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[数论] 若xy=k^2,则x,y一定是完全平方数吗?

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hjfmhh Post time 2024-4-19 17:05 |Read mode
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kuing Post time 2024-4-19 18:20
这不是很显然吗,对于 x 的质因数分解中的任意一个质数 p,由 x, y 互质知 y 的质因数分解必不含 p,于是 xy 中 p 的次数 = x 中 p 的次数,那由 xy=k^2 知 p 的次数必为偶数,所以 x 为完全平方数,同理 y 也是。

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hbghlyj Post time 2024-4-19 18:52
hjfmhh 发表于 2024-4-19 09:05
$\text { 若 } x, y \inN^*, x , y \text {互质 且 } x y=k^2,\left(k \inN^*\right)$ 则$x=m^2,y=n^2(m,n\inN^*)$
在$a^2+b^2=c^2$的通解的证明中,用到了这个

$a = p^2 − q^2 , b = 2pq, c = p^2 + q^2$
math.stackexchange.com/questions/1826816/
The starting point is observe that: $b^2 = c^2-a^2 = (c-a)(c+a)$. Thus if $b$ is even, then since $(c+a) - (c-a) = 2a$ which is even. So $c-a,c+a$ are both even. Then you can put $c-a = 2p^2, c+a = 2n^2$. If $b$ is odd, then both $c-a, c+a$ are odd. Thus you can write $c-a = k^2, c+a=r^2$. In both cases,  you are led to the solution above.

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