kuing的解答
由于 $a$ 与 $a+1$ 互素,故 $a$, $a+1$, $b$ 为素股勾数,所以必有
\[
\left(\begin{aligned}
a&=m^2-n^2,\\
a+1&=2mn,\\
b&=m^2+n^2,
\end{aligned}\right.
~\text{或}~
\left(\begin{aligned}
a+1&=m^2-n^2,\\
a&=2mn,\\
b&=m^2+n^2,
\end{aligned}\right.\]
其中正整数 $m$, $n$ 互质、一奇一偶且 $m>n$。由此可得
\[m^2-n^2-2mn=\pm1,\]
即
\[(m-n)^2=2n^2\pm1,\]
又由 $b<500$ 得
\[500>m^2+n^2\geqslant (n+1)^2+n^2,\]
可解出 $n\leqslant 15$,我们列出 $n$ 由 $1$ 到 $15$ 时 $2n^2$ 的值
\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15\\
\hline
2n^2&2&8&18&32&50&72&98&128&162&200&242&288&338&392&450\\
\hline
\end{array}
可以看出,只有 $n=1$, $2$, $5$, $12$ 才能使 $2n^2+1$ 或 $2n^2-1$ 为平方数,具体地可以解出以下四组解
\[\left(\begin{aligned}
n&=1,\\
m&=2,
\end{aligned}\right.
\quad
\left(\begin{aligned}
n&=2,\\
m&=5,
\end{aligned}\right.
\quad
\left(\begin{aligned}
n&=5,\\
m&=12,
\end{aligned}\right.
\quad
\left(\begin{aligned}
n&=12,\\
m&=29,
\end{aligned}\right.\]
代回去分别又得到
\[\left(\begin{aligned}
a&=3,\\
b&=5,
\end{aligned}\right.
\quad
\left(\begin{aligned}
a&=20,\\
b&=29,
\end{aligned}\right.
\quad
\left(\begin{aligned}
a&=119,\\
b&=169,
\end{aligned}\right.
\quad
\left(\begin{aligned}
a&=696,\\
b&=985,
\end{aligned}\right.\]
除最后一组外其余都符合条件,故此满足条件的 $a$, $b$ 只有三组解。 |