# Modular Arithmetic: Solve (a + b)^5 in Z_5

• Coto
In summary, the task is to compute (a + b)^5 in Z_5 (Z mod 5). The end result is (a^5 + b^5), and to solve this, one can exploit the properties of arithmetic in Z_n. However, simply expanding (a+b)^5 may not work, as shown in the example of (a+b)^2. It is important to note that in general, (a+b)^p = a^p+b^p mod any prime p.
Coto

## Homework Statement

Compute:
$$(a + b)^5$$
in Z_5 (Z mod 5).

## Homework Equations

The end result is apparently:
$$(a^5 + b^5)$$

Intuition would tell me to exploit the properties of arithmetic in Z_n, however I don't see how I can reconcile this solution with just a normal expansion of (a+b)^5 (which seems to be allowed due to the def. of arithmetic for these classes.)

## The Attempt at a Solution

Well, my best guess would be to show that the middle terms go to zero no matter what. But trying this with for example (a+b)^2 (which has the analagous sol'n) would mean that ab + ba = ab + ab = 0 (mod 5). However, choosing for example a = [2], and b = [2], shows that this would actually be [4]+[4] = [8] = [3], which clearly is not 0... so I'm at a loss.

or.. I could have read this question wrong.. or perhaps my world view of modular arithmetic is one big illusion. ugh.

The 5 up there is important. (a+b)^2 doesn't need to be equal to a^2+b^2 mod 5. However, it is equal to that mod 2. In general, (a+b)^p = a^p+b^p mod any prime p.

Gotcha. Thanks for the answer, it makes sense now.

## 1. What is modular arithmetic?

Modular arithmetic is a branch of mathematics that deals with integers and their remainders when divided by a fixed number, called the modulus. It is often used in fields such as cryptography, computer science, and number theory.

## 2. What does Z_5 mean in this context?

Z_5 refers to the set of integers modulo 5, also known as the integers modulo 5 ring. This means that the numbers in this set range from 0 to 4, and any arithmetic operations performed on them are done modulo 5.

## 3. How do you solve (a + b)^5 in Z_5?

To solve (a + b)^5 in Z_5, we can expand the expression using the binomial theorem and then reduce each term modulo 5. This means that we replace any value greater than 4 with its remainder when divided by 5. Finally, we add up all the terms to get the final answer in Z_5.

## 4. Can (a + b)^5 have different solutions in Z_5?

No, in Z_5 there is only one solution for (a + b)^5. This is because the ring of integers modulo a prime number, such as 5, is a field, which means that every non-zero element has a unique multiplicative inverse. This guarantees that there is only one solution for (a + b)^5 in Z_5.

## 5. What are some real-world applications of modular arithmetic?

Modular arithmetic has many real-world applications, including computer science (in coding and error correction), cryptography (in creating secure algorithms and codes), and number theory (in studying patterns and properties of numbers). It is also used in fields such as finance, music theory, and art.

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