FLT and Beal's Conjecture

Algebraic Proof for Fermat’s Last Theorem and Beal’s Conjecture

Abstract:

There are many proofs to prove Fermat’s Last Theorem as well as Beal’s Conjecture. Here is yet another proof, much simpler than any other proofs.

Keywords:

Fermat’s Last Theorem- Beal’s Conjecture- algebraic proof – numerical examples

Algebraic Proof for Fermat’s Last Theorem and Beal’s Conjecture

Introduction:

According to Beal all three member multi-power relations  a^x  +  b^y   = c^z  , where a,b,c and the exponents x,y,z  are all positive integers such that (x,y,z) > 2, the base numbers a,b,c  will invariably have one or more common factors. Conversely the equation  a^x  +  b^y   = c^z has no solutions for any positive integers a,b,c,x,y and z with (x,y,z)  > 2  for any set of co-prime (a,b,c). In such three members like or unlike power relations both the Fermat’s assertion and Beal’s conjecture have some similarities and differences.

The observed similarities are (i)  All the members can be even, then the relation will be reducible (ii) All the members cannot be odd due to non-conservation of oddness (iii) One or two members  may be square or higher power but all the members cannot be square or with the same higher power, (iv) All the three members cannot be simultaneously primes (v) FLT and Beal’s conjecture are mathematically true with fractional and rational or complex numbers, and (vi) In irreducible form, both the FLT and Beal’s conjecture do not have a common prime factor among the members.

Among the differences the most significant are (i) In a^x  + b^y  = c^z , FLT requires that a,b,c cannot be positive integers when the exponents x,y,z are same and greater than 2, but they can be in Beal’s expression, when the exponents are different (ii) In FLT,( a + b) will always be greater than c. but it can be greater, lesser or even be equal depending upon the value and position of the exponents in the multi-power relation. These differences give an explanation why Beal’s conjecture is allowed and Fermat’s last theorem is not for exponents > 2 .

In this paper both the Fermat’s last theorem and Beal’s conjecture  are proved with an algebraic method which is much simpler than any other methods. One and the same expression admits multi-power relations in accordance with Beal’s conjecture and forbids like-power relations in agreement with Fermat’s last theorem.

Algebra of three member power relation

In all the three member like  aⁿ + bⁿ  = cⁿ , and  unlike  (a^x + b^y  = c^z)   power relations, one can establish that  any two members are proportional to the remaining member.

a^x  = c^z – b^y  = (c^z/2 – b^yl2 )(c^z/2 + b^y/2 )

 If c^z/2 – b^yl2  = a^x/2/k  then  c^z/2 + b^y/2 = k a^x/2 , where k is a proportional factor

It gives c^z  = a^x [ (k² + 1)/2k]²= a^x β²  and b^y  = a^x [(k² – 1)/2k]² = a^x α². It clearly indicates that all the three member power relations have the condition  1 +    α²  =  β² as a common feature.  This unique condition clearly predicts the mathematical reason why whole integral solutions are not permitted when all the exponents are same and greater than 2, and permitted when the exponents are different. The difference between the  FLT and Beal’s equation lies in the successful formation of a^x β²  and a^x α² into  single power.

In fact this mutual dependence among the members provides them a common factor ,which is usually one of the members of the Beal’s equation. If we are interested whole integral solutions . the proportional factors α², β² must not only be positive integer, but also its product with the common power  a^x must be expressible as another power. When all are supposed to be expressed  in the same power, the proportional factor must also be  expressible in the same power.   since 1 + αⁿ   = βⁿ is not possible for any integral  values of α and β  when n >2, aⁿ + bⁿ = cⁿ    is possible only for n = 2 and becomes impossible when n > 2,which  is in accordance with FLT. When  the proportional factors obeying the unique condition are expressible as different power and  give different power when multiplied with the common power it becomes a multi-power relation  so-called Beal’s equation.

Forbidden Fermat and allowed Beal

If a,b,c,m and n  are all positive integers, aⁿ + bⁿ  = c^m  is allowed as Beal’s  relation when m ≠ n, and forbidden due to FLT when m = n.

(a^n/3)³ + (b^n/3)³ = c^m

P³ + q³ =  (p+q) (p² –pq +q²) =  c^m , where  p = a^n/3 and q = b^n/3. P and q to be whole integers, a and b must be cubes or n must be some multiples of 3. If p + q = α c (for m >n ) and p+q = c/α (for m <n) where α > 1  may be a fraction or whole integer but not complex .In the case of m >n,  p² –pq +q²  =   c^m-1/α  and  3q² - 3αcq + α²c² – c^m-1/α = 0 by eliminating p. Its solution is q = (αc/2) ± (αc/6)                    √[( 12 c^m-3/α³) – 3]. When m <n ,α <1 and hence  p² –pq +q²  =  α c^m-1  and  3α²q² - 3αcq +c² –α³ c^m-1 = 0 by eliminating p. Its solution becomes q = (c/2α) ± (c/6α) √[( 12α³  c^m-3) – 3].

This solution gives adequate explanation for the FLT and Beal’s conjecture. “q” to be whole integer, the terms inside the square root must be either z.ero or a square number s². For positive roots                4 c^m-3/α³  > 1 and for integral roots c^m-3/α³  = 3, 7,12,19,28,39,52,67,........ so that c is related to α.

When m = 3,  1< α < 1.6, the solutions for p,q and hence a,b are irrational for all fractional values of α  The real solutions are obtainable only  when α is whole integer. When  α ≥ 1.6 , the roots become imaginary . The non-existence of real roots is the clear indication of FLT. However if m is different form n, the solutions are obtainable. For example, when n =3, m = 4  , 12c/α³ = 7, then α = 6, c = 126, which gives 126³ + 630³   = 126⁴, when n = 3 , m = 5, 12c/α³ =28, then α = 21 , c = 147 which gives  4116³  = 147⁵  + 1029³ . Solutions are found when n=3,m = 2, 12/cα³ = 4, α = 1,c=3  which  gives 1³ + 2³ = 3^2.,it is 2³  + 2³  = 4²  for α = 1 , c = 4. For various values of α and c, one can get innumerable relations.

                m = 2                                                        m=4                                                       m=5

α              c            [a,b]³ = c²                                        α           c        [a,b]³ = c⁴                     α       c        [a,b]³ = c⁵

2             32          {8,8]³ = 32²                        2            2       [2,2]³=2⁴                      3        3       [3,6]³  = 3⁵

3            108         [18,18]³ =108²                  3            9      [9,18]³ = 9⁴                             4         4       [8,8]³ = 4⁵

 2                     24         [4,6]³  =24²                                     4           16     [32.32]³ = 16⁴                      12      24      [96.192]³ =24⁵

 7           588        [14,70]³ = 588²                  5           35    [70,105]³ =35⁴               16      32    [256.256]³ =32⁵   

m=6,,9,12......3n is not possible due to FLT.When n = 3,m = 7 ,12c⁴ /α³ =4, α = 27, c = 9 which gives  81³  +  162³  = 9⁷

In terms of α and c, the multi-power relation can be expressed as

[(αc/2) – (αc/6)√ (12c^m-3 /α³  - 3)]³     [(αc/2) + (αc/6)√ (12c^m-3 /α³  - 3)]³  = c^m                         

It shows that (p,q,c) and hence (a,b,c) has a common prime factor and this is in concurrence with Beal’s conjecture. Reducing the expression by dividing with the common factor and keeping α = 1, one can get the irreducible form of Beal’s relation. It is exemplified for a typical set of Beal’s relation  a³   + c⁴  = b³

[(1/2) + (1/6)√(12 c – 3)]³  -   [(1/2) + (1/6)√(12 c – 3)]³  = c

 c                    irreducible à reducible Beal’s relation

7                      2³  - 1³   = 7   à   x 7³  = 7³  + 7 ⁴  = 14³

19                   3³  -  2³  = 19 à   x 19³  = 38³ + 19⁴ = 57³

37                   4³ –  3³ =  37 à  x 37³   = 111³  + 37⁴  = 148³

61                  5³   -  4³    = 61 à x 61³  = 244³ + 61⁴  = 305³

This set of relations can generally be expressed as

(3n²   + 3n + 1) ;      (n+1)³  - n³= 3n²   + 3n + 1  à[(n)( 3n²   + 3n + 1)]³ + [( 3n²   + 3n + 1)]⁴ =  [(n+1)( 3n²   + 3n + 1)]³  .  

Number theory

Number Theory of  a³ + b³  + c³  = d³

Abstract:

The number theory of cubical relations predicts that it is not possible to get  a cube equals to sum of two cubes, but it can be shown to  be equal to sum of three or more cubes. In this article few methods of generating such cubical relations and its general properties are studied. 

Key words:

Diophantine equation -Number theory- Fermat’s Last theorem, Beal conjecture

Introduction:

When a,b,c and  x are positive integers, then the like power relation  a^x + b^x  = c^x  is  not possible for  all values of x ≥3 (Fermat’s Last theorem), since in such relation any two members are proportional to the other member. If b and c are proportional to a, then  b^x = α a^x  and c^x   = β a^x  where the proportionality factors satisfy a condition 1 +α = β, which gives insignificant solutions due to the requirements α = p^x , β = q^x. However multi-power relation a^x + b^y = c^z  with at least one of the exponents is different from the other is possible, where a,b and c  have invariably a common prime factor (Beal conjecture).The missing common prime factor when the exponents become same can be taken as an indirect proof for FLT.

In the like power relations, it is found that  a number with an exponent 3 or more can be expressed as a sum of three or more numbers with the same exponent and such relations are generally called Diophantine equation. In this article the methods of generating a cube equals to sum of three cubes   and its associated mathematical properties are discussed. It gives yet another valid proof for FLT – in  a^x  + b^x  + c^x = d^x  if the number of members with like powers in the LHS reduces to 2, the integral  solutions disappear.

Diophantine  way
One can get a cube equals to sum of three cubes by many different ways using algebraic identities, which one can generate arbitrarily .

Let   (x+ ma)³  - (x- ma)³  = ` [(x-na)/b  + c]³  - [(x-na)/b  - c]³  be our algebraic identity of interest. When m ≠ n  and b >  1, the relation reduces to

m³a³  + 3max²  = c³  + (3c/b²) (x-na)² .

 If c = mab², then  x = (a/6n)[m²b⁶ + (3n² – m²)]. In  terms of m,n and b , the three independent variables, the most general relation becomes

{b[m² b⁶ +  (3n² – m²) + 6mn]}³  + [m² b⁶ +  (3n² – m²) – 6n² – 6mnb³]³  = {b[m² b⁶ +  (3n² – m²) - 6mn]}³  + [m² b⁶ +  (3n² – m²) – 6n² + 6mnb³]³.

The values of the independent variables can be so chosen so that one of the members becomes negative. Under the condition 3n²  + m²  + 6mnb³ > m² b⁶ , the given algebraic identity gives one cube equals to sum of three cubes.

m      n        b                 a³  + b³  + c³  = d³

 1      2        2                [45,126,147=174]³

1       3        2                [108,144,180=216]³

2       3        2                [63,486,513 = 630]

1       4        2                [174,177,207=270]³

2       4        2                [180,504,588 = 696]³

3       4        2                [ 57,1086,1095=1374]³

                    

There are quite a large number of algebraic expressions for a sum of two cubes equal with a sum of two other cubes. When one of the members gives negative value it is then transformed into a cube equal with a sum of three cubes. For example, the algebraic identity  (x-a)³  + (x+a)³  = (x-a +p n)³  + (x+a-q n)³  can be used to generate such relations by solving  for x and a,  and by giving convenient  values for p,q  and conditional value for n. When p = 9 and q = 1 and n = 3k where k = 1,2,3,4,5.......we have  x = [30a – 369)± 3 √ 36 a² – 540 a – 2343 ]/24 for  k = 1 and a =  [45 ± √ 4368 + s² ]/6  where s²  = 36 a² – 540 a – 2343. Few typical solutions are

n            s                a                   x            a³  + b³  + c³  = d³

3            11            56/3            28/3     [28,53,75=84]³

                                                   79/12   [145,179,267 = 303]³

              31            59/3            16/3     [38,43,66=75]³

                                             - 14/5          157/12   [79,245,357=393]³

              64           137/6         127/6     [5,76,123=132]³

                     71             71/3           16/3      [26,55,78=87]³

                                        -26/3         277/12   [7,317,525=561]³                                  

When the value of n is changed, it gives reducible relation with a common prime factor and consequently it reduces to the relation corresponding to n = 3.

Ramanujan’s way

In Ramanujan’s method, if p.q,r are so assumed as p + 3a²  = q + 3ab = r + 3b² = (a+b)²  and m, n are any two independent variables , then

n (mp + nq)³  + m (mq + nr)³  = m (np + mq)³  + n (nq + mr)³ . A particular case of the above theorem is

(3a² + 5ab – 5b²)³ + ( 4a² -4ab + 6b²)³  + (5a² – 5ab – 3b² )³  = (6a² – 4ab + 4b²)³ 

Some irreducible numerical relations are given  below

a          b                  a³  + b³  + c³  = d³

1         0                  [3,4,5=6]³

            2                 [7,14,17=20]³

                3                 [27,30,37=46]³

            4                 [54,57,63=84]³

2          3                 [ 3,36,37=46]³

3          1                    [27,30,37=46]³

           

When a= b, a sum of two cubes equals to sum of two other cubes can be obtained where the members in both sides will be identical. When b= 0, we get irreducible relation for a = 1 and reducible for its higher values, which is reduced to the irreducible one corresponding to a =1.It is interesting to note that the relation remains same when a and b are interchanged. When one of the members is made to be equal to zero, then the value of a for any integral value of b will become irrational and vice-versa, so that a cube cannot be shown to be equal to a sum of two cubes all with integers. Alternatively, it can be stated that it is not possible to get a number which when added with a cube cannot be expressed as another cube and as sum of two cubes simultaneously. 

Meyyappan’s way

In a³  + b³  + c³  = d³ , d is greater than a,b and c but less than a+b+c . Let d = c+n . then

a³  + b³  = 3cn(c+n) + n³ . In all the cubical relations, the difference between the sum of members in both sides of the relation will always be equal to some multiples of 6. i.e.,

a +b = 6m + n, which gives

a³   + b³ = (6m+n)³  - 3ab (6m+n)  =3c²n + 3cn²  + n³

Solving for c,

c = - (n/2)  + (1/6n) √ [ 288 m³n + n⁴  + 144 m² n²  + 24 m n² – 4abn(6m+n) }

      n           m          a+b           a³  + b³  + c³  = d³

     1            1            7                  [1,6,8=9]³

     1            2          13                  [3.10,18 = 19]³

      1             3          19                  [2,17,40 =41]³

      1             4          25                   not possible

     1             5          31                 [12,19,53=54]³

     1             6          37                 [14,23,70 = 71]³

      1             7          43                 [12,31,102=103]³

     1             8          49                   not possible

     1             9          55                   not possible

This method is well suitable to get a³  + b³  + c³  = (c+n)³  for a given value of ”a”  where n is an integer. It predicts that for every value of a (= m), there is a numeral relation  m³ + b³ + c³  = (c+1)³  and for any particular value of a, there would be one or more numeral relations a³  + b³  + c³  =(c+n)³ with different value of n.

When n = 1 and  a= m (m = 1,2,3,4,5,6....)

 b³  =3c²  + 3c + (1- m³ ) and   b = (6k + 1 – m)

 Solving for c ,   c =  ( -1/2 ) + (1/6) √  [12(b³ + m³) – 3]

 a (=m)     b          c            c+1(=d)

1               6           8              9

2             17         40            41

3             34        114          115

4             57        248          249

5             86        460          461

6            121       768          769

7            162      1190        1191

8            209      1744        1745

9            262      2448        2449

10          321      3320        332

It is interesting to note that as “a”  increases with a constant difference , all other members b,c and d  also increase proportionately. The general expression for this type of relation is

                 ^{n                                                                 n-1                                                                                n-1}

n³  + [ ∑ (6m)  - (n-1)]³  + [ 8n + ∑ 6 (3m-1)(n-m)]³   = [ 8n + 1 + ∑ 6 (3m-1)(n-m)]³ ]³

                 ^{m=1                                                          m=1                                                                               m=1}

For a given m (=a) , there  can be one or more relations with d-c = 1 , for example when m = 3.

m (=a)       b            c            c+1 (=d)

3              4             5               6

3             10          18             19

3             34         114          115

For same m (=a) , n ( =d – c) can have different selective values,

m= 3 ; n = 1;  [3,10,18=19]³ 

m= 3;  n = 9 ; [ 3,36,37=46]³

m = 7; n = 1 ; [7,162,1190=1191]³

m= 7 ; n = 3;  [7.14,17=20]³

m= 7 ; n=36 ; [ 7,317,528,561]³

Again it is found that the relation does not exist when either a or b becomes zero.When n= 0 or 1, the relation simply converts into 1³ = 1³ .  b=0  gives a condition   _mⁿ ∑ 6m =  (n-1) .It is possible only when n = 1 and m =0. For any other values of m and n   _mⁿ ∑ 6m ≠  (n-1). This is in support of FLT. Alternatively, it can be stated that it is not possible to get a number which when added with a cube cannot be expressed as another cube and as sum of two  different cubes simultaneously.

Properties of  a³  + b³  + c³  = d³                                 

1.Any four member sum of like power relations can be used to generate six member sum of like power relations. For example, [3,4,5 = 6]³   

[1,4/3,5/3 =2]³ and [ ¾,1,5/4 = 6/4]³  give  24³  + 9³  + 15³  = 18³ + 16³ + 20³

[1,4/3, 5/3 = 2]³  and [ 3/5,4/5,1 = 6/5)³ give 18³ + 20³ + 25³  = 30³ + 9³   + 12³

2.a^x + b^x  + c^x  = d^x has integral solutions for n = 2 and 3, provided a,b,c and d are not squares or higher powers simultaneously. . It predicts the non-existence of a⁶  + b⁶  + c⁶  = d⁶ .

The general form of a² + b²  + c²  = d²  is  a²  + a² [(k² – 1)/2k]² + a² [(k² –1)/2k]² [(k² +1)/2k]²  = a² [k²+ 1)/2k]⁴, where 1 +[(k² – 1)/2k]²  = [(k² + 1)/2k]²   If “a”  is  a square number , then (k² – 1)/2k = p², (k² –1)/2k] [(k² +1)/2k] =p² q²  which gives an unfathomable condition 1 + p⁴   = q⁴   , since the minimum difference between any two fourth power is 15. Similarly one can show that all the members in   a³ + b³  + c³  = d³  cannot be squares or cubes due to non-trivial condition 1 + p⁶  = q⁶

3. It can be shown that all the members cannot be primes simultaneously. a³  + a³ [(k² – 1)/2k]² + a³ [(k² –1)/2k]² [(k² +1)/2k]²  = a³ [k²+ 1)/2k]⁴ . If a = p_{1 ,} a³ (k² – 1)/2k]² = p₂³, a³ [(k² –1)/2k]² [(k² +1)/2k]² = p₃^{3 ,} then  the required condition p₁³ (p₃³ – p₂³ ) = p₂⁶ which predicts that p₂ cannot be a prime.

Beal's conjecture

Link between  Pythagoras’  and  Beal’s  equation

It is found that there is a definite correspondence between the Pythagoras’  equation and Beal’s equation .

Let a² + b² = c²   be a known Pythagoras’ equation. With simple mathematical manipulation, it can be converted  into a  multi-power relation  p^x  + q^y  = r^z  which reveals  the basic characteristics of Beal’s conjecture. It states that if p,q,r,x,y and z are positive integers with (x,y,z) > 2 then p,q,and r have a common prime factor.

The conversion of a Pythagoras’ equation into a Beal’s equation can be done with fundamentals of Algebra . It is exemplified as follows.

a² + b² = c²  à  1 + (b²/a²) = (c²/a²)

Multiply by pⁿ à pⁿ  + pⁿ (b²/a²) = pⁿ(c²/a²)

If b²/a² = qⁿ then c²/a² = qⁿ  + 1 . If p = qⁿ  + 1 we get multi-power relation with exponents [n,n,n+1]

(qⁿ +1)ⁿ  + [q(qⁿ + 1)]ⁿ  = (qⁿ + 1)ⁿ⁺¹

n=2, q =2 gives  5² + 10²  = 5³

n=3,q=2 gives  9³  + 18³  = 9⁴

n=3 , q=3 gives  28³  + 84³ = 28⁴

if b²/a²  = q^n-m/p then c²/a² = [ 1 + q^n-m/p]

pⁿ  + (pq)^n-1    = p^n-1 [p + q ^n-1] = (pr)^n-1  where p = r^n-1 – q^n-1, we get multi-power relation with exponents [n,n-1,n-1]

(r^n-1 – q^n-1 )ⁿ  +  [q (r^n-1 – q^n-1 )]^n-1 = [r(r^n-1 – q^n-1 )]^n-1

When  n= 4, r=5,q=4 ; 61⁴  + 244³  = 305³

When n=5 ,r=4 ,q=2;  240⁵ + 480⁴  = 960⁴

If  b²/a²  =  p/αⁿ⁺¹ then  c² /a²  = [ 1 + p/αⁿ⁺¹] = βⁿ

p = αⁿ⁺¹[βⁿ – 1], we get multi-power relation with exponents [n,n+1,n)

when  n = 3, α = 2,β = 3; p = 416 and 416³ + 208⁴  = 1248³

when  n = 3 α = 1,β =2; p = 7 and  7³ + 7⁴  = 14³

when n = 4 ,  α = 2,β =3; p =2560  and 2560⁴  + 1280⁵ = 7680⁴

extraordinary properties of products of any four successive numbers

The extraordinary properties of products of four successive numbers

The product of any four successive numbers with any given difference d  has some peculiar properties

d = 1

1x 2 x 3 x 4 = 24 + 1 = 25 = 5²

 2 x 3 x 4 x 5 = 120 + 1 = 121 = 11²

 3 x 4 x 5 x 6 = 360 + 1 = 361 = 19²

  4 x 5 x 6 x 7 = 840 + 1 = 841 = 29²

  5 x6 x7 x8 = 1680 + 1 = 1681 = 41²

d=2

1 x 3 x 5 x7 = 105 + 16 = 121 = 11²

2 x 4 x 6 x8 = 384 + 16 =  400 = 20²

3 x 5 x7 x9 = 945 + 16 = 961 = 31²

4 x 6 x 8 x10 = 1920 + 16 = 1936 = 44²

5 x7 x9 x11 = 3465 + 16 = 3481 = 59²

d = 3

1 x 4 x7 x 10 = 280 + 81 = 19²

2 x 5 x 8 x11 = 880 + 81 = 31²

3 x 6 x9 x12 = 1944 + 81 = 45²

4 x7 x 10 x 13 = 3640 + 81 = 61²

5 x 8 x 11 x 14 = 6160 + 81 = 79²

The square root number is found to be the mean of  products of extreme numbers  a (n+3d) and middle numbers  (n +d ) (n + 2d) and is equal to (n² + 3nd + d²)² The number added with the product of four successive numbers is equal to d^4..  It gives the following algebraic relation.

n (n+d) (n+2d) (n+3d) + d⁴  =  (n² + 3nd + d²)²

Pythagorean triples

Inherent Properties of Pythagorean Triples
Abstract:

The Pythagorean triples are interesting to everyone because of its well known geometrically related ( right angle triangle) mathematical properties ( sum of two squares is equal to a square) There are many hidden mathematical properties associated with Pythagorean triples. It is a recreational attempt to find out such hidden properties. 

Key words: Pythagorean triple – numeral relation – multi-coincidence

Introduction

Pythagoras, the Greek philosopher and mathematician  formulated his famous theorem around 520 BC.He was the first to predict the geometrical aspect of the mathematical relation of a square equal to a sum of two squares. This oldest theorem still has many mathematical funs. Some of the fascinating and curious properties  associated with Pythagorean triples are predicted here.

Property-1

 If  (a_1,b₁,c₁) and (a₂, b₂, c₂) are any two set of irreducible Pythagorean triples  then c₁ c_{2 ±} ( a ₁ b_{2  ±}  a₂ b₁ ) is always a perfect square provided the two product terms have the same parity (even).

(3,4,5) and (5,12,13) gives,

 5x13 + (3x12 – 4x5) = 81 = 9²

5x13 -  (3x12 – 4x5) = 49 = 7²

Where the two squares 7² and 9²  are in arithmetic progression with  the product c₁ c₂

5 x 13 + (3x12 + 4x5 ) = 121 = 11²

5x13 – [3x12 + 4x5 } =  9 = 3²

 Where the two squares 3²  and 11² are in arithmetic progression with c₁ c₂

This can be explained with the knowledge of Taylor’s numbers for a prime set of Pythagorean triple. Accordingly the two irreducible sets can be expressed  as [(m₁² – n₁² ), 2m₁n₁ , (m₁² + n₁²)] and [(m₂² – n₂² ), 2m₂n₂ , (m₂² + n₂²)]  respectively Substituting these  algebraic values in c₁ c_{2  ±} ( a ₁ b_{2 ±-}  a₂ b₁ ) we get,

(m₁² + n₁²) (m₂² + n₂²)  ± [2m₁n₁ (m₂² – n₂² ) ± 2m₂n₂ (m₁² + n₁²) }

     = [m₁ (m₂  -/+ n₂ )  ± n₁ ( m_{2 +/-} n₂ )]²

It is a perfect square for all possible values of m_1,m_2, n₁ and n₂

Property-2

If  (a_1,b₁,c₁) and (a₂, b₂, c₂) are any two set of Pythagorean triples  then its linear combination gives a relation where sum of two squares is equal to sum of two other squares.

(a₁a₂  + b₁b₂)² + (a₁b₂ – b₁ a₂)²  = (a₁a₂  - b₁b₂)² + (a₁b₂ +b₁ a₂)²  = (c₁c₂)²

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For example, (3,4,5) and (5.12,13) give

63²  + 16²  =  33²  + 56²  =  65²

When  the same Pythagorean triple is used for linear combination, we get higher order Pythagorean triples. e.g., (3,4,5) gives (7,24,25)which in turn gives (336,527,625), where the greatest number of the resultant triple is always a square of the  greatest number in the initial triple.

The product of squares of the greatest numbers in any two Pythagorean triples can be shown to be a sum of four squares.

(c₁c₂)²   = ( a₁² + b₁² ) (a₂² + b₂² ) = (a₁a₂)²  + (a₁ b₂)²  + (b₁a₂)²  + (b₁b₂)²

For example (3,4,5) and (5,12,13) give  65²  = 15²  + 36²  = 20² + 48²

Instead of splitting both the squares c₁ and c_{2 ,} one can split any one of the squares  to get multi-coincidence in Pythagorean relation

(c₁c₂)²   = ( a₁² + b₁² ) c₂² =   c₁² (a₂² + b₂²) 

With the same set of Pythagorean triples, we have, 65²  = 39²  + 52²   = 25²  + 60²                  

Property-3

One can arrive at the numeral relation with sum of two squares is equal to sum of two other squares instantly from any known Pythagorean triple. If (a,b,c) is a known Pythagorean triple, then (am + cn)² + (bm + cn)²  = (cm +an + bn)² + (an-bn)², where m and n may have any integer value. For example with (3,4,5)

m= 1 , n= 1 ;  9²  + 8²  = 12²  + 1²

m= 2 ,n= 3;  21² + 23²   = 31²  + 3²

m=3 , n = 2; 19² + 22²  = 29²   + 2²

Property-4

If (a,b,c) is a Pythagorean triple, then (a+bc)² + (b-ac)²  =  (a-bc)²  + (b+ac)²  = c²  + (c²)² = c² ( c² + 1) . It gives equal sum of two squares in three different ways..For example

(3,4,5)   gives 23²  + 11² = 17²  + 19² = 5²  + 25²= 25 x 26

 (5,12,13) gives  161²  + 53²  = 151²  + 77²  = 13²  + 169²  = 169 x 170

(8,15,17) gives 263² + 121²  = 247² + 151²  = 17²  + 289²  = 289x290

Property-5

Intramural mixing in a given Pythagorean triple is a rich source of generating numeral relations for equal sum of two squares in two different ways. This technique is valid only for numeral relations where equal number of squares with equal number of digits are equated. If the number of squares in one side is less by

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one or more from that in  the other side, then one or more  0² ( for single digit numbers)  00² (for two digit numbers ) must be added without affecting the balance of the relation. The root numbers having unequal number of digits can be changed to have equal number of digits by annexing 0 in front of the numbers. For example in the case of (3,4,5) we have

                                   3²  + 4²  = 5²   + 0²

In intramural mixing the root numbers in side are suffixed or prefixed with the root number in the other side. On can interchange the pairing of numbers ,but once paired , it must be same for both side.

Suffixing with a pair :           35²  + 40²  = 53²  + 4²

Interchanging the pairing  30²  + 45²  = 54²  + 3²

This in turn can be used to generate more and more numeral relations

35²  + 40²  = 53²  + 04²  gives, 5335²  + 440²  = 3553² + 4004²

                                                                         435²  + 5340² =  4053²  + 3504²

 30²  + 45²  = 54²  + 3² gives  5430² + 345² = 3054²  + 4503²

                                                330² + 5445² = 4554² + 3003²

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