$2 + 2 = 5$ (Fake proof, or ?)

Question

Basically, there is no error in the following steps(as it seems), but there is some error due to which 2 + 2 = 5. What is it?

                      -20 = -20
                    16-36 = 25-45
             16-36+(81/4) = 25-45+(81/4)
(4^2)-(2*4*9/2)+((9/2)^2) = (5^2)-(2*5*(9/2))+((9/2)^2)
              (4-(9/2))^2 = (5-(9/2))^2
                  4-(9/2) = 5-(9/2)
                        4 = 5
                      2+2 = 5

Answer 1

Hint -

$a^2-2ab+b^2=(a-b)^2\ne(a+b)^2$

This is your mistake.

Also $x^2=y^2$ doesn't imply that $x=y$ is the only possibility.

It should be $|x|=|y|$.

Answer 2

$$4-\frac92=\pm\left(5-\frac92\right)$$ so that

$$4=5\lor4+5=\frac92+\frac92$$ and the world is safe...

Answer 3

The problem is when you go from a quadratic to a linear.

$$(4-\frac{9}{2})^2 = (5-\frac{9}{2})^2$$ $$|4-\frac{9}{2}| = |5-\frac{9}{2}|$$ $$|\frac{-1}{2}| = |\frac{1}{2}|$$ $$\frac{1}{2} = \frac{1}{2}$$

Answer 4

Hint :

$$a^2=b^2 \implies |a|=|b|$$

That is :

$$a=b ~~\text{or}~~ a=-b$$

Therefore $$\Big(4-\frac{9}{2}\Big)^2=\Big(5-\frac{9}{2}\Big)^2$$ $$\implies \Big|\Big(4-\frac{9}{2}\Big)\Big|=\Big| \Big(5-\frac{9}{2}\Big) \Big|$$ $$\implies 0.5=0.5$$ Which is absolutely correct.