The problem presented in the image involves proving a modular arithmetic property.

Problem Statement:

Let $n$ be a natural number, and let $a, b, c,$ and $d$ be integers. We are given two modular congruences:

• $a \equiv b \pmod{n}$
• $c \equiv d \pmod{n}$

We are asked to prove that: $(a + c) \equiv (b + d) \pmod{n}$

Proof:

Step 1: Understanding the modular equivalences

From $a \equiv b \pmod{n}$, we know that $a - b$ is divisible by $n$, which means: $a = b + kn \quad \text{for some integer} \, k$ Similarly, from $c \equiv d \pmod{n}$, we know that $c - d$ is divisible by $n$, which means: $c = d + ln \quad \text{for some integer} \, l$

Step 2: Add the two equations

Now, adding these two congruences: $a + c = (b + kn) + (d + ln)$ Simplifying this: $a + c = (b + d) + (k + l)n$

Step 3: Conclusion

From the equation $a + c = (b + d) + (k + l)n$, we see that $(a + c) - (b + d) = (k + l)n$, which means that $a + c \equiv b + d \pmod{n}$.

Thus, we have proven the required result: $(a + c) \equiv (b + d) \pmod{n}$

Do you have any questions or would you like more details on any of the steps? Here are a few related questions for further exploration:

1. How does modular arithmetic apply to cryptography?
2. Can you prove the result for multiplication: $ac \equiv bd \pmod{n}$?
3. What happens if $n$ is not a natural number but an integer?
4. What are the real-world applications of modular arithmetic?
5. Can we extend this property to more than two numbers (i.e., summing multiple congruences)?

Tip: When working with modular arithmetic, always focus on the idea that congruent numbers differ by a multiple of the modulus.