Wednesday, December 30, 2009
If x, y, z and n are integers, there are no solutions of x^n+y^n=z^n with n>2 and x,y,z>0.This is known as Fermat's last theorem.
This problem can be an extension of Fermat's Infinite Descent
Find all solutions to c2 + 1 = (a2 − 1)(b2 − 1), in integers a, b, and c.
Solution Scheme and approach
Clearly a = b = c = 0 is one integer solution.
Also, if c = 0, a^2 − 1 = ±1, and so a = 0; similarly b = 0.
Without loss of generality, we now seek a solution with c > 0.
Assume such a solution exists.
Consider the equation, modulo 4.
Since the square of any integer is either 4k or 4k+1
we have, in mod 4, (a^2 − 1) and (b^2 − 1) = −1 or 0(mod 4),So (a^2-1)(b^2-1)=0 or 1 and (c^2 + 1) =1 or 2(mod 4).
As, L.H.S cant be 2, Hence (a^2 − 1) = (b^2 − 1) = −1 (mod 4), and (c^2 + 1) = 1 (mod 4); that is, a, b, c are even.
Set a = 2x, b = 2y, c = 2z.
Then 4z^2 + 1 = (4x^2 − 1)(4y^2 − 1).
Simplifying, we have z^2 = 4x^2y^2 − x^2 − y^2.
Again considering mod 4, z^2=1 or 0(mod 4) and x^2=y^2=0,1(mod 4) So, R.H.S. can be -2,-1 or 0. Considering R.H.S we can decide z,x,y all are multiple of 4 i.e. even.
Now set z=2s, x=2r, y=2t. Then 16s^2+1=(16r^2-1)(16t^2-1) =>s^2=-16r^2t^2-r^2-t^2. Following the previous modulo 4 method again we can conclude that s,r,t all are even.
And this series continues till infinity. But as c>z>s>...>0 that means c is bounded. Which contradicts. Hence our assumption was wrong. There is no solution exists if c>0.
Therefore the only integer solution to c2 + 1 = (a2 − 1)(b2 − 1) is a = b = c = 0.Q.E.D.