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Last edited by hbghlyj 2023-3-20 09:58Honsberger 1976 Mathematical Gems II，第 33 页
$$f(x, y)=\frac{y-1}{2}\left[\left|B^{2}-1\right|-\left(B^{2}-1\right)\right]+2$$其中$B=x(y+1)-(y !+1)$, 当 $x,y$ 为正整数, $f(x,y)$为素数, 且取遍每个奇素数, 每个奇素数仅取一次.
Proof. For all natural $x$ and $y$, the value of $B$ is an integer. Thus $B^2$ is a nonnegative integer. Accordingly, there are the two cases: (a) $B^2 \geqslant 1$, and (b) $B^2=0$.
(a) $B^2 \geqslant 1$ : If $B^2 \geqslant 1$, then $B^2-1 \geqslant 0$, implying that $\left|B^2-1\right|$ $=B^2-1$. This makes $f(x, y)=2$, a prime.
(b) $B^2=0$ : For $B^2=0$ the value of the function is
$$
\begin{aligned}
f(x, y) & =\frac{y-1}{2}[|-1|-(-1)]+2 \\
& =\frac{y-1}{2}[1+1]+2=y-1+2=y+1 .
\end{aligned}
$$
In this case, however, $B$ itself must also be 0 , implying that the numbers substituted for $x$ and $y$ make
$$
x(y+1)-(y !+1)=0, \text { or } x(y+1)=y !+1 .
$$
Accordingly, the number $y+1$ divides $y !+1$. By Wilson's theorem, then, the number $y+1$ is a prime. Consequently, $f(x, y)$ yields prime numbers exclusively.
We note that $f(1,1)=2$. Let $p$ denote an odd prime number. Then, using
$$
y=p-1 \text { and } x=\frac{1}{p}[(p-1) !+1]
$$
($x$ is guaranteed to be a natural number by Wilson's theorem), we see that $f(x, y)$ yields the prime $p$ : from the definitions of $x$ and $y$,
we have
$$
x p=(p-1) !+1 \text { and } p=y+1
$$
then $x p=x(y+1)=(p-1) !+1=y !+1$, making $B=0$ and $f(x, y)=y+1=p$. Thus $f$ yields every prime number.
Since $f$ yields only the values 2 and $y+1$, an odd prime $p$ can arise only as $y+1$. Thus in every pair $(x, y)$ which makes $f(x, y)$ yield the odd prime $p$, we must have $y=p-1$. However, $f$ will yield the odd number $y+1$, rather than 2 , only when $B=0$. Thus the pair $(x, y)$ must also make $x(y+1)=y !+1$. This means that $x$ must have the value $(y !+1) /(y+1)$. The single choice for $y$, then, leads to a single value for $x$, yielding the unique pair
$$
(x, y) \equiv\left(\frac{(p-1) !+1}{p}, p-1\right)
$$
($x$ is natural by Wilson's theorem).
which makes $f$ take the value $p$. Thus, while $f$ produces the number 2 "most of the time," it yields each odd prime exactly once. |
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Posted 2023-3-19 09:38
