Modular arithmetic How do I prove the following inequality with modular arithmetic? (No use of Fermat's last theorem is allowed.)
$$3987^{12} + 4365^{12} \neq 4472^{12}$$
 A: Fermat's last theorem is the most obvious way.
A more elementary proof could use Fermat's little theorem:
$$(3987)^{12} \equiv (4365)^{12} \equiv 1 \; \bmod 13,$$ but $(4472)^{12} \equiv 0 \; \bmod 13$ because $13$ divides $4472$.
A: Note that $3|3987,3|4365,3\not| 4472$ so we have $3^{12}a^{12}+3^{12}b^{12}=4472^{12}\implies 3|4472,\text{ and }3\not| 4472\implies 3987^{12}+4365^{12}\ne 4472^{12}.$
A: My first intuition is to use modular arithmetic. 
Note that 3986 = 4000 - 14 and so 3987 = 3 mod 4.  Now
$$(3987)^2 \space mod \space 4 = 9 \space mod \space 4 = 1$$
$$\text{and so }3987^{12} = 1 \space mod \space 4$$
Now, we also use the fact that
$$4365 \space mod \space 4 = 1$$
$$\text{and so }4365^12 = 1 \space mod \space 4$$
So, one side is 2 mod 4, but the other side is a square, since
$$4472^{12} = (4472^6)^2$$
Now, if 4472 is 2 mod 4, then
$$4472^2 = 0 \space mod \space 4$$
$$\text{and so }4472^{12} = 0 \space mod \space 4$$
Hence, this proves that the sides are not equivalent mod 4 and we are done. 
A: $3987$ and $4365$ are both divisible by 3, and $4472$ is not.
So the left hand side is divisible by 3, and the right hand side is not.
Thus the two can't be equal.
