$If both $n$ and $\sqrt{n^2+204n}$ are positive integers, find the maximum value of $𝑛$.$

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$If both $n$ and $\sqrt{n^2+204n}$ are positive integers, find the maximum value of $𝑛$.$

If both $n$ and $\sqrt{n^2+204n}$ are positive integers, find the maximum value of $n$.

Number Theory Math Competitions Algebra

Paul F

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Let
\[𝑚^2=𝑛^2+204𝑛\]
\[𝑘^2=𝑛^2+204𝑛+10404=(𝑛+102)^2.\]
Then
\[(𝑘−𝑚)(𝑘+𝑚)=2^2⋅3^2⋅17^2 \]
Since $𝑘−𝑚$ and $𝑘+𝑚$ have the same parity, they must be both even. Let
\[\frac{𝑘−𝑚}{2}\frac{𝑘+𝑚}{2}=3^2⋅17^2\]
Since $𝑘>𝑚$ there are only two possibilities: $𝑘−𝑚=2$ and $𝑘−𝑚=18$. The former gives $𝑚=2600$ and the latter $𝑚=280$. Thus we are interested in the former, which gives $𝑛=2500$.

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$If both $n$ and $\sqrt{n^2+204n}$ are positive integers, find the maximum value of $𝑛$.$

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