MATHS!!!!

all my statements are fixed now, Thanks! xD

I remember for dividing fractions the anagram KFC which is Keep Flip Change

Would it be x^3

Diophantine equations are equations in multiple unknowns, where the only solutions that are of interest are integer solutions (fractional, decimal or irrational solutions are ignored).

Here's an example:

Suppose you were asked to find two numbers, X and Y, such that they add to give you 20 and whose squares will add to 208.

Since we are dealing only with integer solutions, it's pretty easy to just guess the answer... there are only a few possibilities to test. But for a harder problem, can you find a specific PROCEDURE that will solve ANY such Diophantine problem, without any need for guessing?

There's an interesting and rather elegant way to do it:

First set up the two equations:

a) X + Y = 20

b) X^2 + Y^2 = 208

Now let's shift perspective. Any pair of numbers X and Y can be represented in several alternative ways. One way would be to represent them as an AVERAGE number N (exactly halfway between X and Y) and an offset Z (which is how far both X and Y are from the average, one higher, one lower).

For instance, the numbers 100 and 200 could be represented as N = 150 and Z = 50, since the number 100 is 50 less than 150 while the number 200 is 50 more than 150.

In our Diophantine problem given above, this might seem like a pointless complication, representing X as N + Z and representing Y as N - Z, but in fact it simplifies the simultaneous equation that we're going to build.

Now go back to equation a) above, but substitute N + Z for X and N - Z for Y:

a) X + Y = 20

c) (N + Z) + (N - Z) = 20

d) 2N = 20

e) N = 10

So the two numbers we are looking for (by which I mean, X and Y, not N and Z) are symmetrical around the average number 10. One will be Z units greater, the other will be Z units smaller.

Now do the same with equation b) above:

b) X^2 + Y^2 = 208

f) (N + Z)^2 + (N - Z)^2 = 208

But we've already determined (in equation e) that N = 10, so:

g) (10 + Z)^2 + (10 - Z)^2 = 208

h) 100 + 20Z +Z^2 + 100 - 20Z + Z^2 = 208

i) 2Z^2 +200 = 208

j) 2Z^2 = 8

k) Z^2 = 4

So Z equals plus and/or minus 2.

And since X = N + Z, X must be 12, while Y = N - Z must be 8.

Let's check, by substitution into our original two equations:

l) X + Y = 12 + 8 = 20

m) X^2 + Y^2 = 8^2 + 12^2 = 64 + 144 = 208.

The problem is solved, without any guessing... and this same method can be used to solve ANY similar Diophantine problem.

The way I tried to solve it was that I looked at it as d/d*x *x^2 Then I did d/d=1 so 1*x=x So then it is x*x^2 And finally x*x=x^2 and therefore the answer is x^3 I hope this is correct

But then Socks gets none!