Fermat Theorem Puzzle

Here's an other solution. If you simply calculate the first digit and length for N=2, you will see that 2233445566 has a length of 19 and starts with 4, 7788990011 has a length of 20 and starts with 4 and 9988776655 has a length of 20 and starts with 8. If the first number had a length of 20 it could match, but since it has a length of 19 no mater the rest of the digits it can't be equal.

Here's my solution:

  x=2233445566
  y=7788990011
  z=9988776655

  6 * 6 = 36 (last digit stays the same)
  1 * 1 = 1 (last digit stays the same)
  5 * 5 = 25 (last digit stays the same)

so: x^N ends with 6 y^N ends with 1 z^N ends with 5

for just that last digit, 6 + 1 != 5.

qed

It's easy to show, using the binomial theorem, that if x < y and x^n + y^n = z^n, with x, y, z, n positive integers, then we must have n(z-y) <= x.

For the given numbers, this puts an upper limit on n of 1, and so x, y, z cannot be a Fermat counterexample.

Looking at the equation mod 3 you get 1 + 2^N = 1 mod 3 or 2^N = 0 mod 3. Not possible.

Solution: Look at the equation modulo 10.

it's a puzzle for 10yr olds