[mathwiz3502]

Hello, I just created a pair of equations that I believe have no counterexamples.
Do these equations new and do they have no counterexamples?
a+b=2c
ab=c^2-(a*-c)
*or b

[mathwiz3502]

Sorry, (a*-c) should be squared.

[elscarta]

a+b=2c -> c = 0.5a +0.5b

c^2 - (a-c)^2
= (0.5a +0.5b)^2 -(a-(0.5a +0.5b))^2
=(0.5a +0.5b)^2 -(0.5a -0.5b)^2
=((0.5a +0.5b)+(0.5a -0.5b))((0.5a +0.5b)-(0.5a -0.5b)) (difference of squares)
=ab

also

c^2 - (b-c)^2
= (0.5a +0.5b)^2 -(b-(0.5a +0.5b))^2
=(0.5a +0.5b)^2 -(0.5b -0.5a)^2
=((0.5a +0.5b)+(0.5b -0.5a))((0.5a +0.5b)-(0.5b -0.5a)) (difference of squares)
=ba
=ab

therefore no counterexamples.

I have seen these somewhere before, in a maths text book I think.

[mathwiz3502]

Another way is polynomials.
let's say a>b (but it really doesn't matter).
So, the mean of a and b is c.
so, let's have only 2 variables, let a-d=c and a-2d=c
So,
a(a-2d)=(a-d)^2-(a-(a-d))^2
simplify
a^2-2ad=(a-d)^2-(a-(a-d))^2
a^2-2ad=a^2-2ad+d^2-(a-(a-d))^2
a^2-2ad=a^2-2ad+d^2-(-d)^2
0=d^2-(-d)^2
0=0, but that was a little complicated.