Published on 29th March 2018, City: Ahmedabad, Country : India.

\(A^2+B^2+C^2 = D^2\) in Ambalal Cuboid Theory 1.1.

\(A = a\), where \(a\) is any member of \([ N,Z,Q,R ]\)
Put \(B = a+1\), means \(B = A+1\)
\(C = a^2+a\), means \(C = A*B\)
\(D = a^2+a+1\) means \(D=C+1\)

Results:
\(A^2+B^2+C^2=D^2\) becomes true for any value of a where a is any member of \([ N,Z,Q,R ]\)

Proof:
\(D^2-C^2 = (a^2+a+1 )^2 – (a^2+a)^2\)
\(= (1)( 2a^2+2a+1)\)
\(= a^2+a^2+2a+1)\)
\(= a^2+(a+1)^2\)
\(= A^2+B^2\)
∴ \(A^2+B^2+C^2=D^2\)

For Example 1:
\(A= , B= a+1 , C= A*B, D=C+1\)
Put \(a = 5\) in \(A^2+B^2+C^2=D^2\)
Then \((5)^2 + (6)^2 + (30)^2 = (31)^2\)
∴ \(25 + 36 + 900 = 961\)

Example 2:
Put \( a = -7\)
Then \((-7)^2 + (-6)^2 + (42)^2 = (43)^2\)
∴ \(49 + 36 + 1764 = 1849\)

Example 3:
Put \(a =1.7\)
Then \((1.7)^2 + ( 2.7)^2 + (4.59)^2 = (5.59)^2\)
\(2.89 + 7.29 + 21.0681 = 31.2481\)

Example 4:
Put $$ a = { 3 \over 5 }$$
Then $$ ({3 \over 5})^2 + ({8 \over 5})^2 + ({24 \over 25})^2 = ({49 \over 25})^2$$
∴$$ {9 \over 25} + {64 \over 25} + {576 \over 625} = {2401 \over 625}$$
∴ $$ {9*25 \over 25*25} + {64*25 \over 25*25 } + {576 \over 25*25} = {2401 \over 25*25}$$
∴$$ {225 \over 625} + {1600 \over 625 } + {576 \over 625 } = {2401 \over 625}$$

Now for your satisfaction, take A= any value \(a\), then take \(B = a+1 , C = A*B, D = C+1\)
and put in formula \(A^2+B^2+C^2 = D^2\) to get successful result.

For any doubt and additional details please contact me via Email : ambalalparmar40@yahoo.com

---