How many possible value of \( c <100\), so that

\[ \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b} \]

only have \( 9 \) solutions for \( (a,b) \).

Detail : \( ( a,b,c ) > 0 \)

How many possible value of \( c <100\), so that

\[ \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b} \]

only have \( 9 \) solutions for \( (a,b) \).

Detail : \( ( a,b,c ) > 0 \)

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TopNewestHere is another hint:

The number of positive integer solutions for \(\frac {1}{a}+\frac {1}{b}=\frac {1}{n}\) is the number of factors of \(n^2\) – Sathvik Acharya · 1 month, 2 weeks ago

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Hint: Show that a divides b (or b divides a) first.

And since there's 9 (odd number) solutions, then there's a solution for which a=b, so c is even. – Pi Han Goh · 1 month, 3 weeks ago

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