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

2008/014) Solve the equation 4xABCD=DCBA

A has to be < 3 because 3* 4 = 12 so RHS is a 5 digit number
A cannot be 1 as from RHS A has to be even.
So A has to be 2.
now B can be either 1 or 3 or 5 or 7 but B < 5 because 4*25 = 100 that is 5 digit
So AB = 21 or 23
if AB = 23 DC >= 92do D = 9 which is not possible as 4*8 is 2 ending but 4*9 is notAB = 21
So D = 8
so the number = 4*(2108+10C) = 8032+100C
or 8432+40C = 8012+ 100C
or 60C = 420
so C =7
so number = 2178*4 = 8712

2008/013) 2 is the only prime sum of 2 positive cubes

We know a^3 + b^3 = (a+b)(a^2-ab+b^2)
if a= b then a^3+b^3 = 2 a^3 which not a prime unless it is 2
with out loss of generality we can assume a> b
now a+b >=2
a^2+b^2-ab = a(a-b) + b^2 > b^2 so
a^2 + b^2-ab > 1
as it has 2 factors and both are >2 a^3+b^3 cannot be prime or in other words a prime number > 2 cannot be sum of 2 positive cubes