 7*3 + 4 or4= (25-7*3) = (25-3*(82-3*25) = 10 * 25 - 3*82now knowing 1 = 2*4 - 7 = 2(10*25-3*82) - (82-3*25)= 23*25 - 7*82
so reciprocal of 25 is 23so raise the number to the power 23x= 37^23 mod 83= 37* (37*2)^11 mod 83= 37 * (1369)^11 mod 83= 37 * 41^11 mod 83= (37*41)* 41 ^10 mod 83= 23 *41 ^10 mod 83= 23 *(41^2)^5 mod 83= 23 *21^5 mod 83= 23*21 * 21^4 mod 83= 483 * 21^4 mod 83= 68 * 21^4 mod 83= 68 * 441 * 441 mod 83= 69