To prove $n \equiv 1 \pmod{4}$ if and only if $n \equiv 1 \pmod{8}$ or $n \equiv 5 \pmod{8}$, we need to show both directions:

Forward Direction: $n \equiv 1 \pmod{4} \Rightarrow (n \equiv 1 \pmod{8} \text{ or } n \equiv 5 \pmod{8})$

Assume $n \equiv 1 \pmod{4}$. This means that $n = 4k + 1$ for some integer $k$.

Now, we need to consider this modulo 8:

$n = 4k + 1$

Case 1: $k$ is even

Let $k = 2m$ for some integer $m$. $n = 4(2m) + 1 = 8m + 1$ Thus, $n \equiv 1 \pmod{8}$.

Case 2: $k$ is odd

Let $k = 2m + 1$ for some integer $m$. $n = 4(2m + 1) + 1 = 8m + 4 + 1 = 8m + 5$ Thus, $n \equiv 5 \pmod{8}$.

So, if $n \equiv 1 \pmod{4}$, then $n \equiv 1 \pmod{8}$ or $n \equiv 5 \pmod{8}$.

Reverse Direction: $(n \equiv 1 \pmod{8} \text{ or } n \equiv 5 \pmod{8}) \Rightarrow n \equiv 1 \pmod{4}$

We need to show this in two cases:

Case 1: $n \equiv 1 \pmod{8}$

This means $n = 8m + 1$ for some integer $m$.

Now, consider this modulo 4: $n = 8m + 1$ $8m + 1 \equiv 0 + 1 \pmod{4} \Rightarrow n \equiv 1 \pmod{4}$

Case 2: $n \equiv 5 \pmod{8}$

This means $n = 8m + 5$ for some integer $m$.

Now, consider this modulo 4: $n = 8m + 5$ $8m + 5 \equiv 0 + 5 \pmod{4} \Rightarrow 5 \equiv 1 \pmod{4} \Rightarrow n \equiv 1 \pmod{4}$

Thus, if $n \equiv 1 \pmod{8}$ or $n \equiv 5 \pmod{8}$, then $n \equiv 1 \pmod{4}$.

Conclusion

We have shown both directions of the if and only if statement: $n \equiv 1 \pmod{4} \iff n \equiv 1 \pmod{8} \text{ or } n \equiv 5 \pmod{8}$

Do you want details or have any questions? Here are some questions you might want to ask:

1. How do I solve other modular arithmetic problems?
2. Can you explain more examples of modular equivalence?
3. How does this relate to solving congruences in general?
4. What are some applications of modular arithmetic in computer science?
5. How do we handle modular arithmetic with negative numbers?

Tip: When dealing with modular arithmetic, breaking down the problem into smaller congruences can simplify your calculations.