1.solve a² + b²  = c² with the help of algebra where the factors of one of the square number is kown 

 With single variable a = α²

a²  = c^{2  -} b²  

α⁴ = (c-b)(c+b) .If c - b = α and c + b = α³ , which give c = α (α² +1)/2 and b = α (α^{2  -} 1)/2 . The solution then becomes (2α²)² +  [ α (α^{2  -} 1) ]² = [ α (α² +1) ]²

with two variables a = αβ      

α² β²  = (c-b) (c+b) . If c-b = α  and c+b = αβ² .By solving for b and c, we get a solution as

(2αβ)² + [α(β² – 1)]²  = [α(β² + 1)]²

The distribution of prime factors of a  to (c-b) and (c+b) may have different values ,however its product must be equal to the square of the prime factors of a. If If c-b = α²  and c+b = β² .By solving for b and c, we get a solution as

(αβ)² + (β² – α²⁾²  =  (β² + α²⁾²

(2) Solve a³ + b³ = c³ by using prime factor method

With single variable . let a = α^2. Then α^{6 =} (c – b) [(c-b)² + 3cb]. Under one possible distribution of prime factors (c-b) = α and [(c-b)² + 3cb].=  α² + 3cb = α⁵ or 3cb = α² (α³ -1) or b[= [α² (α³ -1)]/3c .Substituting this value in c-b ,  c -  [α² (α³ -1)]/3c = α  or  3c² – 3cα -[α² (α³ -1)] = 0. Solving this quadratic equation for c, we get

c =[3α ± √ (9α² + 12 α^5- 12 α² ) ]/6 = α/2 + (1/6) √ (12 α^3- 3 5² ) ] and b = c – α = - α/2 + (1/6) √ (12 α^{5 -} 3 α² ) ]. It gives its solution as (α ²)³ + [- α/2 + (1/6) √ (12 α^3- 3 α²)]³ = [ α/2 + (1/6) √ (12 α^5- 3 α²)]³ . Whatever may be the prime factors of a and its distribution among b and c, ,there will be integral solution. Ie., all a,b,c cannot have positive integers. This is called Fermat’s Last theorem.

With two varibales. Let a = αβ. Then  α³ β³ =   (c – b) [(c-b)² + 3cb].. If c-b = α² ,  then   [(c-b)² + 3cb]. = α⁴  + 3cb =  α β³ or b = [α (β³ – α³ )/3c. Substituting this value in c-b , c - [α (β³ – α³ )/3c.]^- = α²  or  3c² – 3c α² - α (β³ – α³ ) = 0. The solution of this quadratic equation becomes c = [3 α^{2 ±} √ 9 α⁴ + 12 α (β³ – α³ )]/6 . It is reduced to c = (α²/2) + (1/6) √  12 α β³ –3 α⁴ )  and b = c- α^{2 = -} (α²/2) + (1/6) √12 α β³ –3 α⁴ ) ^-.It gives the solution as (αβ)³ + [ (α²/2) + (1/6) √12 α β³ –3 α⁴)]³ = [(α²/2) + (1/6) √12 α β³ –3 α⁴)]³

(3) The prime factor method is application to solve a^x + b^y = c^z.,where the exponents are not same..S=olve a² + b³ = c^{3 .}for the integral values of a,b and c.

Let  a = =   α², then α⁴ = (c-b) [ (c-b)² + 3cb]. One possible distribution  is c-b = α and  (c-b)² + 3cb] = α² + 3cb = α³  or b = [α²  (α -1)]/3c  .Substituting this value of b in c-b  = α we get a quadratic equation 3c² – 3cα - α² (α -1) = 0, Solving for c, we get its two roots c =[(α/2) + (1/6)√ (12 α³ - 3 α² )]  and b = [ - α/2) + (1/6)√ (12 α³ - 3 α² )]]. The general form of the relation becomes ( α² )² + [(- α/2) + (1/6)√ (12 α³ - 3 α)]³ = [ (α/2) + (1/6)√ (12 α³ - 3 α)] . It is seen that when α = 1,7,19,37……. The root numbers become whole integers. When α = 7 ;  49² + 7³ = 14³

 (4) Solve a³  + b² = c² for integral values of a,b,and c.

 a³   = c²  - b²= (c-b) (c+b)

With single variable  a,  c-b =a and c+b = a² which give the value for b and c in terms of a.

c = a(a+1)/2 and b = a(a-1)/2. The relation in terms of a single variable a is a³ + [a(a-1)/2]² = [a(a+1)/2]²

 a=2 ; 2³ + 1²= 3²

a=3 ; 3³ + 3² = 6²

with two variables a = αβ; (αβ)³ = (c-b)(c+b) . One possible distribution of the prime factors is  c-b = α² β  and c+b = αβ^{2 .} It gives c = αβ ( β + α) / 2 and b = αβ (β - α) / 2. In terms of these variables, the relation becomes  (αβ)³ + [αβ (β - α) / 2]² = [αβ (β + α) / 2]²

α = 2,β = 3;  6³ + 3² = 15²

α = 2,β = 5; 10³ + 15² = 35²