Analysis using Diopphantine Equations

Diophantine Equations are used to remove as many variables as possible and write the remaining unknowns in terms of the other unknowns. By analyzing the remaining terms as whole numbers, we can decided whether there are infinite number of solutions or zero solutions.

Because we have the given equation X^n + Y^n = Z^n.

From this equation we can derive to associated equations

1) X + Y = c/d * Z where c/d > 1. c,d are rational also c/d < 2^1/2

2) X^2 + Y^2 = e/f * Z^2 where e/f > 1 and c/d > e/f e,f are rational


from 1) we get d*X + d*Y = C*Z

d*X = c*Z - d*Y

or finally X = c*Z/d - Y


If we substitute value for X into 2) we get

( c^2/d^2 * Z^2 - 2c/d * Z * Y + Y^2) + Y^2 = e/f * Z

rearranging

c^2/d^2 * Z^2 - 2c/d * Z * Y - e/f*Z = -2Y^2

or

Z *(c^2/d^2 * Z - 2c/d * Y - e/f) = -2Y^2

Now let Z = h *Y where 2 > h > 1 and substitute into equation

h * Y *((c^2/d^2 * h *Y) - 2c/d * Y - e/f) = -2Y^2

factor Y out of both sides of equation


and you get


h *((c^2/d^2 * h *Y) - 2c/d * Y - e/f) = -2Y

Now move -2Y to left hand side and extend h across members of brackets

c^2/d^2 * h^2 *Y - 2c/d * h * Y + 2Y - e/f * h = 0

move e/f * h to right hand side. Now left hand side has the Y come out of all terms on left hand side


(c^2/d^2 * h^2 - 2c/d * h + 2) * Y = e/f * h

or finally

Y = (e/f * h)/(c^2/d^2 * h^2 - 2 * c/d * h + 2)

We do not know what the exact value of y is, but we can determine the range

when Y is close to Z, e/f approximates 1, h approximates 1, c^2/d^2 approximates 1, but is always less than 1

In that case

Y is approximately (1 * 1)/((1 * 1^2) - 2 * 1 * 1 + 2) = 1/1 = 1, but less than 1




When Y is close to X, e/f approximates 1.3, h approximates 2, c^2/d^2 approximates 1.4

Then
Y is approximately (1.3 * 2)/(1.4^2 * 2^2 - 2 * 1.4 * 2 + 2)= 2.6/7.84 - 5.6 + 2

or Y is approximately 2.6/4.24 or approximately 0.6

These approximations could be made more defined, but the real issue has already been decided.

The range of values Y takes on is between 0.6 and 1. There is no way Y could be equal to 2, even if the numbers are refined. It should be noted that Y is the larger of the supposed values X,Y. This only leaves an integer value of less than 0 for X, which is not allowed.

This completes the proof for Fermat's Last Theorem using Diophantine Equations.