Suppose we call numbers of the form $a+b\sqrt{10}$ (where $a$ and $b$ are integers) extended integers.
Both $1$ and $-1$ divide every integer. We call these numbers units. In the set of integers, there are only these two values which are units -- but what about the extended integers? Prove that $3+\sqrt{10}$ is a unit for the extended integers. What characteristic must units of extended integers have?
Hint: when one divides an extended integer by a unit, and tries to write it as an extended integer (possibly involving a multiplication by a conjugate), what must be true of the resulting denominator to be assured the coefficient on $\sqrt{10}$ turns out to be an integer?
In the set of integers, one can define the primes to be those integers that can't be written as the product of two non-units. One might wonder which numbers in the set of extended integers have this same property. Prove that $2$, $3$, and $4 \pm \sqrt{10}$ are all extended integers that can't be written as the product of two non-units.
Hint: argue indirectly -- assume they can be written as the product of two non-units and attempt to show that no matter what happens, one of the factors ends up having the defining characteristic of units found above. To do this, try to put things back into the realm of integers by considering the product of the extended integers in question and their conjugates:
$$\begin{array}{lll}
m+n\sqrt{10} &= (a+b\sqrt{10})(c+d\sqrt{10}) &\textrm{, then}\\\\
m-n\sqrt{10} &= (a-b\sqrt{10})(c-d\sqrt{10}) &\textrm{, and so}\\\\
m^2-10n^2 &= (a^2-10b^2)(c^2-10d^2)\\
\end{array}$$
Now we can think about the possible factorizations of $m^2-10n^2$ over the integers, as $m$ and $n$ are chosen so that $m+n\sqrt{10}$ yields $2,3,4+\sqrt{10}$ and $4-\sqrt{10}$, respectively.
Use these factorizations to show if $a+b\sqrt{10}$ and $c+d\sqrt{10}$ are not units (remember the condition you found in problem 2 above), then in each of these four cases:
$$a^2-10b^2 \quad \quad \textrm{and} \quad \quad c^2-10d^2$$
must be $2, -2, 3$, or $-3$. Finally, show that for any $a$,
$$a^2 \equiv 0,1, \textrm{ or } 4 \pmod{5}$$ and therefore,
$$a^2-10b^2=2,-2,3, \textrm{ or } -3$$
has no integral solutions.
Suppose we call extended integers that can't be written as the product of two non-units indivisibles. Certainly, $6$ can be written as the product of two indivisibles, as $6 = 2 \cdot 3$. Can you find two different indivisibles $m$ and $n$, such that $6=mn$?
We know that $2$ is prime and $2 \mid 6$. Does $2$ have to divide either $m$ or $n$? What might you conclude about the statement "if $p$ is prime and $p \mid mn$, then $p \mid m$ or $p \mid n$" and the unique factorization of integers?
Prove that $\log_5 7$ is irrational.
Prove that every root of an equation of the form $x^n + c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + \cdots + c_1 x + c_0 = 0$ (with each $c_i \in \mathbb{Z}$) is either an integer or irrational, and then use this to show $\sqrt[3]{17}$, $\sqrt[5]{11}$, and $\sqrt[11]{13}$ are all irrational.
Hint: argue the main result indirectly. Assume there is a rational root (i.e., a root expressable as a fraction of integers in lowest terms $a/b$) and try to contradict the lowest terms nature of the fraction in question.
Prime numbers are defined as those numbers that have exactly two divisors. What type of numbers have exactly three divisors? ...exactly four divisors?
Prove that every integer $n \ge 1$ can be written as a product of an integer squared and an integer that has no divisors that are perfect squares (i.e., a square-free integer).