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]