1. ## A^2=B^2, C^2=D^2......does (A+C)^2=(B+D)^2?

Given
A^2=B^2, C^2=D^2

......does (A+C)^2=(B+D)^2?

I got down to this

AC=BD

I then divived both sides by "A"

C=BD/A

My proof goes as following....

It is possible that B=A.....eg A=6....B=6.....A^2=B^2.......36=36

So if this is possible. this equation breaks down to

C=D.... However it is clear that C is not always equal to D

Can anyone recommend a better proof?

Thankyou

2. What about A=B=1, C=1, D=-1?

3. extra note:

i ended up with

AC=BD

+-A/+-B=+-D/C
+-1=+-1

Which is not always TRUE.... all i have to do is show one case where it is not true.....

Defunkt. For A=B=1,C=-1,D=1

it becomes

1=1/-1....which is false. Do you think this proof is ok?

All im proving is that it doesnt work for all numbers of A and B... shouldnt all
i have to show is one FALSE? thanks for the input

4. Originally Posted by sebko
extra note:

i ended up with

AC=BD

+-A/+-B=+-D/C
+-1=+-1

Which is not always TRUE.... all i have to do is show one case where it is not true.....

Defunkt. For A=B=1,C=-1,D=1

it becomes

1=1/-1....which is false. Do you think this proof is ok?

All im proving is that it doesnt work for all numbers of A and B... shouldnt all
i have to show is one FALSE? thanks for the input
That is correct! (It is false because $A^2 = B^2 = 1 \ , \ C^2 = D^2 = 1$ but $(A+C)^2 = 2^2 = 4 \neq 0 = (1 + (-1))^2 = (B + D)^2$)