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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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