# modular calculus, find all solutions in ℤ

#### _rob

##### New member
How do I go about solving something like this:

Find all the solutions in ℤ / 80ℤ of the equation
4x + 5 ≡ 1 mod 80

Now obviously x must be -1 for the equation to hold true, but I don't know what to do about the "Find all the solutions in ℤ / 80ℤ" part.

Another similar one:
Find all the solutions in ℤ / 28ℤ of the equation.
4x – 24 = 0

Here x must obviously equal 6, but again I don't know what to do about "Find all the solutions in ℤ / 28ℤ".

#### ksdhart2

##### Senior Member
It seems like it would be wise for you to brush up on the idea of what modular arithmetic is and what it really means. It's true that x = -1 would be the only solution to the equation $$4x + 5 = 1$$, and likewise x = 6 would be the only solution to the equation $$4x - 24 = 0$$ if you were working under the "normal" rules of arithmetic, but that's not what's being asked here.

The notation $$\displaystyle \frac{\mathbb{Z}}{N\mathbb{Z}}$$ means you're working in a ring where you only have access to the integers $$0, 1, 2, \cdots, N-1$$ and when you would reach $$N$$, you "roll over" and begin again at 0. So, if you were doing some workings and you found an answer of 27 but you were working in $$\displaystyle \frac{\mathbb{Z}}{8\mathbb{Z}}$$, you'd have to find what element 27 is congruent to... 27 - 8 = 19, 19 - 8 = 11, 11 - 8 = 3, so we say that $$27 \equiv 3 \: (\text{mod} \: 8)$$

As a full example of this concept, consider finding all solutions to $$3x + 2 = 2$$ in $$\displaystyle \frac{\mathbb{Z}}{6\mathbb{Z}}$$. As a visual means of further demonstrating the idea, let's manually substitute in each value $$0, 1, 2, \cdots, 5$$, and see what we get:

$$\displaystyle 3(0) + 2 \equiv 2 \: (\text{mod} \: 6)$$
$$\displaystyle 3(1) + 2 \equiv 5 \: (\text{mod} \: 6)$$
$$\displaystyle 3(2) + 2 \equiv 2 \: (\text{mod} \: 6) \: \: \: {\color{red} \textit{3(2) + 2 = 8, but 8 is not an element of the ring, so we "reduce" it to one that is}}$$
$$\displaystyle 3(3) + 2 \equiv 5 \: (\text{mod} \: 6) \: \: \: {\color{red} \textit{3(3) + 2 = 11, but 11 is not an element of the ring, so we "reduce" it to one that is}}$$
$$\displaystyle 3(4) + 2 \equiv 2 \: (\text{mod} \: 6)$$
$$\displaystyle 3(5) + 2 \equiv 5 \: (\text{mod} \: 6)$$

Now you give it a try. It happens to be true in both cases that the solution under "normal" arithmetic is also a solution in the new ring, but are there other solutions to these modular equations? What might they be?

#### pka

##### Elite Member
Find all the solutions in ℤ / 80ℤ of the equation
4x + 5 ≡ 1 mod 80
Now obviously x must be -1 for the equation to hold true, but I don't know what to do about the "Find all the solutions in ℤ / 80ℤ" part.
Here are hints by way of additional solutions: $$\displaystyle x=79~\&~x=-81$$ HERE is a tool that might help.
Now think about why that is.

#### _rob

##### New member
Here are hints by way of additional solutions: $$\displaystyle x=79~\&~x=-81$$ HERE is a tool that might help.
Now think about why that is.
Thanks, so while the upper bound is limited to 80-1, the lower bound is not? So you can keep recursively subtracting 80 from -1 to negative infinity? So -241 and -161 are also solutions?

#### Dr.Peterson

##### Elite Member
Actually, the numbers -1 and 79 and so on are all the same solution in Z/(80Z) -- they are congruent mod 80, so they are not considered different.

The more important thing is that there are other solutions mod 80 -- numbers not congruent to -1 that are also solutions. This comes about because there are numbers other than -1 that you can multiply by 4 to get -4 (mod 80). (Can you find them?) Your claim that it is "obvious" that x must be -1 was wrong.

What have you learned about modular arithmetic and solving equations?