Harristotelian Logic

Arguing with Tim Peters gave me an idea, as in one reply he deleted out my equations to put in something with T, and that got me to thinking.

What if I used

S = (k_1*sqrt(x) + k_2*sqrt(y))*(k_3*sqrt(x) + k_4*sqrt(y))

and

T = (k_1*sqrt(x) - k_2*sqrt(y))*(k_3*sqrt(x) - k_4*sqrt(y))?

Multiply them out, subtract one from the other, and I think the correct way to get the k's might be there, but I STILL think there is a way going from just the solution to the equation with S if you actually try to get a solution versus win points in stupid newsgroup arguments.

If this idea works, then people can see how dangerous it can be when political minded people can hold so much leeway on math newsgroups, and lie freely, with no fear of consequences.

I say, if this idea works, make the people who distracted, pay.

Make them pay enough that the next time someone thinks it's fun and harmless to go after someone in a public forum, they think twice, knowing that if they're wrong, the weight of the world can fall on them.

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Well, I didn't work everything out, but just came up with this idea.

So let's do so now.

S = k_1*k_3*x + (k_2*k_3 + k_1*k_4)*sqrt(xy) + k_2*k_4*y

and

T = k_1*k_3*x - (k_2*k_3 + k_1*k_4)*sqrt(xy) + k_2*k_4*y

So

S - T = 2*(k_2*k_3 + k_1*k_4)*sqrt(xy)

and

S+T = 2*k_1*k_3*x + 2*k_2*k_4*y

and it looks like between the two of them you CAN in fact, define the k's, where it looks like you're want to choose squares for x and y, like maybe dividing off some factors of

S-T?

Then you have two equations with 4 unknowns, so you'd still have to pick two of the k's and then that's it.

So no, I don't know what you're griping about this time Silver as that looks like a route to a solution to me, which completely refutes Tim Peters and Rick Decker, as well as any other people who have been working to shoot down this idea.

It looks completely viable at this point. God help us.

So simple too. It's so easy once you see how it all works out.