Divisible or not by 7?

5555^2222 + 2222^5555 = (5555^2222 - 4^2222) + (2222^5555 + 4^5555) - (4^5555 - 4^2222)

The first "term", (5555^2222 - 4^2222), is divisible by 5555 - 4 = 5551 = 7 * 793, since a^n - b^n is always divisible by (a - b).

The second "term", (2222^5555 + 4^5555), is divisible by 2222 + 4 = 2226 = 7 * 318, since a^n + b^n is divisible by (a+b) if n is odd.

And the third term may be written as 4^2222 * (4^3333 - 1) = 4^2222 * (64^1111 - 1), which is clearly divisible by (64 - 1) = 63, and hence by 7 .

Let $N=5555^{2222} + 2222^{5555}$ . Note that $5555 \bmod 7 = 4$ and $2222 \bmod 7 = 3$ .

$\begin{aligned} N & \left(\equiv (5551+4)^{2222} + (2219+3)^{5555}\right) \text{ (mod 7)} \\ & \equiv \left(4^{2222} + 3^{5555}\right) \text{ (mod 7)} \\ & \equiv \left(4^{3\times 740+2} + 3^{3\times 1851+2}\right) \text{ (mod 7)} \\ & \equiv \left(64^{740}\times 4^2 + 27^{1851}\times 3^2\right) \text{ (mod 7)} \\ & \equiv \left((63+1)^{740}\times 16 + (28-1)^{1851}\times 9\right) \text{ (mod 7)} \\ & \equiv \left(1^{740}\times 2 + (-1)^{1851}\times 2\right) \text{ (mod 7)} \\ & \equiv 2-2 \equiv \boxed{0} \text{ (mod 7)} \end{aligned}$

Yes , $N$ is divisible by 7.