Fun with Mathematics

One more general proof for Fermat's Last theorem
The equal power relation in the form a^n + b^n = c^n cannot be existing with integral values for a,b and c and for any value of n except 1 and 2.
This relation ,if it exists ,can then be expressed in the Pythagorean form as
(a^n/2)^2 +(b^n/2)^2 = (c^n/2)^2
where the Pythagorean triple is a power of numbers.
All Pythagorean triples (x,y,z) must satisfy a condition xy = 2n(z+n)
Applying this condition to our equal power relation in the converted form ,we have,
ab(ab)^1/2 = 2n [c(c)^1/2 +n]
When a and b are odd ab(ab)^1/2 will be odd, but cannot be equated with 2n[c(c)^1/2+n] which is even for all values of c and n.

If a is even and b is also even, then c must be even. It is reducible without violating the equality of the
power relation . The reduction is stopped only when one of the members of the triple becomes odd.
If a is even, b is odd, c should necessarily be odd, then n must be either even or odd.
In (x,y,z) if x is even and y and z are odd it can be written as
z^2 -y^2= x^2
then z-y = x-2n
squaring and simplification gives,
y(z-y) = 2n(x-n)
Applying this condition to our equal power relation in the converted form,
b(b)^1/2 [c(c)^1/2 - b(b)^1/2] = 2n [a(a)^1/2- n]
when c and b are odd  LHS will be odd but RHS is even for all odd and even values of n and a.
As it contradicts , we can conclude that the equal power relation cannot be existng

Fun with Mathematics

More about the Pythagorean triples

1.In all Pythagorean triples (a,b,c) with c>b>a ,both a and b together cannot be odd.
Proof:
If a^2+b^2 = c^2, then a+b = c+2n where n = 1,2,3,4....
Squaring both sides and after simplification
ab = 2n(c+n)
That is the product of a and b is even which means that both a and b cannot be odd.

2.           2n^2 + 2cn -ab = 0
solving for n, we get,
n = (-1/2) + [1/2(c^2+2ab)^1/2]
It implies that c^2 +2ab must be a square number and if c is even c^2+2ab will be an even square and
if c is odd, it will be an odd square.

3. Again 2n^2= [an-2cn]
c+2n = a+b
 cubing both sides and after simplification
c^3-a^3-b^3 = 3ab(a+b)-8n^3-6cn(c+2n)
3(a+b)(ab-2cn)- 8n^3
(a+b)6n^2-8n^3
It shows that c^3-a^3-b^3 is divisible by 2 and also by a square number n^2 where n = (1/2)(a+b-c) 

Fun with Mathematics

Yet another proof for Fermat's last theorem
Dr.M.Meyyappan,Dean of Science
Sri Raja Raajan College of Engineering and Technology
Amaravathipudur-630301, Tamilnadu,India

The numeral relation in the form a^n +b^n = c^n is found to exist only when a,b,c and n are all whole
integers provided n = 1 or 2.
When n=1; a+b=c ,then a^n+b^n < c^n for n > or = 2
When n =2 ; a^2+b^2 = c^2  then a^n+b^n> c^n for n <2
                                                     a^n+b^n < c^n for n >2
The problem in the conditional relation a^n+b^n=c^n begins from n = 3 and it prevails for all higher
values n >3 .
a^n+b^n =/=c^n when n > or = 3
Let us assume that a^3 +b^3 = c^3 exists with a,band c are all whole numbers. Since sum of two
cubes is equal to a cube, c>a,c>b, but c <a+b .
The cube of a number x can be expressed as x^3 = (x-1)x(x+1) +x
For any product of three successive numbers ,6 will invariably be a factor, and hence,
x^3 = 6n(x) + x
Applying this fact to the cubical relation assume,we have
a+b-c = 6[n(c)-n(a)-n(b)] = 6n
or a+b = c+6n
For a given a and b, that is for given digital endings of a and b, the digital ending of c cannot
have all possible value, but is fixed ,because the digital ending of c^3 must be equal to the
sum of the digital ending of a^3 and b^3.
Table given below shows that digital ending of c for all possible digital endings of a and b
Digital ending of c in a^3 +b^3 = c^3

a       /    b        0     1     2     3     4     5     6     7     8      9

0                      0     1     2     3     4     5     6     7     8     9
1                      1     8     9     2     5     6     3     4     7     0
2                      2     9     6     5     8     7     4     1     0     3
3                      3     2     5     4     1     8     7     0     9     6
4                      4     5     8     1     2     9     0     3     6     7
5                      5     6     7     8     9     0     1     2     3     4
6                      6     3     4     7     0     1     8     9     2     5
7                      7     4     1     0     3     2     9     6     5     8
8                      8     7     0     9     6     3     2     5     4     1
9                      9     0     3     6     7     4     5     8     1     2

According to the requirement of digital endings of the equal power relation, one can assume the
following typical relation as an example.
(10x+7)^3 + (10y+4)^3 = (10z+3)^3
Fermat's last theorem demands that x,y,z all cannot be whole numbers simultaneously to satisfy the
relation. Since c>b>a (or c>a>b) ,one can assume that c =b+m, where m = 1,2,3,......
The acceptable and permitted values of m according to digital ending requirement are
9,19,29..... or in general (10k +9) with k =0,1,2,3,.......
It demands that z = y +1
Since a+b = c+6n
 10(x+y)+11 = 10z +3 + 6n
The digital ending requirement allows only certain quantized value n = (3+5m)
where m = 0,1,2,3,......It demands that x+y= z+1 or x = 2
By introducing these derived conditions, we can express the typical cubical relation in terms of one
variable y.
27^3 + (10y+4)^3 = (10y+13)^3
90y^2 + 153y - 585 = 0
Solving this quadratic equation for y,
y = [-153 + (2340090)^1/2]/180
   = 1.8374708.....   or -3.5374708.....
It gives
27^3 + (22.374708...)^3 = (31.374708...)^3
It clearly shows that when a is taken as whole number, then both b and c cannot be whole
integers in the relation a^3+b^3 = c^3. Hence Fermat's last theorem is true for n = 3.

Fun with Mathematics

Digital ending method of getting conditional Pythagorean triples

A Pythagorean triple (a,b,c) must satisfy the relation a^2 + b^2 = c^2
where c >a and c > b but a+b > c.
The values of a,b and c must be such that the digital endings of both sides must be same..
Table given below shows the digital ending of c for all possible digital endings of a and b.
Table.: Digital ending of c in a^2 + b^2 = c^2
a       /     b        0      1     2      3      4      5      6      7      8      9

0                       0      1      2     3      4      5      6      7      8      9
1                       1      x      5     0       x    4,6     x      0      5      x
2                       2      5      x     x       0    3,7     0      x      x      5
3                       3      0      x     x       5    2,8     5      x      x      0
4                       4      x      0     5       x    1,9     x      5      0      x
5                       5    4,6   3,7  2,8    1,9    0     1,9   2,8   3,7   4,6
6                       6     x       0     5       x    1,9     x      5      0      x
7                       7     0       x     x       5    2,8    5       x      x      0
8                       8     5       x     x       0    3,7    0       x      x      5
9                       9     x       5     0      x     4,6    x       0      5      x

It is noted that certain digital ending combinations for a and b are forbidden. For example both a,and b
cannot be ending with identical digits except 5
It is found that a^2 - a = a(a-1) is always equal to some multiples of 2,hence
a^2 = a+ an(a)
b^2 = b + 2n(b)
c^2 = c+2n(c)
It gives a condition
a +b = c + 2n
According to the requirement of digital endings of the equal power relation ,one can assume the following
typical relation as an example,
(10x+7)^2 + (10y+4)^2 = (10z+5)^2
Since a+b = c+2n, 10(x+y)+11 = 10z+5 + 2n
It demands that n = (3+5k) with k = 0,1,2,3......and x+y = z+k
With k = 0 , x+y = z
(10x+7)^2 + (10y+4)^2 = [10(x+y)+5]^2
it gives y = 2(x+1)/10x+1
When x = 0; y=2,z=2 ; 7^2+24^2=25^2
when x=1 ,y=4/11,z=15/11; 187^2+84^2 = 205^2
when x=2 , y = 6/21,z=48,21; 567^2 +144^2= 585^2
when x = 3, y=8/31,z=101/31 ; 1147^2+204^2 = 1165^2
When k = 1, x+y= z+1
(10x+7)^2 + (10y+4)^2 = [10(x+y)-5]^2
It gives y = 12x+2/10x-9
when x=1 ,y=14,z=14 ;17^2+144^2=145^2
when x=2,y=26/11,z=37/11;  297^2+304^2 = 425^2

Fun with mathamatics

More properties of Pythagorean triples

It is found that cⁿ  - a ⁿ   - bⁿ    is always divisible by (n/2)^{3  ,} where n

is even and (a,b,c)  are any Pythagorean triples.

It is found that cⁿ  - a ⁿ   - bⁿ     is  divisible by a² and  b²    for all

even values of n .>2

e.g., c⁴  - a⁴  - b⁴  = 2a² b²  and c⁶ – a⁶ – b⁶  = 3a² b² c² 

All the members of a Pythagorean triple cannot be odd, since sum

of two odd squares is even which cannot be equal to square of the

other odd number.

It implies that all the members of any triple cannot be prime.

All of them can be even. Such Pythagorean triples will always be

reducible.

All of them cannot be square numbers, cube numbers or in general

equal power numbers with power n >1. If (a² , b² , c² ) exists as a

 possible set of Pythagorean triple, then the Pythagorean relation

demands that  a⁴  + b⁴  =  c⁴  which is not possible as it is against

the Fermat’s last  theorem.

1 cannot be a member in any Pythagorean triples. If 1 is a member

( lowest) then 1 =  c²  - b²   or 1 = (c+b)(c-b). The mathematical

description says that the product of a number and its reciprocal will

always give unity. Since c-b cannot be a fraction and  c² – 1 cannot

be a square number , for any integral values of c and b  the product

cannot be made to unity.

        

Arika ariviyal

சூரியக் குடும்பத்தில் பூமி 

 

 

 

சூரியக்குடும்பத்தில் புதன்,வெள்ளி ,பூமி,செவ்வாய் ,

வியாழன்,சனி,யுராநெஸ் ,நெப்டியூன் என்ற வரிசையில்

8  கோள்கள் உள்ளன. (புளுட்டோ நீங்கலாக ).இவற்றுள்

பூமியில் மட்டும் உயிரினங்கள் வாழ்கின்றன . இதற்கு

என்ன காரணம் ?

                       ***************

 

நீரானது பனி,நீர்,நீராவி என மூன்று நிலைகளிலும்

இருக்கக் கூடியவாறு சூரியனிலிருந்து அளவான

வெப்பத்தைப் பெற சரியான தொலைவில் பூமி 

இருப்பதே இதற்குக்  காரணம்.

 

வியாழன்,சனி,யுரனேஷ் ,நெப்டியூன் போன்ற

பெரிய கோள்களில் ஈர்ப்பு அதிகம்,வெப்பநிலைக்

குறைவு.இதனால் ஹைட்ரஜன் நீங்கலான

வளிமங்கள் உறைந்து இருக்கின்றன. மீத்தேன்,

அமோனியா  போன்ற வளிமங்கள் தெவிட்டிய

நிலையில் உள்ளன .இவை நச்சுத் தன்மை

கொண்டவை. இவ் வளிமண்டலத்தில் உயிரினம்

தோன்றி பரிணாம வளர்ச்சியில் ஈடுபடவே முடியாது.

புதனில் சிறிதளவு கூட காற்றோ,வளிமண்டலமோ

இல்லை. வெள்ளியில் உள்ள வளி மண்டலத்தில் கார்பன்டை ஆக்சைடின்  செறிவு 

மிக அதிகம். இங்கு முற்றிலும் முரண்பாடான 

தட்ப வெப்ப நிலை நிலவுகிறது. செவ்வாயில் 

வளி மண்டலம் மிகவும் மெல்லியது. பூமியைவிடக் 

குளிர்ச்சியாக இருக்கிறது.நீர் உறைந்து பனிக்கட்டியாக 

இருப்பதால் உயிரினம் எல்லா கால நிலைகளிலும் 

இயல்பாக வாழக்கூடிய சூழ்நிலை  இல்லை. 

Fun with Mathematics

New correlations with Pythagorean triples

If (a,b,c) is a Pythagorean triples,then  a² + b²  = c²  is uniquely true,

where c>a,c>b but c < a+b. Let a + b = c+k where k is an equalizer..

Property.1

 a+b= c+k

Squaring both sides, we have,

a²+b² +2ab = c² +k²+2ck

2(ab-ck) =k²

It shows that both a and b cannot be odd. If a and b are odd,

then c must be even since a² + b²  = c²  and k must also be even

since a+b = c+k.  It says that k²   is even but k² /2

must be odd in nature as ab-ck is odd which contradicts the

previous statement.

If a is odd and b is evedn then c must be odd as a² + b²  = c²    

and k must also be even as a+b= c+k. Since (ab-ck) is even it

can be equal with k ² /2 with proper values of a,b and c

 Property-2

a+b= c+k

(ab-ck) = k²  /2   

Which necessitates ab>ck or ab/c >k

This can be rewritten as

k²  + 2ck – 2ab = 0

Solving the quadratic equation for the positive value of  k,

k  =  -c + √ c²    +2ab

or a+b = c+k = √ c²    +2ab   

 c²    +2ab   must be a square number

Property-3

a³    +b³  + 3ab(a+b) = c³ + k³   + 3ck(c+k)

3(a+b)[ab-ck]-k³ = c³ –a³- b³

Substituting the value for ab-ck = k²/2

c³ –a³- b³  =

 = 3k² /2[a+b]-k³  = 3k² /2(c+k)- k³   

On simplification we get

c³ –a³- b³  = k²/2 [3c+k]

k=a+b-c,hence

c³ –a³- b³  = k²  [c + (a+b)/2]

where  (a+b)/2  and c  are left and right mean values respectively

of the Pythagorean relation

Property-4

(a+b)⁴ = (c+k)⁴

  a⁴  + b⁴   + 4ab( a²  +  b²  )+6a² b²   = c⁴   + k⁴  +4ck (c²  +k² ) + 6c² k²

c⁴  -a⁴-b⁴ =  4c²  (k²  /2) + 6 k² /2 (ab+ck) – 4ck³ – k⁴

On simplification we have,

c⁴  -a⁴-b⁴ =    (k²  /2) (2c+k)²   = (k²  /2)(a+b+c)²

= [(a+b)²- c²]

   

Fun with Mathematics

Creative thoughts

From   R ^2   to   R ^2 – relation-II

          2  1           2  2

By taking different Pythagorean triples, one can construct

more such general relations for  R^2- relations  

                                               2  2

(5,12,13); (5m+13n)^2 +(12m+13n)^2

                                                  = (13m+17n)^2+(7n)^2

(8,15,17; (8m+17n)^2+(15m+17n)^2

                                                   =(17m+23n)^2+(7n)^2

(12,35,37); (12m+37n)^2+(35m+37n)^2

                                                 =(37m+47n)^2+(23n)^2

In an another method, for one smaller number, the biggest

number of the triple is multiplied with the other smaller

number and added with it, for the other smaller number ,

the biggest number is multiplied with the other smaller

number and subtracted with it. The sum of the squares

of them is equal to the sum of the squares of the biggest

number of the triple and its square. The general form of

this relation can be written as,

If a^2+b^2=c^2 than

(a+bc)^2+(b-ac)^2 = c^2 + (c^2)^2= c^2(c^2+1)

That is the sum, in this case, is equal to a product of

square of the biggest number of the triple and its next

higher number in the natural series.

It is interesting to know that a  R^2 is a self generator

of more                                    2  2                                                 

                                                      

such relations. If a^2+b^2=c^2+d^2, then

 (a+b)^2 +(a-b)^2= (c+d)^2+(c-d)^2

For example, 13^2+ 14^2 = 2^2+19^2

It gives, 27^2+1^2 = 21^2+ 17^2

Which in turn,

26^2+28^2 = 4^2+ 38^2, which is twice of the first new

relation generated.

                                       

To make  R^2 –relations with bigger and biggest numbers,one       

              2  2

can make use of two different Pythagorean triples instead of one.

With (a,b,c) and (x,y,z), we can generate,

(cz)^2= c^2(x^2+y^2) = (a^2+b^2)z^2

or, (cx)^2+(cy)^2 = (az)^2+(bz)^2

e.g., with (3,4,5) and (8,15,17) we get,

(3x17)^2+(4x17)^2 = (8x5)^2 +(15x5)^2

51^2+68^2=40^2+75^2

A general proof for Fermat’s last theorem

(Dr.M.Meyyappan,Professor of Physics,Sri Raaja Raajan

College of Engineering and Technology,

Amaravathipudur-630301)

Equations of the form a^n + b^n = c^n  have no solutions,

when n is a whole number greater than 2 and when a,b and c

are positive whole numbers.

When n = 1, a+b will always be equal to c, but when n ≥ 2,

a+b will always be greater than c. Let

                         a+b = c+k

where k is a small quantity, added with c to $balance  a+b.

Greater the value of n ,smaller will be c  and greater will be k.

By manipulating this logical relation, one can argue the

non-realistic nature of the equal power relation

a^3 +b^3 = c^3 .

From the above logical relation

                       a+b = c[1 + k/c]

 or                   (a+b)/c = [1+ k/c]

cubing both sides, we have

  [ a^3+b^3 +3ab(a+b)]/c^3 = [1+k/c]^3

If a^3+b^3 = c^3 is true, then the relation

1 + 3ab/c^2 [1+k/c]  =  [1+k/c]^3,

will be true. If the later is not true, the former will also

be not true.By rearranging the final relation, we get,

        1 = [1+k/c][(1+k/c)^2 – 3ab/c^2],

where the product of two factors is equal to 1.When one

of the factors [1+k/c] is greater than unity, the other

factors must necessarily be lesser then unity say (1-y),

where y is positive.

(1+x) (1-y) = 1 +x-y-xy = 1

or,  x= y(x+1)

When y is positive, it is an impossible relation.

Taking the final relation,

1 = [1+k/c][ 1 – (3ab/c^2- k^2/c^2 – 2k/c)]

 Comparing with the general relation,

x=k/c  and y = 3ab/c^2 – k^2/c^2-2k/c

k/c = [(3ab/c^2-k^2/c^2-2k/c)][(k/c)+1]

Since 3ab/c^2 –k^2/c^2-2k/c is taken as a positive quantity,

the above relation cannot be existing. In such relations

3ab/ck > [(k/c)+2]

A similar argument can be given for odd power relations.

For example, when n = 5,

a^5 + b^5 = c^5

a+b > c or a+b = c+ k

(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+ 10a^2b^2(a+b)

                                           = c^5(1+k/c)^5

Where

a^3+b^3 = c^3(1+k/c)[(1+k/c)^2- 3ab/c^2]

and

a+b = c(1+k/c)

substituting these values and dividing by c^5, we have,

1       + (5ab/c^2)(1+k/c)[(1+k/c)^2-3ab/c^2] +

                       (10a^2b^2/c^4)(1+k/c)

                        = (1+k/c)^5

or, 1 = (1+k/c) {(1+k/c)^4-(5ab/c^2)[(1+k/c)^2-3ab/c^2]

                 - 10a^2b^2/c^2]}

1 = (1+k/c) { 1- [(5ab/c^2)(1+k/c)^2+ 5 a^2b^2/c^4-

                         k^4/c^4+4k^3/c^3+6k^2/c^2+ 4k/c]}

In the product of two factors, since one is greater than 1,

the other one must be less than one.

1 = (1+x)(1-y)= 1 +x – y - xy

                     x = y(x+1)

More details can be obtained from

meydhanam@gmail.com

Fun with mathematics

Creative thought-

From   R ^2   to   R ^2 – relation

          2  1           2  2

The Pythagorean triples are having its own importance in

recreational mathematics. It is a source of generating

higher order like power numeral relations,particularly

with squares. In fact there is a close relationship between 

   R ^2   and   R ^2 –relations

2   1            2   2

In   R ^2 . a square is equated with a sum of two squares  and in 

    2   1

 R ^2  , a sum of two squares is equated with a sum of two other

2  2     

squares. By using a Pythagorean triple or two sets of triples,

one can generate     R^2

        2  2

 With one Pythagorean triple (x,y,z) we can generate

(x+z)^2 + (y+z)^2 = (z+x+y)^2 + (y-x)^2

For example (3,4,5) yields,

8^2 + 9^2 = 12^2 + 1^2

It is curious to  know that a single R^2 relation can generate
                                                                     2  1

a series of R^2 relations . For example the above numeral

relation can be written with numbers existing and number

added

(3+5)^2 + (4+5)^2 = (5+7)^2 + (0+1)^2

If all the existing numbers only or if  all the added numbers

only are multiplied with a number n , the equality of the relation

is not affected.

 (3n+5)^2 + (4n+5)^2 = (5n+7)^2 + 1^2

or

(3+5n)^2 +(4+5n)^2 = (5+7n)^2 + n^2

Again if all the existing numbers are multiplied with m and

all added numbers with n, the equality of the relation n is

still preserved.

(3m+5n)^2 + (4m+5n)^2 = (5m+7n)^2 + n^2  

Fun with Mathematics

Pythagorean triples with one same number

All the numbers in the Pythagorean triple cannot be odd,

because Sum of two odd squares will be even and cannot

be an odd square.Knowing the odd-even nature of the two

numbers of the Pythagorean triple, the odd-even nature of

the other number can be predicted.The permitted and

forbidden set of triples in terms of its odd-even

characteristics are given below                                  

                 permitted                                forbidden

        x        y        z                            x      y       z

      even  even  even                     even  even  odd

      even  odd    odd                      even  odd   even

      odd   odd   even                      odd   odd    odd

There are many puzzles associated with Pythagorean

triples. One among them is to find out two or more set

of triples with one identical number in them. This

identical number may be either the smallest or the

biggest of the set.

Factorisation technique is advantageous to generate such

sets.Any number (N), composite or prime can be expressed

as the product of its two factors (x,y). For primes it will the

1 and the prime number itself.

                                  N = x y

The  number N can be expressed in term of its factors as

N = [(x+y)/2]^2 – [(x-y)/2]^2

If N happens to be a square number N = n^2, then

n^2 + [(x-y)/2]^2 = [ (x+y)/2]^2

By using different pairs of two even or two odd factors for n^2

Pythagorean triples with one identical number n can be formed.

For example,

64 = 4x16 = 2x 32

which give (8,6,10) and (8,15,17)

From

 81 = 3x27 = 1x81 we have

(9,12,15) and (9,40,41)

If the square number has many pair of such factors

(both even or  both odd) ,then we can get three or

 even more sets with an identical number in them. For example,

144 = 2x72 =4x36=6x24=8x18

It gives,(12,35,37),(12,16,20),(12,9,15),(12,5,17)

Another way of getting such set of triples with identical bigger

number is linear combination of numbers in the given

Pythagorean triples(x,y,z) or with another

Pythagorean triples (a,b,c)

(ax+by)^2 + (ay-bx)^2 = (a^2+b^2)(x^2+y^2)= (cz)^2

For example

      x   y   z       a      b      c 

      3   4   5       7    24     25  gives 117^2 + 44^2 = 125^2

      3   4   5      24    7      25  gives 100^2 + 75^2 = 125^2

      3   4   5      20   15     25  gives 120^2 + 35^2 + 125^2

                                                                         (reducible)

It is very useful to generate double or triple coincidence in

numeral relations where sum of two squares is equated

with sum of two other squares. It is a different puzzle

which we will see in the next blog.

Fun with mathematics

Pythagoras theorem with different statement

Pythagoras theorem states that if square of a number is equal

to sum of squares of two other numbers ,then these three

numbers represent the three sides of a right-angled triangle.

Conversely, if (x,y,z) denotes the length of opposite side,

adjacent side and hypotenuse of a right-angled triangle

respectively, then

                           x^2 + y^2 = z^2

Where z>x and z>y but x+y >z.(3,4,5) and (5,12,13)  are the

two well-known examples for Pythagorean triples.

                          3^2 + 4^2 = 5^2 ;

                          5^2 + 12^2 = 13^2

 Since x^2 represents a square of a side x, we usually consider

 squares in each side of a right-angled triangle  to show that

the area of the square drawn on the hypotenuse is equal to the

sum of the areas of the squares drawn on other two sides.

It is curious to know that the Pythagoras theorem holds good

for many geometrical figures such as rectangle, circle,

semi-circle, equilateral triangle,

isosceles triangle etc.,

Squares need one variable- the side length, which is taken

as the length of each side of the given right-angled triangle.

But rectangle or isosceles triangle requires two variables,

the length and breadth  or base  and height.  The length of the

side of the triangle is taken as one of the variables for the

rectangles/isosceles triangles to be drawn on each side of the

triangle. The other variable required is derived from the other

sides of the given right-angled triangle . For example, the

length of the rectangle or the height of the isosceles triangle

drawn on the hypotenuse  is equal to the sum of the length

of the other two sides of the triangle  and for  other sides 

 it is the length of the hypotenuse.  It can be shown that the

areas of the rectangle/isosceles triangle drawn on hypotenuse

is equal to the sum of areas of the rectangles/isosceles

triangles drawn on other two sides.

               For rectangle :     xz + yz = z(x+y)

              For isosceles triangle :  ½ xz + ½ yz = ½ z(x+y)

                                                           Or xz + yz = z(x+y)

 If the length of the rectangle is taken as its breadth which is

 equal to the length of the side of the triangle over which it is

 drawn, it becomes a square. It holds good even in the case

of isosceles triangle. If the height of the isosceles triangle

is equal to the length of the side of the given right-angled triangle,

their areas make the very same Pythagoras expression

x^2+y^2= z^2.

If semi-circles/circles are drawn by keeping the sides of the

given right-angled triangle as its diameter, the n the area of

 the semi-circle/circle drawn on hypotenuse will be equal to

the sum of the areas of the semi-circles/circles drawn on

other two sides.   

Funwith mathematics-16

There are quite a large number of mathematical puzzles related with prime numbers. Some of them are illustrated below.

1. For example, one can fix up a target sum and try to represent it as the sum of least and largest primes. It is very simple and simplest among prime number puzzles. At first get the list of prime numbers upto the target sum of interest. Then, subtract the largest primes in the list one by one from the target sum and see that the resultant is also existing in the list. It is the least prime required. If the target sum is 100, then the solution is 97 + 3, and it is 997 + 3, for the target sum 1000 and so on.

The target sum can be arrived by the addition of two primes by many ways.

                        100 = 89 + 11 = 83 + 17 = 71 + 29 = 59 + 41 = 51 + 49

It is interesting to note that 100 can be expressed as the sum of the first nine primes starting from2

                                    100 = 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23

With the smallest 3 primes ending with 3 and 3 primes ending with its complementary number 7,it can be shown as

                                    100 = 3 + 13 + 23 + 7 + 17 + 37

It is noted that 100 can never be expressed as the sum of 3 primes. If 2 is included, an even number with two odd primes cannot be summed up to 100.Again, if all are  odd primes, which are ending  with 1,3,7 and 9, any choice of three primes cannot never yield 100 ,an even number, on summation.

2. The alternating sums of the factorials seem to be prime.

                                    3! – 2! + 1!  = 5

                                    4! – 3! + 2! – 1! = 9

                                    5! - 4! + 3! – 2! + 1! = 101

                                    6! – 5! + 4! – 3! + 2! – 1! = 619

                                    7! -6! + 5! – 4! + 3! – 2! + 1! = 4421

                                    8! – 7! + 6! – 5! + 4! – 3! + 2! – 1! = 35899

This is not true continuously. 9! - 8! + 7! – 6! + 5! – 4! + 3! – 2! + 1! = 326981 = 79 x 4139.

3. The study of primes having same digit in all of its decimal place is also interesting. The multi-digit primes can never be even. Hence primes with an even digit in all of its decimal place are ruled out. If the equal digits are any number except 1, then it will have that digit as its factor, But the numbers with “1” in all of its decimal place are interesting.

           

                                                11 is prime

But, 111 (= 3 x 37), 1111 ( =11 x 101), 11111 (=41x271), 111111 (=11x10101),1111111 (239x4649), 11111111111 (=21649x513239), 1111111111111 (53x79x265371653), 11111111111111111 (=2071723 x 5363222357) are all not primes. Surprisingly the number with 19 ones is found to be prime. Instead of ‘1’ we can now introduce ‘0’ in between the two extreme 1’s. If we go on adding more and more zeros, will the resultant number be prime? The study of primes of the form 10ⁿ + 1 is also interesting.

With even number of zeros                                              with odd number of zeros

  10³ + 1 = 1001 = 11 x 91= 7 x 143                          10⁴ + 1 = 10001 = 73 x 137

  10⁵ + 1 = 100 001 = 11 x 9091                                10⁶ + 1 = 1000 001 = 101 x 9901

  10⁷ + 1 = 100 00 001 = 11 x 90 9091                      10⁸ + 1 = 1000 00 001 =17 x 5882353

  10⁹ + 1 = 100 00 00 001 = 11 x 9090 9091             10¹⁰ + 1 = 1000 00 00 001 = 101 x 9900                                                                                                                                                                                                                                                                                                            

                                          = 7x 142857143                                                                          9901

  10¹¹+1 =100 00 00 00 001 =11x 90 9090 91          10¹² + 1 =1000 00 00 00 001= 73 x                                                                                                                                                                                                13698630137

  10¹³ + 1 = 11 x 9090 9090 9091                                           10¹³ + 1 = 29 x 344 827 586 2069

  10¹⁴ + 1 = 7 x 142857 142857 143  

For all numbers 10²ⁿ⁺¹ (n > 1), 11 is invariably a factor, and in the other factor a block of ‘90’ is increasing for every unit increase in n.

If the extreme digits in these set of numbers are other than 1, whatever may be the number of intermittent zeros, it will be divisible by the replaced digit itself. The primality nature of the numbers having general forms 10ⁿ + m, where m takes values 3, 7 and 9 only and n may be any integer.

 13,103 are primes, but not all the larger numbers belonging to this group are primes.

            10³  + 3 = 1003 = 17 x 59                                10⁴  + 3 = 10003 = 7 x 1429

            10⁵  + 3 = 100003 = prime                               10⁶   + 3 = 1000003 = prime

            10⁷  + 3 = 10000003 = 13 x 769231               10⁸  + 3 = 100000003 = 643 x 155521

            10⁹  + 3 = 1000000003 = 23 x 43478261       10¹⁰ + 3 = 10000000003 = 7 x

                                                                                                                        1428571429                                                                                                            

            10¹¹ + 3 = 100000000003 = prime                  10¹² + 3 =1000000000003 = 61 x

                                                                                                                       16393442623

            10¹³ + 3 = 13 x 769230769231                       10¹⁴ + 3 = 19 x 5263157894737

            10¹⁵ + 3 = 14902357 x 67103479                   10¹⁶  + 3 = 7 x 142857 142857 1429

 The next two numbers under this series are found to be composite. If the digital order of 13 is reversed we get 31, which is also a prime. Such primes are called emirps (‘prime’ spelled backwards).An emirp is defined as a prime whose reversal is also prime, but which is nota palindromic prime. We get 301 on the introduction of ‘0’ in between 1 and 3, which is not a prime. It develops a curiosity to make primality test over the numbers of the form 3 x 10ⁿ + 1.The pair of factors, if they have, they must end with either (3, 1) or (7, 9), otherwise the ending 3 in the numbers examined cannot be accounted for.

          3001             = prime                                    30001     = 19 x 1579

          300001         = 13 x 23077                           3000001 = 853 x 3517

          30000001    = prime                                      300000001 = 7 x 42857143

          3000000001 = 7589 x 395309                      30000000001 = prime

          300000000001 = 13 x 23076923077            3000000000001 = 67 x 44776119403

          30000000000001 = 17 x 1764705882353    300000000000001 = 7 x 42857142857143

          3000000000000001 = 29 x 103448275862069           

The pair of factors, if they have, they must end with either (3, 7), (9, 9) or (1, 1) otherwise the ending 1, in the numbers examined cannot be accounted for.

             

            17 = prime                                                107 = prime

            1007 = 19 x 53                                         10007 = prime

            100007 = 97 x 1031                                 1000007 = 29 x 34483

            10000007 = 941 x 10627                         100000007 = prime

            1000000007 = prime                                10000000007 = 23 x 434782609

            100000000007 = 353 x 283286119         1000000000007 = 34519 x 28969553

            1000000000000= 167 x 59880239521    100000000000007= 43 x 2325581395349                1000000000000007 = 47 x 21276595744681

The emirp of 17 is 71. 71,701,7001,70001,700001 are all primes. The primality of the other higher numbers under this series are given below.

            7000001 = 197 x 35533                                  70000001 = 43 x 1627907

            700000001 = prime                                         7000000001 = prime

            70000000001 = 53 x 1320754717                  700000000001 =41149 x 17011349

            7000000000001= 23 x 304347826087           70000000000001 = 67 x 104477611                                                                                                                                        9403 

4. See the sequence of the number – 31, 331, 3331, 33331, 333331, 3333331, 33333331. All of them are prime numbers. If we assume logically that the next number in the sequence 333333331 will also be prime, it will be a wrong conclusion. It has factors 17 and 19607843, so that

                                                333333331 = 17 x 19607843

3333333331 is not a prime as it is divisible by 673.

In this sequence of numbers, in between the fixed digits at the edges, the number of decimals with a particular digit is increased in step. Another example in which the decimal numbers in between the fixed edge numbers are introduced from 1 in the same order as in natural series is given below.

                                                                        17

                                                                        127

                                                                        1237

                                                                        12347

                                                                        123457

All are primes, but 1234567 and 123456787 are not primes, since they can be factorized.

                                                          1234567   = 127 x 9721

                                                        12345677   =   29 x 425713

                                                  123456787   =   31 x 31 x 128467

                                                  1234567897    = 17 x 73 x 004817

It is a kind of deletable prime which can be defined as prime that remains as prime when the digits are deleted in some chosen manner. In the above cited example, the digits are deleted one by one from the right and not from the left.  Chris Caldwell has shown one such a deletable prime- 41056793, where the digits are removed in some chosen order. Here 410256793, 41256793, 4125673, 415673, 45673, 4567, 467, 67, 7 are all the primes in a sequence.

5. Among the two-digit primes, there are only four pairs, in which one of the primes is obtained simply by reversing the order of digits of the other. They are (13,31), (17,71) ,(37,73) and (79,97). It is found that 11 is a factor for the sum of the primes in a pair and 9 is the factor for the difference between them. The difference between the squares of the primes in a pair has factors, 8, 9 and 11 uniquely.

                                   

31² – 13² =  8 x 9 11

                                    71² – 17² = 6 x 8 x 9 x 11

                                    73² -  37² = 5 x 8 x 9 x 11

                                    97² – 79² = 4 x 8 x 9 x 11

Among the three-digit primes, there are 15 pairs. They are (107,701), (113,311), (149,941), (157,751), (163,361), (167,761), (179,971), (199,991), (337,733), (347,743), (359,953),(389,983), (709,907), (739,937),(769,967).These pairs also exhibit similar properties.

6. A palindromic primes, sometimes called a palprime, is a prime number that is also a palindromic number. The first few palindromic primes are (excluding the single digit primes) 11,101,131,151,181,191,313,353,373,383,727,757,787,797,919,929,10301,10501,10601,11311, ...........Except 11, all palindromic primes have an odd number of digits, because the divisibility test for 11 tells us every palindromic number with an even number of digits is a multiple of 11.

7. It is impossible to show that a sum of prime and its emirp is equal to a product of two consecutive primes. Because the sum of two primes will be even while its product will be odd. For the same reason the possibility of getting prime + emirp to be equal to a product of more than two consecutive primes is also ruled out. But it is possible when the sum is odd, that is, if one number is odd (may be prime or composite), the other number obtained by the reversal of digits must be even. It is possible if the number taken begins with an even digit.

For example,

                                                   251 + 152 = 403 = 13 x 31

If the product of consecutive primes begins with prime number 2, one can have a solution.

                                             1009 + 9001 = 10010 = 2x5x7x11x13

8. The sum a prime and its emirp can be shown to be equal to twice the product of a prime and its emirp,

                                             1006441 + 1446001 = 2x1021x1201

or to be equal to twice a cube of a prime,

                                                1061 + 1601 = 2662 = 2 x 11³

                                                1151 + 1511 = 2662 = 2 x 11³