User:Marty/UCSC Math 110 Fall 2008/Homework 7

This homework assignment can be handwritten. You do not need to LaTex your answers (though you certainly are allowed to, if you wish).

=Translation Exercises, 2 Points=

We have discussed four ways of "saying the same thing", related to congruences. For example, the following four sentences are equivalent:
 * 1)  $20 - 11$ is a multiple of $3$.
 * 2)  If you divide $20$ by $3$, and you divide $11$ by $3$, you get the same remainder.
 * 3)  $20 \equiv 11$, mod $3$.
 * 4)  $\overline{20} = \overline{11}$ (modulo $3$).

Each of the following five sentences is written in one of the four forms above. Translate each of the following four sentences into the other three forms.


 * $17 - (-3)$ is a multiple of $5$.
 * If you divide $x^2$ by $6$, and you divide $y$ by $6$, you get the same remainder.
 * $m^2 \equiv 0$, mod $m$.
 * $\overline{3x + 2} = \overline{5}$ (modulo $7$).

=Computation Exercises, 3 Points=

In each of the following problems, carry out a computation in modular arithmetic. Your answer should have the form $\bar a$, where $a$ is a natural number between $0$ and the modulus. After carrying out the computation, interpret your result using a sentence involving the words "multiple" or "divide" and "remainder". The following is an example: Computation: $$\overline{11}^{12} = \overline{-2}^{12} = ((\overline{-2})^4)^3 = \overline{16}^3 = \overline{3}^3 = \overline{27}=\overline{1}.$$ Interpretation: If you divide $11^{12}$ by $13$, you get a remainder of $1$. Alternatively, $11^{12} - 1$ is a multiple of $13$.
 * $\overline{11}^{12}$, modulo $13$.

Here are the problems:
 * $\overline{11} \cdot \overline{13}$, modulo $15$.
 * $\overline{9}^{1020}$, modulo $40$.
 * $\overline{3}^{10} - \overline{2}^{10}$, modulo $11$.

=Solving a Linear Congruence, 3 Points=

Find an integer $x$, between $0$ and $46$, such that: Show your work, including the Euclidean algorithm.
 * $\overline{36} \overline{x} + \overline{11} = \overline{35}$, modulo $47$.

=No Solutions, 2 Points=

Prove that there are no solutions to the Diophantine equation $x^6 + y^6 = 700003$. Use the following strategy:
 * What are the possible remainders if you divide $x^6$ or $y^6$ by $7$?
 * What is $\bar 1^6$? What is $\bar 2^6$? etc..
 * What are the possible remainders if you divide $x^6 + y^6$ by $7$?
 * What is the remainder if you divide $700003$ by $7$?