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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