## Counterexample to Fermat's Last Theorem

Fermat's Last Theorem states that there is no solution to a^{n} +
b^{n} = c^{n} for a,b,c,n all integers and n>2.
We will give a counterexample to this.

First we will discuss how to approximate the sum and product of two large
integers if we're only given the n right-most digits of each. We will use the
ellipses, ..., to the left of an integer to indicate that there are more digits to
the left that aren't specified.

To add ...5972286 + ...9421769, simply follow the normal right-to-left algorithm
for adding numbers together.

1 111
...5972286
+...9421769
__________
...5394055

Clearly you can see that if we're given the n right-most digits of two integers,
the sum of these digits matches the sum of the actual integers.

Multiplication is a little more involved, but leads to the same general
conclusion.

...5972286
x...9421769
__________
...3750574
...5833716
...1806002
...5972286
...1944572
...3889144
...3750574
...
_________________
...9093934

We can see that when we multiply the n right-most digits of an integer against a
digit, the product agrees with the product of the actual integer and the digit to
at least n digits. When we add these numbers up, the n right-most digits of this
sum will agree with the n right-most digits of the product of the actual integers.

Thus if you have the n right-most digits of an integer, you can
square this approximation and the n right-most digits of the product will agree with
the n right-most digits of the actual square. The same result applies if you cube
this number.

So let's try cubing a large integer by using just the right-most digits.

...51543^{3} = ...00007

...2217051543^{3} = ...0000000007

...58470473184453160041581408885453706924422217051543^{3}

= ...00000000000000000000000000000000000000000000000007

No matter how many digits you want of this integer, let's call it s, you can see
that the cube of its right-most digits will always give you 7.
So 1^{3} + s^{3} = 2^{3}.

s is called a leftin.

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© Scott Sherman 1999