More fun with Gauss

Since $10=(2+i)(2-i)2$ , the ring $\mathbb{Z}[i]/(10)$ is isomorphic to $\mathbb{Z}[i]/(2+i)\times\mathbb{Z}[i]/(2-i)\times \mathbb{Z}[i]/(2)$ $=R_1\times R_2\times R_3=R$ . Now $R_1$ and $R_2$ , both being fields, contain 4 units each, while $R_3$ contains 2 units, the congruency classes of 1 and $i$ , for a total of $4\times 4\times 2 =32$ units in $R$ . Since the (cyclic) multiplicative groups of $R_1,R_2,R_3$ all have an order that divides 4, we have $z^4\equiv 1$ for all units, so that there are $\boxed{32}$ solutions.

If $z^4 \equiv 1 \pmod{10}$ in $\mathbb{Z}[i]$ , then $(z-1)(z+1)(z-i)(z+i)$ is divisible by $(1+i)(1-i)(2+i)(2-i) = -i(1+i)^2(2+i)(2-i)$ in the Gaussian integers, and $1+ i$ , $2 \pm i$ are all prime.

Since $2 = (1+i)(1-i)$ , we see that $1+i$ divides $z-a$ if and only if it divides $z+a$ . If we consider the issue of divisibility by $1+i$ , there are three cases to consider; we could have $(1+i)^2$ dividing $z-1$ , we could have $1+i$ dividing both $z-1$ and $z-i$ , or we could have $(1+i)^2$ dividing $z-i$ . Checking these three cases tells us that either $z \equiv 1 \pmod{2}$ or else $z \equiv i \pmod{2}$ .

Looking at divisibility by $2+i$ and $2-i$ , note that these two Gaussian integers are coprime, and each can divide one of $z-1$ , $z+1$ , $z-i$ , $z+i$ . Moreover, neither $2+i$ nor $2-i$ divide $2$ , $2i$ , or $1 \pm i$ , and hence each of $2+i$ and $2-i$ can divide exactly one of these factors. There are thus $16$ possibilities (four choices for which factor is divisible by $2+i$ , and four choices for which factor is divisible by $2-i$ ). For each of these $16$ choices, the Chinese Remainder Theorem gives us a unique solution modulo $5$ .

Thus, to be a solution of the original congruence, there are two values that $z$ can take modulo $2$ , and $16$ values that $z$ can take modulo $5$ , Using the Chinese Remainder Theorem one more time, we deduce that the original equation has $2\times16 = \boxed{32}$ distinct solutions modulo $10$ .

Let $z = a + bi$ , where $a,b \in \{0,1,2,3,4,5,6,7,8,9\}$ . Then $(a+bi)^4 = (a^4 - 6a^2 b ^2 + b^4) + 4ab(a^2-b^2)i$ . So we need to find all $z$ such that $a^4 - 6a^2 b ^2 + b^4 \equiv 1 \pmod{10}$ and $4ab(a^2-b^2) \equiv 0 \pmod{10}$ .

The second equation simplifies to $ab(a+b)(a-b) \equiv 0 \pmod{5}$ . Thus one of $a, b, a+b, a-b$ must be a multiple of $5$ .

Case 1: a = 0 or b = 0

If $a = 0$ then we need $b^4 \equiv 1 \pmod{10}$ . This has 4 solutions: $1i, 3i, 7i, 9i$ . Similarly there are 4 solutions when $b = 0:$ $1, 3, 7, 9$ .

Case 2: a = 5 or b = 5

If $a = 5$ then the first equation becomes $625 - 150b^2 + b^4 \equiv 1 \pmod{10}$ . Thus $b^4 \equiv 6 \pmod{10}$ . This has 4 solutions: $5+2i, 5+4i, 5+6i, 5+8i$ . Similarly there are $4$ solutions when $b = 5:$ $2+5i, 4+5i, 6+5i, 8+5i$ .

Case 3: a+b = 5

$a^4 - 6a^2 b ^2 + b^4 = a^4 - 6a^2(5-a)^2 + (5-a)^4 = -4a^4 + 40a^3 - 500a + 625 \equiv 1 \pmod{10}$ $-4a^4 + 624 \equiv 0 \pmod{10}$ $a^4 \equiv 156 \equiv 1 \pmod{5}$ Since $a,b \le 5$ , there are $4$ solutions: $1+4i,2+3i,3+2i,4+i$ .

Case 4: a + b = 10

$a^4 - 6a^2 b^2 + b^4 \equiv a^4 - 6a^2 (-a)^2 + (-a)^4 = -4a^4 \equiv 1 \pmod{10}$

This case clearly has no solutions.

Case 5: a + b = 15

This is similar to case $3$ , and has $4$ solutions: $6+9i, 7+8i, 8+7i, 9+6i$ .

Case 5: a = b

$a^4 - 6a^2 b^2 + b^4 = -4a^4 \equiv 1 \pmod{10}$

This case has no solutions.

Case 6: a - b = 5

$a^4 - 6a^2 b^2 + b^4 = a^4 - 6a^2 (a+5)^2 + (a+5)^4 = -4a^4 - 40a^3 + 500a + 625 \equiv 1 \pmod{10}$ $a^4 \equiv 1 \pmod{5}$ This gives us $4$ solutions: $6+i, 7+2i, 8+3i, 9+4i$ .

Case 7: b - a = 5

This is similar to case 6, and has $4$ solutions: $1+6i, 2+7i, 3+8i, 4+9i$ .

This covers all the cases and there are no repeats. Therefore the total number of solutions is $32$ .