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The Bezout identity states that for integers $a,b$, not both zero, there are integers $r,s$ such thatIf the prime number $p$ divides $ab$ but does not divide $a$, then $\gcd (p,a)=1$. Thus $rp+sa=1$ for some integers $r,s$, and multiplication by $b$ givesBoth terms on the right are divisible by $p$, so $p\mid b$. Hence

If $\gcd (m,n)=1$, choose $r,s$ with $rm+sn=1$. Thensatisfies $x\equiv a(\mathrm{mod}m)$ and $x\equiv b(\mathrm{mod}n)$. If $x,x^{\prime }$ are two simultaneous solutions, both $m$ and $n$ divide $x-x^{\prime }$. Coprimality implies $mn\mid x-x^{\prime }$, so the solution is unique modulo $mn$. This proves the two-modulus Chinese remainder theorem.

Let $p$ be odd and supposeThen $p^d\mid (x-1)(x+1)$. Since $\gcd (x-1,x+1)$ divides $2$, the odd prime $p$ cannot divide both factors. The full power $p^d$ must therefore divide one of them, givingThese are distinct, so there are exactly two solutions.

For an odd integerthe Chinese remainder theorem identifies a solution modulo $n$ with independent solutions modulo the $k$ prime powers. Each component has two choices, hence

For $n=2^d$, there is one solution when $d=1$ and two when $d=2$. If $d\ge 3$, a solution $x$ is odd, and of the consecutive even integers $x-1,x+1$, exactly one is divisible by $4$. The 2-adic valuations therefore have minimum one; for their sum to be at least $d$, the other must be at least $d-1$. Thus $x\equiv \pm 1(\mathrm{mod}{2}^{d-1})$, giving the four distinct classesConsequently the answer is

LetIf $x$ were rational, thenwould make $\sqrt[3]{6}=(x^3-5)/(3x)$ rational. But if $\sqrt[3]{6}=a/b$ in lowest terms, then $a^3=6b^3$, which is impossible by unique prime factorization: the exponent of $2$ on the two sides is respectively a multiple of three and one more than a multiple of three. Therefore

Use the given convergent seriesSuppose this number were $a/b\in \mathbb{Q}$. Choose $N$ large enough that $b\mid (2N+1)!$. Multiplication by $(2N+1)!$ makes both the rational number and the partial sum through $n=N$ integers. Their differencewould therefore be an integer. It is positive, whilefor sufficiently large $N$, a contradiction. Hence

A transcendental number is a complex number that is not a root of any nonzero polynomial with rational, equivalently integer, coefficients. LetIf $a\ne 0$ and $y$ were an algebraic number, then $e$ would satisfyover the algebraic extension $\mathbb{Q}(y)$. By transitivity of algebraic extensions, $e$ would be algebraic over $\mathbb{Q}$, contradicting its transcendence. If $a=0$, then $b\ne 0$ and $e=b/y$ would again be algebraic. Thus

Suppose there were only finitely many prime numbers congruent to $2$ modulo $3$, and list them as $p_1,\dots ,p_k$. ConsiderNo $p_i$ divides $N$, and $3$ does not divide $N$. In the prime factorization of $N$, not every prime factor can be congruent to $1$ modulo $3$, since their product would then also be congruent to $1$, whereasThus some prime factor is congruent to $2$ modulo $3$, contradicting the completeness of the list. Hence

Now let $p$ be prime. If $p\ne 3$, then either $p=2$, givingor $p$ is not divisible by $3$, so $p^2\equiv 1(\mathrm{mod}3)$. In the latter caseand is greater than $3$, hence is composite. For $p=3$ the number is $19$, which is prime. Therefore