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1. Prove that: for any given positive integer $n $, there is an infinite number of positive integers $a $, so that $n ^4 + a$ is composite.
Answer:
Make $a = 4m^4$, $m \in\mathbf{z}$, $ $n ^4 + 4m^4 = (n^2 + 2m^2) ^2-4m^2n^2$$ $$= (n^2 + 2m^2 + 2mn) (n^2 + 2m^2-2mn) $$ easy to know, $ M > 1$ when $n ^2 + 2m^2-2mn = (n-m) ^2 + m^2 > 1$, that is, there is an infinite number of positive integers $a $, so that $n ^4 + a$ is composite.
2. Proof: $n \in\mathbf{n^*}$, $n ^2 + n + 1$ is not a full square number.
Answer: $ $n ^2 < n^2 + n + 1 < (n+1) ^2$$ two consecutive complete squares without a complete square number.
3. Proof: The product plus $1$ of four consecutive positive integers must be a total square number.
Solution: $ $n (n+1) (n+2) (n+3) + 1 = (n^2 + 3n) (n^2 + 3n + 2) + 1$$ $$= (n^2 + 3n) ^2 + 2 (n^2 + 3n) + 1 = (n^2 + 3n + 1) ^2 $$
4. Set $p $ is the minimum prime factor of composite $n $, proving that $\dfrac{n}{p}$ is a prime number if $p > n^{\frac{1}{3}}$.
Answer:
Suppose $\dfrac{n}{p} = p_1 \cdot k$, where $p _1$ is prime and $p <\le p_1 \le k$. $$\rightarrow n = P\cdot p_1 \cdot k \ge p^3$$ $$\rightarrow p \le \sqrt[3]{n}.$$ with known contradiction.
5. Find the $x $, $y $, $z $ for all primes that satisfy the equation $x ^y + 1 = z$.
Answer:
$x ^y$ differs from $z $ parity, so $x = 2$. $$\rightarrow 2^y + 1 = z$$ If $y $ is odd, then $z $ is composite, so $y = 2$, $z = 5$.
6. Proof: $n > 2$, there must be a prime number between $n $ and $n!$.
Answer:
Easy to know $ (n!, n!-1) = 1$. The primes in $1\sim n$ can be evenly divisible $n!$ but not divisible $n! -1$, so $n! -1$ must contain a prime number that is different from all primes in $1\sim n$ $p $, and $p > n$.
That is, there must be a prime number between $n $ and $n!$.
2017 Winter Monkey Tutoring Elementary Number theory-3: "Prime number and the only decomposition theorem (i)" solution to the work problem