A lot of you might have made this mistake
before: (a+b)^2 = a^2+b^2, or in general,
(a+b)^m=a^m+b^m. This is famously coined Freshmen's
dream.
Now, we of course know that this is wrong
in general, but it isn't completely wrong.
In some cases, it could be correct. One very
easy example is where a or b is 0,
then both sides are a^m or b^m, so the equality
obviously holds. Mathematicians like to call
those trivial, or degenerate cases, because
they are not interesting.
The more interesting case involves modular
arithmetic. Technically, we don't have equality,
but a congruence relation. However,
since the congruence works pretty much like
equality in modular arithmetic, we count it
as a special case of the Freshman's dream.
There is another video, where I explained
many elementary modular arithmetic concepts.
It will be shown in the card above. Here is
a crash course: if you see a congruent to
b modulo m, that means a-b is divisible by
m. The congruence is in the sense that when
a and b are divided by m, their remainders
are the same.
We now consider a positive integer a, a prime
number p, which is coprime with a, and this
list of multiples of a. Consider the remainders
when these multiples of a are divided by p.
It could only be 0 to p-1, but 0 is not possible
here. Why? If, say, i times a is leaves remainder
0 when divided by p, that means p is a factor
of i times a. a does not contribute the factor
p, as they are coprime, and neither does i,
because it is less than p. So, 0 is not a
possible remainder here. In addition, no two
numbers on the list should leave the same
remainder when divided by p. Let’s say i
times a and j times a are the two numbers.
Then (j-i) times a is divisible by p, which
contradicts what we discussed a moment ago.
So every number on this list has a unique
remainder from 1 to p-1 when divided by p.
So now, if we consider the product of these
numbers, their remainders should also multiply,
so we have (p-1)!*a^(p-1) (wait) congruent
to (wait) (p-1)! modulo p. Since (p-1)! is
coprime with p, we can cancel them from both
sides (wait) This end result is called Fermat’s
little theorem. In this result, we can multiply
both sides by a, to get a congruence relation
that is true even if a and p are not coprime.
So (a+b)^p is congruent to a+b modulo p, and
a^p + b^p is also congruent to a+b modulo
p, so (a+b)^p is congruent to a^p+b^p modulo
p, which is the correct version of the Freshmen’s
dream. This mistake committed by a lot of
people could in some sense be right.
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on. See you in the next video. Bye!
