more about Pythagorean triples

More about Pythagorean  triples

In  the Pythagorean triple (a,b,c) ,if all of them are even, it be a reducible set.  And if ‘a’ and ‘b’ are even ‘c’ will also be even. Hence In all the irreducible sets either ‘a’ or ‘b’ must  necessarily  be odd..All squares will have only certain number digits-1,4,7 and 9. .Since the sum of two squares  is equal to a square,the sum of number digits of a² and b² must be equal to the number digit of c². It requires that the number digit of either a² or b² must be 9 so that the number digit of the remaining square in one side will ne equal to the number digit of c². All numbers which are multiples of three will give number digit 9 for its square. Hence either ‘a’ or ‘b’ must be a multiple of three.If ‘a’ is multiple of three, then a² will be divisible by three.It means that c² – b² is divisible by three.that either (c-b) or (c+b) must be divisible by three.

Pythagorean triples with given endings

Here is yet another method to identify Pythagorean triples with given endings.

The Pythagorean triples (a,b,c) satisfy the relation a² + b² = c² .Let us try to find a series of Phythagorean triples ending with 3,4 and 5 (greatest)

The most general form for this set is

(10x+3)² + (10y+4)² = (10z+5)². It gives a condition for the suitable values of x,y,and z.- 5(x² + y²) +3x +4y = 5z(z+1)

The other requirement for the Pythagorean triples is a+b = c+2α ,where α may be either odd or even. When we apply this condition to our general form ,it predicts α takes only certain allowed values 1,6,11,16,21… and so on.

 α = 1 ; x+y = z

α = 6 ; x+y-1= z

α = 11 ; x+y-2 = z

α = 16 ; x+y -3 = z

When α = 1

5(x² + y²) +3x +4y = 5 (x+y)(x+y+1) ,it is reduced to 10xy+2x+y = 0. The oly possibility is x=0;y=0 and z=0.

When α = 6

5(x² + y²) +3x +4y = 5(x+y-1)(x+y) ,the required condition becomes x = 9y/[2(5y-4)].If x and y are positive integers,the pre determined endings will not altered.For example when y=1 then x = 9/2 which gives

Irreducible triple 7² + 24² = 25^2; When y=3 ;x=27/22  which gives 84² + 187² = 205²

 When α = 11 ,the required condition becomes x = (19y-10)/(10y-18) with y being even.When y=2 ,x=14.It gives (143,24,145)

Y=4 gives (3,4,5)

Y=6 gives (583,1344,1465)

Y=8 gives (803,2604,2725)

New relations

New Reations with Pythagorean triples

A "Pythagorean Triple" is a set of positive integers, a, b and c that fits the rule:

a² + b² = c²

And when we make a triangle with sides a, b and c it will always be a right angled triangle

List of the First Few

Here is a list of the first few Pythagorean Triples (not including "scaled up" versions mentioned below):

(3,4,5)	(5,12,13)	(7,24,25)	(8,15,17)	(9,40,41)
(11,60,61)	(12,35,37)	(13,84,85)	(15,112,113)	(16,63,65)
(17,144,145)	(19,180,181)	(20,21,29)	(20,99,101)	(21,220,221)
(23,264,265)	(24,143,145)	(25,312,313)	(27,364,365)	(28,45,53)
(28,195,197)	(29,420,421)	(31,480,481)	(32,255,257)	(33,56,65)
(3,544,545)	(35,612,613)	(36,77,85)	(36,323,325)	(37,684,685)

c is the longest side of the triangle, called the "hypotenuse"

a and b are the other two sides such that a< c ,b< c but a+b > c.  It is noted that one of the numbers a or b is 3 or mutiplles of 3. Let

 a+b = c + α, where α is also a positive integer. Squaring both sides, we have

 α² +2cα – 2ab = 0 Solving for α, we get α = - c + [c² + 2ab]^1/2. i.e, c² + 2ab must be a square number.

It is again noted that the product of ab i.e., twice the area of the right angled triangle is equal to a multiple (β) of the circumference (a+b+c) of the  right angled triangle such that β = α/2 .Since β is also a positive integer, α must be even.

Puzzle

Fun with  Mathematics

Puzzle

A smart boy used to play computer game ‘ Free cell’  every day  before he used his laptop for assigned work. The percentage of win is displayed after the game is over.The fraction in the percentage is not considered by the computer.. He aimed 100 % win but he scored only 50 % after playing 300 games.

To make it to be 51% how many wins successively  he has to score?  He was unable to win all the games then played. If he got this 51 % after playing 20 games , then how many games he won in 20 ?.

Answer

In the beginning games he won 150 out of 300.  Suppose he played x games and won in all the games,

Then the new winning percentage becomes  150 + x/(300 +x) = 51/100 .X must be at least 7.

In the additional 20 games he won x games to make  the winning percentage 51. Then

150+x /320 = 51/100  , x must be 14.

Remainder of power numbers on division

Remainder of power numbers on division

The remainder of a number n to the power of p when divided by a divisor d is cyclically varying not only for p but also for n . For example when d = 7 ,the cyclic variation of the remainder is same for n^p  and

(n+ 7x)^p

Remainder of n^p when divided by 7

n/p         0              1              2              3              4              5              6              7              8              9              10

1              1              1              1              1              1              1              1              1              1              1              1

2              1              2              4              1              2              4              1              2              4              1              2             

3              1              3              2              6              4              5              1              3              2              6              4

4              1              4              2              1              4              2              1              4              2              1              4

5              1              5              4              6              2              3              1              5              4              6              2

6              1              6              1              6              1              6              1              6              1              6              1

7              1              0              0              0              0              0              0              0              0              0              0

8              1              1              1              1              1              1              1              1              1              1              1             

9              1              2              4              1              2              4              1              2              4              1              2

10           1              3              2              6              4              5              1              3              2              6              4

It is found that the remainder of (1+7x)^p when divided by 7 is 1,for all values of x and p..The remainder of  n⁶ when divided by 7  is 1 for all values of n except n = 7x. It is 0 for n = 7x

power numbers and its divisibility

Power numbers and its divisibility

When a seventh power of a number ‘a’ is divided by 7, the remainder will be equal to that number itself when a < 7 and (a-7n) when a > 6 where n is the  whole number of 7’s in ‘a’ . e.g.,

1⁷ = 1 = 7(0) + 1

2⁷ = 128 = 7(18) + 2

3⁷ = 2187 = 7(312) + 3

4⁷ = 16384 = 7(2340) + 4

5⁷ = 78125 = 7(11160) + 5

6⁷ = 279936 = 7(39990) + 6

When a = 7 or > 7, a= (7n+d)

a⁷ = (7n+d)⁷ and the remainder is determined by d⁷ / 7 and it is equal to d itself.

7⁷ = 823543 = 7 (117648) + 7

                      = 7 (117649) + 0

8⁷ = 2097152 = 7 (299593) + 8

                        = 7 (299594) + 1

 When a fourth power of a number ‘a’ is divided by (a+1) ,the remainder is 1 for all values  of a.

a=1; a⁴ = 1 = 2(0) + 1

a=2; a⁴ = 16 = 3(5) + 1

a=3; a⁴ = 81 = 4(20) + 1

a=4; a⁴ = 256= 5(51) + 1

a=5; a⁴ = 625 = 6(104) + 1

a=6; a⁴ = 1296 = 7(185) + 1

a=7; a⁴ = 2401 = 8(300) + 1

a=8; a⁴ = 4096 = 9(455) +1

The quotient is equal to a³ – a² + a - 1

divisibility of cube numbers (cont.,)

Divisibility of cube numbers (cont.,)

When a³ is divided by (a+4),,the remainder will be 4(a-12),that is for all ‘a’ which are in multiples of 12 ,the remainder will be zero.

The general form correlating a³, the divisor (a+4),the quotient and the remainder is

  a³ = [8+ (a-2)² ] (a+4) + 4(a-12)

If the  divisor is (a+5),  it becomes,

a³ = [14 + (a-2)(a-3)] (a+5) + 5(a-20)

For the divisor (a+6),

a³ = [21 + (a-3)² ] (a+6) + 6(a-30)

for (a+7),

a³ = [ 30 + (a-3)(a-4)](a+7) + 7(a-42)

divisibility of cube numbers (cont.,)

Divisibility of cube numbers (cont.,)

When a³ is divided by 4 ,for all even numbers the remainder is zero where as for odd numbers it is 1 for a in the form of (4n+1) and 3 for (4n-1).

When a³ is divided by (a+4)

1= 5(9) -44

8= 6(8) – 40

27 = 7(9) – 36

64= 8(12) – 32

125 = 9(17) – 28

216 = 10(24) – 24

343= 11(33) – 20

512 = 12(44) – 16

729 = 13(57)- 12

1000 = 14(72) – 8

 In  general it takes a form

a³ = [8+(a-2)²](a+4) + 4(a-12)

divisibility of cube numbers (cont.,)

Divisibility of cube numbers (cont.,)

When a³  is divided by 3, the remainder will not only be same as the remainder for a but also it is periodically varying as 1,2,0.

When a³ is divided by (a+3)

a=1, 4(4) – 15

a=2,5(4) – 12

a=3, 6(6) - 9

a=4,7(10) - 6

a=5,8(16) - 3

a=6,9(24) + 0

a=7,10(34) + 3

a=8,11(46) + 6

a=9,12(60) + 9

                                                                            a-2

In general ,the quotient is in the form [4 +  Σ 2n ] and the remainder is 3(a-6)

                                                                                     0        

a-2

 Σ 2n  = (a-1)(a-2)

 0

It shows that a³ = [4+(a-1)(a-2)](a+3) - 3(a-6)

divisibility of cubes

Divisibility of cubes

We know that a³ – a is equal to a product of three successive numbers (a-1),a,and (a+1).

a³ – a = (a-1)a(a+1). Since the product of any three successive numbers is invariably divisible by 6,

(a-1)a(a+1) = 6n

i.e., a³ – a = 6n  or a³ = a+6n or a² = 1 + [6n/a] i.e., 6n is divisible by a

It is noted that there is some kin d of regularity in the quotient and the remainder when a³ is divided by (a+1),(a+2),(a+3)…..

When a cube a³ is divided by (a+1) or (a-1) ,the quotient will be a(a-1) or a(a+1) with remainder a.

When a³ is divided by (a+2)

a=1 ,2(3) – 5

a=2, 3(4) – 4

a=3, 6(5) -3

a=4,11(6) – 2

a=5,18(7) – 1

a=6,27(8) + 0

a=7, 38(9) + 1

a=8,  51(10) +2

a=9, 66(11) +3

In general, it is in the form,

               a-2

a³ = [ 2+ Σ (2n+1)](a+2) + (a-6)

           0

But               n-2

Σ (2n+1)] is a square number.It is equal to square of a number equal tp the

0

  number of odd numbers added together.i.e, (n-1)² .

a³ = [2+(a-1)²](a+2) + (a-6)

It shows that when a³ is divded by (a+2) , the quotient will be [2+(a-1)²] and the remainder will be (a-6)