Solve this equation modulo 16 : x^8 - x^4 = 0 mod 16

0 obviously works: 0^8 = 0 and 0^4 = 0.

Since 2^4 = 16, every even number raised to either the 4th or 8th power = 0 (mod 16),
since it will have 4 factors of 2.

So for starters, all the even numbers (0, 2, 4, 6, 8, 10, 12, 14) are solutions.

For the odd numbers,
odd numbers are of the form 2k + 1
(2k + 1)^4 = 16 k^4 + 32 k^3 + 24 k^2 + 8k + 1

16k^4 + 32 k^3 + 16 k^2 = 0 (mod 16),
leaving
8 k^2 + 8k + 1

For x = ?, we have k = ...
x: 1 3 5 7 9 11 13 15
k: 0 1 2 3 4 . 5 . 6 . 7

If k = 0, 2, 4, 6, then 8k^2 + 8k = 0 (mod 16)
and x^8 - x^4 = 1 - 1 = 0 (mod 16)
So we can add x = 1, 5, 9, 13 to the solutions

For k = 1 we have
8k^2 + 8k + 1 = 16 + 1 (mod 16) and we can add that one, too.
k = 3, we have
72 + 24 + 1 = 96 + 1 = 1 (mod 16) ... another one
k = 5, we have
8 * 25 + 8 * 5 + 1 = 241 = 1 (mod 16) ... another one
k = 7, we have
8 * 49 + 8 * 7 + 1 = 449 = 1 (mod 16) ... and that's all of them

All integers x satisfy x^8 - x^4 = 0 (mod 16)

There is (no doubt) a shorter proof.
probably based on x^8 - x^4 = x^4 ( x^4 - 1)

if x=2k (x even)
(2k)^8 - (2k)^4 = 0 mod 16
256k^8-16k^4=0 mod 16
16(16k^8-k^4)=0 mod 16 which is true
So the equation is valid if x is even.

Lets see the requirements for x odd.
x=2k+1
x^8 - x^4 = 0 mod 16
(x^4)(x^4-1)=0mod 16
[(2k+1)^4][(2k+1)^4-1]=0 mod16 since gcd(16,any odd number)=1
our equation becomes

(2k+1)^4-1=0 mod16 expanding
16k^4+32k^3+24k^2+8k+1-1=0 mod 16
16k^4+16*2k^3+24k^2+8k=0 mod 16

24k^2+8k=0 mod 16
3k^2+k=0 mod 2
k(3k+1)=0 mod 2
if k is odd, 3k+1 is even so k(3k+1) will be even making it 0 in mod2
if k is even, k(3k+1) will be even making it 0 in mod2

so (x^4)(x^4-1)=0 mod 16 for any odd number
The equation is valid for any integer

Mod 16 solutions:
x = 0 - 15
solution:
x^8 - x^4 = 0
x^4 (x^4-1) = 0
x^4 (x^2 + 1) (x-1)(x+1) = 0
real roots:
x = - 1, 0 , 1
complex roots:
x = -i , i