Proof of any number is equal to any different number and therefore math is pointless

a, b distinct

then t = a + b
then t(a - b) = a² - b²
then ta - tb = a² - b²
then a² - ta = b² - tb
then complete the square on each side
and a² - ta + t²/4 = b² - tb + t²/4
and (a - t/2)² = (b - t/2)²
and a - t/2 = b - t/2
and a = b
QED

QUESTION: What is the earliest reference to this 'proof'?

(a-t/2)is not equal to (b-t/2).the truth is one of the is the positive while the other is negative.

I don't know, but your proof falls down (at least) because when you square root your completed squares either a-t/2 is negative or b-t/2 is negative.
so instead of a-t/2 = b-t/2
you should have
a - t/2 = t/2 - b

Don't forget a (non-zero) number will have two square roots!

it starts going wrong when you go from

(a-t/2)²=(b-t/2)² TO (a-t/2)=(b-t/2)

ex. suppose a=2 & b=-1 ==> t=1

(a-t/2)²=(2-1/2)²=9/4=(-1-1/2)²=(b-t/2)²

But

(a-t/2)=3/2 NOT= -3/2=(b-t/2)