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

In Ambalal Cuboid Theory - `1b`

If `A=1`, `B=b` where `b` is any member of `[N,Z,Q,R]`

Put `A=1` given

`B=b` given

Then `C=b^2/2`

& `D=b^2/2+1` mean `C+1=D`

Results `A^2+B^2+C^2=D` becomes true for any value

Of `b` where `b` is any member of `[N,Z,Q,R]`

Proof `A^2+B^2+C^2= [1]^2+[b]^2+[b^2/2]^2`

`=1+b^2+b^4/4`

`=(4+4b^2+b^4)/4`

`=((b^2+2)/2)^2`

`=(b^2/2+1)^2`

`∴ A^2+B^2+C^2=D^2`

Example 1:
Let `A=1, b=6, C=(6)^2/2=18, D=19`

Then `A^2+B^2+C^2+D^2`

`∴ (1)^2+(6)^2+(18)^2=(19)^2`

`∴ 1+36+324=361`

Example 2:
`A=1, b=-9, C=(-9)^2/2=40.5,D=41.5`

Then `A^2+B^2+C^2=1^2`

`∴ (1)^2+(-9)^2+(40.5)^2=(41.5)^2`

`1+81+1640.25=1722.25` > `A^2+B^2+C^2=D^2`
Example 3:

Let `A=1, b=2.3, C=(2.3)^2/2=5.29/2=2.645, D=3.645`

Then `A^2+B^2+C^2=D^2`

`∴ (1)^2+(2.3)^2+(2.645)^2=(3.645)^2`

`∴ 1+5.29+6.996025=13.286025`

Example 4:

Let `A=1,b=2 1/3 = 7/3, C=(7/3)^2/2=49/18,D=1+49/18=67/18`


Then `A^2=B^2+C^2=d^2`

`(1)^2+(7/3)^2+(49/18)^2=(67/18)^2`


`(1*324)/324+49/9*36/36+2401/324=4489/324`

`324/324+1764/324+2401/324=4489/324`



Now for your satisfaction, take `A=1, b=b`

where `b` is any member of `[N,Z,Q,R]`

`c=b^2/2`& `D=b^2/2+1=C+1`.


And put in formula `A^2+B^2+C^2=D^2` to get

a successful result.

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

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