14 lines
184 B
Markdown
14 lines
184 B
Markdown
$$
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m^3=n^3+n^2p
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$$
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let $m=n+k$ , then
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$$
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3n^2k+3nk^2+k^3=n^2p\\
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(p-3k)n^2-3k^2n-k^3=0
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$$
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witch $p>3k$ . solving this function, we get
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$$
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n = \frac{3k^2 \pm k\sqrt{4kp-3k^2}}{p-3k}
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$$
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